RS Aggarwal Solutions for Class 11 Chapter 12 Geometrical Progressions

Question 1. Find the 6th and nth terms of the GP 2, 6, 18, 54….
Answer: The sequence 2, 6, 18, 54… follows the pattern a, ar, ar², ar³…, where r denotes the common ratio. The opening term is a₁ = a = 2. The second term is a₂ = 6. The common ratio is r = a₂ ÷ a₁ = 6 ÷ 2 = 3. Using the nth term formula aₙ = ar^(n-1), the 6th term equals a₆ = ar⁵ = 2 × 3⁵ = 486. The nth term equals aₙ = ar^(n-1) = 2 · 3^(n-1). Therefore, the 6th term is 486 and the nth term is 2 · 3^(n-1).
In simple words: Multiply the opening number by the common ratio repeatedly. For the 6th term, multiply 2 by 3 five times to get 486. For any position n, use the formula 2 · 3^(n-1).

Exam Tip: Identify the first term and common ratio immediately. Always verify your answer by checking that your calculated term fits the pattern.

Question 2. Find the 17th and nth terms of the GP 2, 2√2, 4, 8√2….
Answer: The sequence 2, 2√2, 4, 8√2… has the form a, ar, ar², ar³…. The opening term is a₁ = a = 2. The second term is a₂ = 2√2. The common ratio is r = a₂ ÷ a₁ = 2√2 ÷ 2 = √2. Using aₙ = ar^(n-1), the 17th term is a₁₇ = ar¹⁶ = 2 × (√2)¹⁶ = 512. The nth term is aₙ = ar^(n-1) = 2(√2)^(n-1) = (√2)^(n+1). Therefore, the 17th term is 512 and the nth term is (√2)^(n+1).
In simple words: The common ratio √2 grows powers of 2 in the exponent. Raise √2 to the power (n+1) to find any position.

Exam Tip: When the common ratio involves square roots, simplify radical powers carefully using exponent rules.

Question 3. Find the 7th and nth terms of the GP 0.4, 0.8, 1.6….
Answer: The sequence 0.4, 0.8, 1.6… follows the pattern a, ar, ar², ar³…. The opening term is a₁ = a = 0.4. The second term is a₂ = 0.8. The common ratio is r = a₂ ÷ a₁ = 0.8 ÷ 0.4 = 2. Using aₙ = ar^(n-1), the 7th term is a₇ = ar⁶ = 0.4 × 2⁶ = 25.6. The nth term is aₙ = ar^(n-1) = (0.4)(2)^(n-1) = (0.2) · 2ⁿ. Therefore, the 7th term is 25.6 and the nth term is (0.2) · 2ⁿ.
In simple words: Double each term to get the next one. For the 7th term, start with 0.4 and multiply by 2 six times. For position n, use (0.2) · 2ⁿ.

Exam Tip: Decimal progressions follow the same rules as whole-number progressions. Simplify the final formula by combining numerical coefficients.

Question 4. Find the 10th and nth terms of the GP \( -\frac{3}{4}, \frac{1}{2}, -\frac{1}{3}, \frac{2}{9}\ldots \)
Answer: The sequence follows a, ar, ar², ar³…. The opening term is \( a_1 = a = -\frac{3}{4} \). The second term is \( a_2 = \frac{1}{2} \). The common ratio is \( r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{-\frac{3}{4}} = -\frac{2}{3} \). Using aₙ = ar^(n-1), the 10th term is \( a_{10} = ar^9 = \left(-\frac{3}{4}\right)\left(-\frac{2}{3}\right)^9 = \frac{128}{6561} \). The nth term is \( a_n = ar^{n-1} = \left(\frac{9}{8}\right)\left(-\frac{2}{3}\right)^n \). Therefore, the 10th term is \( \frac{128}{6561} \) and the nth term is \( \left(\frac{9}{8}\right)\left(-\frac{2}{3}\right)^n \).
In simple words: The common ratio is a negative fraction, so signs alternate and values shrink. Raise the ratio to the needed power and multiply by the base term.

Exam Tip: With negative ratios, track sign changes carefully. Simplify all fraction calculations before finalizing.

Question 5. Which term of the GP 3, 6, 12, 24… is 3072?
Answer: The sequence 3, 6, 12, 24… has the form a, ar, ar², ar³…. The opening term is a₁ = a = 3. The second term is a₂ = 6. The common ratio is r = a₂ ÷ a₁ = 6 ÷ 3 = 2. To find which position holds 3072, use aₙ = ar^(n-1). Setting aₙ = 3072 gives 3072 = 3 · 2^(n-1). Divide both sides by 3 to obtain 1024 = 2^(n-1). Since 1024 = 2¹⁰, we have 2^(n-1) = 2¹⁰, so n - 1 = 10, which means n = 11. Therefore, 3072 is the 11th term in the GP.
In simple words: Use the nth term formula, plug in 3072 as the answer, and solve for n. Express both sides as the same base (2 in this case) to find the exponent.

Exam Tip: When finding which term equals a given value, always express both sides using the same base if possible to compare exponents directly.

Question 6. Which term of the GP \( \frac{1}{4}, -\frac{1}{2}, 1\ldots \) is -128?
Answer: The sequence \( \frac{1}{4}, -\frac{1}{2}, 1\ldots \) follows a, ar, ar², ar³…. The opening term is \( a_1 = a = \frac{1}{4} \). The second term is \( a_2 = -\frac{1}{2} \). The common ratio is \( r = \frac{a_2}{a_1} = \frac{-\frac{1}{2}}{\frac{1}{4}} = -2 \). To find which position holds -128, use aₙ = ar^(n-1). Setting aₙ = -128 gives \( -128 = \frac{1}{4} \cdot (-2)^{n-1} \). Multiply both sides by 4 to get \( -512 = (-2)^{n-1} \). Since \( (-2)^9 = -512 \), we have n - 1 = 9, so n = 10. Therefore, -128 is the 10th term in the GP.
In simple words: Set up the equation with -128 equal to the nth term formula. Isolate the power and express both sides as the same base to match exponents.

Exam Tip: With negative bases and negative targets, carefully track signs when solving for the exponent.

Question 7. Which term of the GP √3, 3, 3√3… is 729?
Answer: The sequence √3, 3, 3√3… has the form a, ar, ar², ar³…. The opening term is a₁ = a = √3. The second term is a₂ = 3. The common ratio is \( r = \frac{a_2}{a_1} = \frac{3}{\sqrt{3}} = \sqrt{3} \). To find which position holds 729, use aₙ = ar^(n-1). Setting aₙ = 729 gives \( 729 = \sqrt{3} \cdot (\sqrt{3})^{n-1} \). This simplifies to \( (\sqrt{3})^n = 729 \). Since \( (\sqrt{3})^{12} = 3^6 = 729 \), we have n = 12. Therefore, 729 is the 12th term in the GP.
In simple words: Each term multiplies the previous one by √3. Combine the opening term with the power of √3 and match it to 729 expressed as a power of √3.

Exam Tip: Convert all radical expressions to the same base. Use exponent rules to match powers and solve for position.

Question 8. Find the geometric series whose 5th and 8th terms are 80 and 640 respectively.
Answer: The nth term formula for a GP is aₙ = ar^(n-1). Given that the 5th term is 80 and the 8th term is 640, we have a₅ = ar⁴ = 80 → (1) and a₈ = ar⁷ = 640 → (2). Dividing equation (2) by equation (1) yields \( \frac{ar^7}{ar^4} = r^3 = \frac{640}{80} = 8 \). Therefore r = 2. Substituting r = 2 into equation (1) gives ar⁴ = a · 16 = 80, so a = 5. The required GP is a, ar, ar², ar³, ar⁴… with a = 5. The opening term is 5. The second term is ar = 5 × 2 = 10. The third term is ar² = 5 × 4 = 20. The fourth term is ar³ = 5 × 8 = 40. The fifth term is ar⁴ = 5 × 16 = 80. The required sequence is 5, 10, 20, 40, 80…
In simple words: Divide the two given equations to cancel the first term and solve for the ratio. Then use either original equation to find the first term. Write out the full sequence by multiplying by the ratio each time.

Exam Tip: When given two terms, always divide the larger term by the smaller to eliminate the first term and isolate the ratio power.

Question 9. Find the GP whose 4th and 7th terms are \( \frac{1}{18} \) and \( -\frac{1}{486} \) respectively.
Answer: The nth term formula is aₙ = ar^(n-1). Given that the 4th term is \( \frac{1}{18} \) and the 7th term is \( -\frac{1}{486} \), we have \( a_4 = ar^3 = \frac{1}{18} \) → (1) and \( a_7 = ar^6 = -\frac{1}{486} \) → (2). Dividing equation (2) by equation (1) yields \( \frac{ar^6}{ar^3} = r^3 = \frac{-\frac{1}{486}}{\frac{1}{18}} = -\frac{1}{27} \). Therefore \( r = -\frac{1}{3} \). Substituting this into equation (1): \( ar^3 = a \cdot \left(-\frac{1}{27}\right) = \frac{1}{18} \), so \( a = -\frac{3}{2} \). The required GP has the form a, ar, ar², ar³, ar⁴…. The opening term is \( a = -\frac{3}{2} \). The second term is \( ar = -\frac{3}{2} \times \left(-\frac{1}{3}\right) = \frac{1}{2} \). The third term is \( ar^2 = -\frac{3}{2} \times \frac{1}{9} = -\frac{1}{6} \). The fourth term is \( ar^3 = -\frac{3}{2} \times \left(-\frac{1}{27}\right) = \frac{1}{18} \). Therefore the GP is \( -\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, -\frac{1}{54}\ldots \)
In simple words: Divide the two term equations to find the ratio cubed. Take the cube root to get the ratio. Substitute back to find the first term, then multiply by the ratio to build the sequence.

Exam Tip: Track negative signs carefully when finding cube roots of negative ratios. Verify your sequence by checking both given terms.

Question 10. The 5th, 8th and 11th terms of a GP are a, b, c respectively. Show that b² = ac
Answer: Given that the 5th, 8th and 11th terms of a GP equal a, b, and c respectively. Let the GP be A, AR, AR², AR³…. The nth term is aₙ = AR^(n-1). The 5th term is a₅ = AR⁴ = a → (1). The 8th term is a₈ = AR⁷ = b → (2). The 11th term is a₁₁ = AR¹⁰ = c → (3). Dividing equation (2) by equation (1): \( \frac{AR^7}{AR^4} = R^3 = \frac{b}{a} \) → (4). Dividing equation (3) by equation (2): \( \frac{AR^{10}}{AR^7} = R^3 = \frac{c}{b} \) → (5). From equations (4) and (5), both equal R³, so \( \frac{b}{a} = \frac{c}{b} \). Cross-multiplying yields b² = ac. Hence proved.
In simple words: Find the ratio of consecutive given terms using the nth term formula. Both ratios equal R³, so equate them and simplify to get the required relationship.

Exam Tip: This classic property holds for any three equally-spaced terms in a GP. Memorize: the middle term squared equals the product of the outer two.

Question 11. The first term of a GP is - 3 and the square of the second term is equal to its 4th term. Find its 7th term.
Answer: It is given that the opening term of the GP equals -3, so a = -3. It is also given that the square of the second term equals the 4th term. The second term is a₂ = ar and the fourth term is a₄ = ar³, so (ar)² = ar³. This simplifies to a²r² = ar³. Dividing both sides by ar (assuming a ≠ 0 and r ≠ 0) yields ar = r², which gives a = r. Since a = -3, we have r = -3. Now the 7th term is a₇ = ar⁶ = (-3)(-3)⁶ = (-3)⁷ = -2187. Therefore, the 7th term equals -2187.
In simple words: Use the condition that (second term)² = fourth term to find a relationship between a and r. Substitute your known first term to get the ratio, then calculate the 7th term.

Exam Tip: When given a special condition linking terms, set up an equation and simplify to discover the ratio. Check your final answer by verifying the given condition.

Question 12. Find the 6th term from the end of GP 8, 4, 2… \( \frac{1}{1024} \)
Answer: The given GP is \( 8, 4, 2, \ldots, \frac{1}{1024} \). The opening term is a₁ = a = 8. The second term is a₂ = ar = 4. The common ratio is \( r = \frac{a_2}{a_1} = \frac{4}{8} = \frac{1}{2} \). The ending term is \( \frac{1}{1024} \). When counting from the end, the series reverses to form a new GP: \( ar^{n-1}, ar^{n-2}, ar^{n-3}, \ldots, ar^3, ar^2, ar, a \). This new GP has first term \( ar^{n-1} = \frac{1}{1024} \) and common ratio \( r' = \frac{1}{r} = 2 \). The 6th term from the end using this reversed GP is \( a_6 = ar^{n-1} \cdot (r')^5 = \frac{1}{1024} \cdot 2^5 = \frac{1}{1024} \cdot 32 = \frac{1}{32} \). Therefore, the 6th term from the end is \( \frac{1}{32} \).
In simple words: When counting from the end, reverse the sequence direction and use the reciprocal of the original ratio. Count forward in this reversed sequence to find your target term.

Exam Tip: Terms from the end form their own GP. The new ratio is always the reciprocal of the original ratio. Count carefully to avoid off-by-one errors.

Question 13. Find the 4th term from the end of the GP \( \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162 \)
Answer: The given GP is \( \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162 \). The opening term is \( a_1 = a = \frac{2}{27} \). The second term is \( a_2 = \frac{2}{9} \). The common ratio is \( r = \frac{a_2}{a_1} = \frac{\frac{2}{9}}{\frac{2}{27}} = 3 \). The ending term is aₙ = 162. When counting from the end, the sequence reverses to form: \( ar^{n-1}, ar^{n-2}, ar^{n-3}, \ldots, ar^3, ar^2, ar, a \). The new first term is \( ar^{n-1} = 162 \) and the common ratio is \( r' = \frac{1}{r} = \frac{1}{3} \). The 4th term from the end using this reversed GP is \( a_4 = 162 \cdot \left(\frac{1}{3}\right)^3 = 162 \cdot \frac{1}{27} = 6 \). Therefore, the 4th term from the end is 6.
In simple words: Start from the last term and count backward. Each step divides by 3 (the reciprocal ratio). After 3 steps from the end, you reach the 4th term from the end.

Exam Tip: Label the reversed sequence carefully. The 1st term from the end is the last term, the 2nd is the second-to-last, etc. The new ratio is always 1/r.

Question 14. If a, b, c are the pth, qth and rth terms of a GP, show that (q - r) log a + (r - p) log b + (p - q) log c = 0.
Answer: Given that a, b, and c are the pth, qth and rth terms of a GP respectively. Let the GP be A, AR, AR², AR³… The nth term is aₙ = AR^(n-1). The pth term is a_p = AR^(p-1) = a → (1). The qth term is a_q = AR^(q-1) = b → (2). The rth term is a_r = AR^(r-1) = c → (3). Dividing (1) by (2): \( \frac{AR^{p-1}}{AR^{q-1}} = R^{p-q} = \frac{a}{b} \). Taking logarithms: (p - q) log R = log a - log b → (4). Dividing (2) by (3): \( \frac{AR^{q-1}}{AR^{r-1}} = R^{q-r} = \frac{b}{c} \). Taking logarithms: (q - r) log R = log b - log c → (5). Dividing (3) by (1): \( \frac{AR^{r-1}}{AR^{p-1}} = R^{r-p} = \frac{c}{a} \). Taking logarithms: (r - p) log R = log c - log a → (6). Multiply (4) by log c: (p - q) log c · log R = (log a - log b) log c → (7). Multiply (5) by log a: (q - r) log a · log R = (log b - log c) log a → (8). Multiply (6) by log b: (r - p) log b · log R = (log c - log a) log b → (9). Adding (7), (8), and (9) and simplifying yields (p - q) log c + (q - r) log a + (r - p) log b = 0. Hence proved.
In simple words: Express the condition using the nth term formula. Take logarithms of the ratio equations. Multiply each by an appropriate log term and add them up to eliminate log R.

Exam Tip: Work systematically by forming ratios of terms. Logarithm properties turn division into subtraction, making elimination easier.

Question 15. The third term of a GP is 4; Find the product of its five terms.
Answer: Given that the third term of the GP equals 4, so a₃ = 4. Consider a GP of the form \( \frac{A}{R^2}, \frac{A}{R}, A, AR, AR^2 \) with opening term \( \frac{A}{R^2} \) and common ratio R. The third term of this sequence is the middle term, which is A. Since a₃ = A = 4, we have A = 4. The five terms are \( \frac{4}{R^2}, \frac{4}{R}, 4, 4R, 4R^2 \). The product of these five terms is \( \frac{4}{R^2} \times \frac{4}{R} \times 4 \times 4R \times 4R^2 = \frac{4 \times 4 \times 4 \times 4 \times 4 \times R^2}{R^2 \times R} \times R = 4^5 = 1024 \). Therefore, the product of its five terms is 1024.
In simple words: When you know the middle term of five consecutive GP terms, write them symmetrically around it. The middle term is 4, so the product simplifies to 4 raised to the power 5.

Exam Tip: For any odd number of consecutive GP terms, the product equals (middle term)^(count of terms). This symmetry eliminates the ratio from the calculation.

Question 16. In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms.
Answer: We must show that in a finite GP, the product formed by any two terms that are equally distant from the start and finish equals the product of the opening and closing terms.

Consider a finite GP: \( A, AR, AR^2 \ldots AR^{n-1}, AR^n \)

The product of the first and last terms is \( A \cdot AR^n = A^2R^n \) → (a)

The \( n \)th term counting from the beginning is \( AR^{n-1} \) → (1)

Starting from the end, the first term is \( AR^n \) and the last term is \( A \). The common ratio becomes \( \frac{1}{R} \).

The \( n \)th term counting from the end is \( A \cdot \left(\frac{1}{R}\right)^{n-1} = AR^{1-n} \) → (2)

The product of the \( n \)th terms from both the beginning and end (from equations 1 and 2) is \( (AR^{n-1})(AR^{1-n}) = A^2R^n \) → (b)

From (a) and (b), we have verified that the product of terms equidistant from the start and finish is the product of the first and last terms in a finite GP.

Exam Tip: Always identify the general term formula for a GP and apply it symmetrically from both ends - this symmetry is key to proving such results efficiently.

Question 17. If \( \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} \) (where \( x \neq 0 \)), then show that a, b, c, d are in GP.
Answer: Given: \( \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} \) → (1)

Cross multiplying and expanding equation (1):
\( (a+bx)(b-cx) = (b+cx)(a-bx) \)
\( ab - acx + b^2x - bcx^2 = ba - b^2x + acx - bcx^2 \)
\( 2b^2x = 2acx \)
\( b^2 = ac \) → (i)

By the geometric mean principle, if three numbers form a GP, the middle term is the geometric mean of the first and last. Since \( b^2 = ac \), we see that \( b \) is the geometric mean of \( a \) and \( c \). Therefore, \( a, b, c \) are in GP.

Now, given: \( \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} \) → (2)

Cross multiplying and expanding equation (2):
\( (b+cx)(c-dx) = (c+dx)(b-cx) \)
\( bc - bdx + c^2x - cdx^2 = cb - c^2x + bdx - dcx^2 \)
\( 2c^2x = 2bdx \)
\( c^2 = bd \) → (ii)

From equation (ii), \( c \) is the geometric mean of \( b \) and \( d \). Therefore, \( b, c, d \) are in GP.

Since \( a, b, c \) are in GP and \( b, c, d \) are in GP, we conclude that \( a, b, c, d \) are in GP.

Exam Tip: The geometric mean condition (\( \text{middle}^2 = \text{first} \times \text{last} \)) is the quickest way to identify GP progressions - use it to avoid lengthy ratio checks.

Question 18. If a and b are the roots of \( x^2 - 3x + p = 0 \) and c and d are the roots of \( x^2 - 12x + q = 0 \), where a, b, c, d form a GP, prove that \( (q + p) : (q - p) = 17 : 15 \).
Answer: From the given equations:

\( x^2 - 3x + p = 0 \) has roots \( a \) and \( b \)

By Vieta's formulas: \( a + b = 3 \) and \( ab = p \) → (2)

\( x^2 - 12x + q = 0 \) has roots \( c \) and \( d \)

By Vieta's formulas: \( c + d = 12 \) and \( cd = q \) → (4)

Since \( a, b, c, d \) form a GP (given), we may write them as \( A, AR, AR^2, AR^3 \) with common ratio \( R \).

From equation (2): \( A + AR = 3 \)
\( A(1 + R) = 3 \)
\( \frac{3}{A} = 1 + R \) → (5)

From equation (4): \( AR^2 + AR^3 = 12 \)
\( AR^2(1 + R) = 12 \) → (6)

Substituting the value of \( (1 + R) \) from (5) into (6):
\( AR^2 \cdot \frac{3}{A} = 12 \)
\( 3R^2 = 12 \)
\( R^2 = 4 \)
\( R = 2 \) (taking positive value)

Substituting \( R = 2 \) back into (5):
\( \frac{3}{A} = 1 + 2 = 3 \)
\( A = 1 \)

Therefore, the GP is \( 1, 2, 4, 8 \), so \( a = 1, b = 2, c = 4, d = 8 \)

From (2) and (4):
\( p = ab = 1 \times 2 = 2 \)
\( q = cd = 4 \times 8 = 32 \)

\( \frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15} \)

Therefore, \( (q + p) : (q - p) = 17 : 15 \)

Exam Tip: When four unknowns form a GP, always express them as \( A, AR, AR^2, AR^3 \) and use the sum conditions from Vieta's formulas to find both \( A \) and \( R \) - this method is far faster than trying to solve four separate equations.

Exercise 12C

Question 1. Find the sum of the GP: \( 1 + 3 + 9 + 27 + \ldots \) to 7 terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \).

Here:
\( a = 1 \) (first term)
\( r = 3 \) (common ratio, found by dividing any term by the preceding term)
\( n = 7 \)

\( S_7 = 1 \cdot \frac{3^7 - 1}{3 - 1} = \frac{2187 - 1}{2} = \frac{2186}{2} = 1093 \)

Exam Tip: Always verify the common ratio by taking the ratio of consecutive terms; for large exponents, calculate the power separately before substituting into the formula to avoid arithmetic errors.

Question 2. Find the sum of the GP: \( 1 + \sqrt{3} + 3 + 3\sqrt{3} + \ldots \) to 10 terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \).

Here:
\( a = 1 \)
\( r = \sqrt{3} \) (common ratio)
\( n = 10 \)

\( S_{10} = 1 \cdot \frac{(\sqrt{3})^{10} - 1}{\sqrt{3} - 1} = \frac{3^5 - 1}{\sqrt{3} - 1} = \frac{243 - 1}{\sqrt{3} - 1} = \frac{242}{\sqrt{3} - 1} \)

Rationalizing the denominator:
\( S_{10} = \frac{242}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{242(\sqrt{3} + 1)}{3 - 1} = \frac{242(\sqrt{3} + 1)}{2} = 121(\sqrt{3} + 1) = 121\sqrt{3} + 121 \approx 330.5 \)

Exam Tip: When the common ratio involves a square root or irrational number, always rationalize the denominator in the final answer - examiners expect simplified form.

Question 3. Find the sum of the GP: \( 0.15 + 0.015 + 0.0015 + \ldots \) to 6 terms
Answer: The sum formula for a GP is \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \).

Here:
\( a = 0.15 \)
\( r = 0.1 \) (common ratio)
\( n = 6 \)

\( S_6 = 0.15 \times \frac{1 - 0.1^6}{1 - 0.1} = 0.15 \times \frac{1 - 0.000001}{0.9} = 0.15 \times \frac{0.999999}{0.9} \)
\( = 0.15 \times 1.111110 \approx 0.166667 \) or \( \frac{1}{6} \)

Exam Tip: For GPs with \( |r| < 1 \), the sum converges quickly; even small powers of \( r \) become negligibly small, so approximate answers are often acceptable.

Question 4. Find the sum of the GP: \( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots \) to 9 terms
Answer: The sum formula for a GP is \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \).

Here:
\( a = 1 \)
\( r = -\frac{1}{2} \) (common ratio)
\( n = 9 \)

\( S_9 = 1 \times \frac{1 - (-\frac{1}{2})^9}{1 - (-\frac{1}{2})} = \frac{1 - (-\frac{1}{512})}{1 + \frac{1}{2}} = \frac{1 + \frac{1}{512}}{\frac{3}{2}} = \frac{\frac{513}{512}}{\frac{3}{2}} = \frac{513}{512} \times \frac{2}{3} = \frac{1026}{1536} = \frac{171}{256} \approx 0.668 \)

Exam Tip: When the common ratio is negative, pay special attention to the sign of each term when calculating the sum; alternating signs often lead to partial cancellation.

Question 5. Find the sum of the GP: \( \sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + \ldots \) to 8 terms
Answer: The sum formula for a GP is \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \).

Here:
\( a = \sqrt{2} \)
\( r = \frac{1}{2} \) (common ratio)
\( n = 8 \)

\( S_8 = \sqrt{2} \times \frac{1 - (\frac{1}{2})^8}{1 - \frac{1}{2}} = \sqrt{2} \times \frac{1 - \frac{1}{256}}{\frac{1}{2}} = \sqrt{2} \times \frac{\frac{255}{256}}{\frac{1}{2}} = \sqrt{2} \times \frac{255}{128} = \frac{255\sqrt{2}}{128} \)

Exam Tip: Simplify the common ratio early - recognizing that \( \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \) prevents unnecessary calculation errors.

Question 6. Find the sum of the GP: \( \frac{2}{9} - \frac{1}{3} - \frac{1}{2} - \ldots \) to 6 terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \).

Here:
\( a = \frac{2}{9} \)
\( r = \frac{3}{2} \) (common ratio)
\( n = 6 \)

\( S_6 = \frac{2}{9} \times \frac{(\frac{3}{2})^6 - 1}{\frac{3}{2} - 1} = \frac{2}{9} \times \frac{\frac{729}{64} - 1}{\frac{1}{2}} = \frac{2}{9} \times \frac{\frac{665}{64}}{\frac{1}{2}} = \frac{2}{9} \times \frac{665}{32} = \frac{1330}{288} = \frac{665}{144} \)

Simplifying: \( \frac{665}{144} \approx 4.62 \)

Exam Tip: When the common ratio is greater than 1, use the formula with \( r^n - 1 \) in the numerator; always compute the power first, then substitute into the formula.

Question 7. Find the sum of the GP: \( 1 + \frac{2}{3} + \frac{4}{9} + \ldots \) to n terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \).

Here:
\( a = 1 \)
\( r = \frac{2}{3} \)
\( n \text{ terms} \)

\( S_n = 1 \times \frac{(\frac{2}{3})^n - 1}{\frac{2}{3} - 1} = \frac{(\frac{2}{3})^n - 1}{-\frac{1}{3}} = -3[(\frac{2}{3})^n - 1] = 3[1 - (\frac{2}{3})^n] = 3 - 3(\frac{2}{3})^n \)

Alternatively: \( S_n = \frac{3[1 - (\frac{2}{3})^n]}{1} = 3\left[1 - \left(\frac{2}{3}\right)^n\right] \)

Exam Tip: Leave answers in general form for "n terms" unless asked to simplify further; factoring out common factors from numerator and denominator reduces errors.

Question 8. Find the sum of the GP: \( \sqrt{7} + \sqrt{21} + 3\sqrt{7} + \ldots \) to n terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \).

Here:
\( a = \sqrt{7} \)
\( r = \sqrt{3} \) (common ratio, since \( \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \))
\( n \text{ terms} \)

\( S_n = \sqrt{7} \times \frac{(\sqrt{3})^n - 1}{\sqrt{3} - 1} \)

Rationalizing the denominator:
\( S_n = \sqrt{7} \times \frac{(\sqrt{3})^n - 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \sqrt{7} \times \frac{[(\sqrt{3})^n - 1](\sqrt{3} + 1)}{3 - 1} = \frac{\sqrt{7}[(\sqrt{3})^n - 1](\sqrt{3} + 1)}{2} \)

Exam Tip: Always factor out radicals early to identify patterns; \( \sqrt{21} = \sqrt{7 \times 3} \) reveals that \( \sqrt{3} \) is the common ratio.

Question 9. Find the sum of the GP: \( 1 - a + a^2 - a^3 + \ldots \) to n terms (where \( a \neq 1 \))
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \).

Here the first term is 1 (not to be confused with the variable a):
\( \text{First term} = 1 \)
\( r = -a \)
\( n \text{ terms} \)

\( S_n = 1 \times \frac{(-a)^n - 1}{-a - 1} = \frac{(-a)^n - 1}{-(a + 1)} = \frac{1 - (-a)^n}{a + 1} \)

Therefore: \( S_n = \frac{1 - (-a)^n}{1 + a} \)

Exam Tip: Distinguish clearly between the variable names in the series and the formula parameters; here 'a' is a series element, not the 'a' in the sum formula.

Question 10. Find the sum of the GP: \( x^3 + x^5 + x^7 + \ldots \) to n terms
Answer: The sum formula for a GP is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \).

Here:
\( a = x^3 \)
\( r = x^2 \) (common ratio)
\( n \text{ terms} \)

\( S_n = x^3 \times \frac{(x^2)^n - 1}{x^2 - 1} = x^3 \times \frac{x^{2n} - 1}{x^2 - 1} = \frac{x^3(x^{2n} - 1)}{x^2 - 1} \)

Factoring further:
\( S_n = \frac{x^3(x^n - 1)(x^n + 1)}{(x - 1)(x + 1)} \)

Exam Tip: Recognize that exponents add when multiplying powers of the same base; here the pattern \( x^3, x^5, x^7 \) indicates the exponent increases by 2 each time, so \( r = x^2 \).

Question 11. Find the sum of the GP: \( x(x + y) + x^2(x^2 + y^2) + x^3(x^3 + y^3) + \ldots \) to n terms
Answer: Expand the given expression:
\( = (x^2 + xy) + (x^4 + x^2y^2) + (x^6 + x^3y^3) + \ldots \text{ to } n \text{ terms} \)
\( = (x^2 + x^4 + x^6 + \ldots \text{ to } n \text{ terms}) + (xy + x^2y^2 + x^3y^3 + \ldots \text{ to } n \text{ terms}) \)

The first group is a GP with \( a = x^2, r = x^2 \):
\( S_{n,1} = x^2 \times \frac{(x^2)^n - 1}{x^2 - 1} = x^2 \times \frac{x^{2n} - 1}{x^2 - 1} \)

The second group is a GP with \( a = xy, r = xy \):
\( S_{n,2} = xy \times \frac{(xy)^n - 1}{xy - 1} = xy \times \frac{x^ny^n - 1}{xy - 1} \)

Total sum:
\( S_n = x^2 \times \frac{x^{2n} - 1}{x^2 - 1} + xy \times \frac{x^ny^n - 1}{xy - 1} = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{xy(x^ny^n - 1)}{xy - 1} \)

Exam Tip: Split complex series into simpler GP components; identify each component's first term and common ratio independently to avoid mixing the parameters.

Question 12. Find the sum to n terms of the sequence: \( \left(x + \frac{1}{x}\right)^2, \left(x^2 + \frac{1}{x^2}\right)^2, \left(x^3 + \frac{1}{x^3}\right)^2, \ldots \) to n terms
Answer: Expanding each term:
\( = \left(x^2 + 2 + \frac{1}{x^2}\right) + \left(x^4 + 2 + \frac{1}{x^4}\right) + \left(x^6 + 2 + \frac{1}{x^6}\right) + \ldots \text{ to } n \text{ terms} \)
\( = (x^2 + x^4 + x^6 + \ldots \text{ to } n \text{ terms}) + \left(\frac{1}{x^2} + \frac{1}{x^4} + \frac{1}{x^6} + \ldots \text{ to } n \text{ terms}\right) + 2n \)

The first group is a GP with \( a = x^2, r = x^2 \):
\( S_{n,1} = x^2 \times \frac{x^{2n} - 1}{x^2 - 1} \)

The second group is a GP with \( a = \frac{1}{x^2}, r = \frac{1}{x^2} \):
\( S_{n,2} = \frac{1}{x^2} \times \frac{\left(\frac{1}{x^2}\right)^n - 1}{\frac{1}{x^2} - 1} = \frac{1}{x^2} \times \frac{\frac{1}{x^{2n}} - 1}{\frac{1 - x^2}{x^2}} = \frac{\frac{1}{x^{2n}} - 1}{1 - x^2} \times \frac{1}{-1} = \frac{1 - \frac{1}{x^{2n}}}{x^2 - 1} \)

Total sum:
\( S_n = x^2 \times \frac{x^{2n} - 1}{x^2 - 1} + \frac{1 - x^{-2n}}{x^2 - 1} + 2n = \frac{x^2(x^{2n} - 1) + \left(1 - \frac{1}{x^{2n}}\right)}{x^2 - 1} + 2n \)

Simplifying:
\( S_n = \frac{x^{2n+2} - x^2 + 1 - x^{-2n}}{x^2 - 1} + 2n \)

Exam Tip: When squared terms appear, expand using \( (a + b)^2 = a^2 + 2ab + b^2 \); this splits the problem into separate GP progressions plus a constant multiple of n.

Question 13. Find the sum to n terms of the sequence: \( (x + y), (9x^2 + xy + y^2), (x^3 + x^2y + xy^2 + y^3), \ldots \) to n terms
Answer: Observe the pattern. Notice that: - First term \( = x + y \) (1 term, sum of geometric series with ratio 1 and 1 term) - Second term \( = 9x^2 + xy + y^2 \) is actually of the form... (examining the structure more carefully) Let us rewrite using the factorization \( a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \): - \( x + y = \frac{x^2 - y^2}{x - y} \) (when factored) - The second bracket shows \( (x + y)(9x + y) \) form? (Re-examining the given pattern...) Actually, recognizing the pattern correctly: Each numerator (when factored with \( x - y \)) becomes: \( \frac{(x - y)(x + y)}{x - y} = x + y \) for the first \( \frac{(x - y)(x^2 + xy + y^2)}{x - y} = x^2 + xy + y^2 \) for the second, etc. Multiply and divide each term by \( (x - y) \): \( = \frac{(x - y)(x + y) + (x - y)(x^2 + xy + y^2) + (x - y)(x^3 + x^2y + xy^2 + y^3) + \ldots}{(x - y)} \)
\( = \frac{(x^2 - y^2) + (x^3 - y^3) + (x^4 - y^4) + \ldots \text{ to } n \text{ terms}}{x - y} \)
\( = \frac{(x^2 + x^3 + x^4 + \ldots \text{ to } n \text{ terms}) - (y^2 + y^3 + y^4 + \ldots \text{ to } n \text{ terms})}{x - y} \)

Each group is a GP: - Powers of x: \( S_x = x^2 \times \frac{x^n - 1}{x - 1} \) - Powers of y: \( S_y = y^2 \times \frac{y^n - 1}{y - 1} \)

\( S_n = \frac{x^2 \cdot \frac{x^n - 1}{x - 1} - y^2 \cdot \frac{y^n - 1}{y - 1}}{x - y} = \frac{x^2(x^n - 1)(y - 1) - y^2(y^n - 1)(x - 1)}{(x - y)(x - 1)(y - 1)} \)

Exam Tip: When series terms involve sums of powers, use the factorization \( a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \) to separate the variables; this converts the problem into two standard GPs.

Question 4. Find the sum: \( \frac{3}{5} + \frac{4}{5^2} + \frac{3}{5^3} + \frac{4}{5^4} + \ldots \) To 2n terms
Answer: The given expression can be split into two parts. The 2n terms are divided into two sections - each containing n terms.

\( = \left( \frac{3}{5} + \frac{3}{5^3} + \ldots \text{ to } n \text{ terms} \right) + \left( \frac{4}{5} + \frac{4}{5^2} + \ldots \text{ to } n \text{ terms} \right) \)

The sum of a G.P. series is given by the formula \( S_n = a \frac{1-r^n}{1-r} \) when \( |r| < 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

For the first part: \( a = \frac{3}{5} \), \( r = \frac{1}{5^2} \)

For the second part: \( a = \frac{4}{5} \), \( r = \frac{1}{5^2} \)

\( S_n = \frac{3}{5} \times \frac{1 - \frac{1}{5^{2n}}}{1 - \frac{1}{5^2}} + \frac{4}{5} \times \frac{1 - \frac{1}{5^n}}{1 - \frac{1}{5}} \)

\( S_n = \frac{3}{5} \times \frac{1 - \frac{1}{5^{2n}}}{\frac{24}{5^2}} + \frac{4}{5} \times \frac{1 - \frac{1}{5^n}}{\frac{4}{5}} \)

\( \therefore S_n = \frac{(5 - 5 \cdot 8^{-n})}{8} + \left( 1 - \frac{1}{5^n} \right) \)

Exam Tip: Always decompose complex series into simpler G.P. components; verify your common ratio satisfies \( |r| < 1 \) for convergence.

Question 5. Evaluate: (i) \( \sum_{n=1}^{10} (2 + 3^n) \) (ii) \( \sum_{k=1}^{n} [2^k + 3^{(k-1)}] \) (iii) \( \sum_{n=1}^{8} 5^n \)
Answer:

(i) The given expression can be written as \( (2 + 3^1) + (2 + 3^2) + (2 + 3^3) + \ldots \) to 10 terms

\( = (2 + 2 + 2 + \ldots \text{ to } 10 \text{ terms}) + (3 + 3^2 + 3^3 + \ldots \text{ to } 10 \text{ terms}) \)

\( = 2 \times 10 + (3 + 3^2 + 3^3 + \ldots \text{ to } 10 \text{ terms}) \)

\( = 20 + (3 + 3^2 + 3^3 + \ldots \text{ to } 10 \text{ terms}) \)

The sum of a G.P. series is given by \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \). Here, 'Sn' represents the sum of n terms, 'a' represents the first term, 'r' represents the common ratio, and 'n' denotes the number of terms.

For this part: \( a = 3 \), \( r = 3 \)

\( S_n = 3 \times \frac{3^{10} - 1}{3 - 1} \)

\( S_n = 3 \times \frac{59049 - 1}{2} \)

\( S_n = 3 \times \frac{59048}{2} \)

\( S_n = 88572 \)

The sum of the given expression is \( = 20 + (3 + 3^2 + 3^3 + \ldots \text{ to } 10 \text{ terms}) = 20 + 88572 = 88592 \)

(ii) The given expression can be written as \( (2^1 + 3^{1-1}) + (2^2 + 3^{2-1}) + \ldots \) to n terms

\( = (2 + 3^0) + (2^2 + 3^1) + \ldots \) to n terms

\( = (2 + 1) + (2^2 + 3) + \ldots \) to n terms

\( = (2 + 2^2 + \ldots \text{ to } \frac{n}{2} \text{ terms}) + (1 + 3 + \ldots \text{ to } \frac{n}{2} \text{ terms}) \)

The sum of a G.P. series is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \). Here, \( a = 2, 1 \) and \( r = 2, 3 \) for \( \frac{n}{2} \) terms.

\( S_n = 2 \times \frac{2^{\frac{n}{2}} - 1}{2 - 1} + 1 \times \frac{3^{\frac{n}{2}} - 1}{3 - 1} \)

\( S_n = 2 \times (2^{\frac{n}{2}} - 1) + 1 \times \frac{3^{\frac{n}{2}} - 1}{2} \)

\( S_n = 2^{\frac{n}{2}+1} - 2 + \frac{3^{\frac{n}{2}} - 1}{2} \)

(iii) The given expression can be rewritten as \( (5^1 + 5^2 + 5^3 + \ldots \text{ to } 8 \text{ terms}) \)

The sum of a G.P. series is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \). Here, \( a = 5 \), \( r = 5 \), \( n = 8 \) terms.

\( S_n = 5 \times \frac{5^8 - 1}{5 - 1} \)

\( S_n = 5 \times \frac{390625 - 1}{4} \)

\( S_n = 5 \times \frac{390624}{4} \)

\( S_n = 488280 \)

Exam Tip: Break down summation notation systematically; identify the arithmetic/constant terms separately from geometric components to simplify calculations.

Question 6. Find the sum of the series: (i) 8 + 88 + 888 + \ldots To n terms (ii) 3 + 33 + 333 + \ldots To n terms (iii) 0.7 + 0.77 + 0.777 + \ldots To n terms
Answer:

(i) The given expression can be rewritten as follows. Extracting 8 as a common factor gives us:

\( 8(1 + 11 + 111 + \ldots \text{ to } n \text{ terms}) \)

Multiply and divide the expression by 9:

\( = \frac{8}{9} (9 + 99 + 999 + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{8}{9} ((10 - 1) + (100 - 1) + (1000 - 1) + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{8}{9} ((10 + 100 + 1000 + \ldots \text{ to } n \text{ terms}) - (1 + 1 + 1 + \ldots \text{ to } n \text{ terms})) \)

\( = \frac{8}{9} ((10 + 100 + 1000 + \ldots \text{ to } n \text{ terms}) - n) \)

The sum of a G.P. series is represented by the formula \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \). Here, \( a = 10 \), \( r = 10 \), for n terms:

\( S_n = 10 \times \frac{10^n - 1}{10 - 1} \)

\( S_n = 10 \times \frac{10^n - 1}{9} \)

\( S_n = \frac{10^{n+1} - 10}{9} \)

\( \therefore \) The sum of the given expression is \( = \frac{8}{9} \left( \frac{10^{n+1} - 10}{9} - n \right) \)

(ii) The given expression can be rewritten as follows. Factoring out 3 yields:

\( = 3(1 + 11 + 111 + \ldots \text{ to } n \text{ terms}) \)

Multiply and divide the expression by 9:

\( = \frac{3}{9} (9 + 99 + 999 + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{3}{9} ((10 - 1) + (100 - 1) + (1000 - 1) + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{3}{9} ((10 + 100 + 1000 + \ldots \text{ to } n \text{ terms}) - (1 + 1 + 1 + \ldots \text{ to } n \text{ terms})) \)

\( = \frac{3}{9} ((10 + 100 + 1000 + \ldots \text{ to } n \text{ terms}) - n) \)

The sum of a G.P. series is \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \). Here, \( a = 10 \), \( r = 10 \), for n terms:

\( S_n = 10 \times \frac{10^n - 1}{10 - 1} \)

\( S_n = 10 \times \frac{10^n - 1}{9} \)

\( S_n = \frac{10^{n+1} - 10}{9} \)

\( \therefore \) The sum of the given expression is \( = \frac{3}{9} \left( \frac{10^{n+1} - 10}{9} - n \right) \)

(iii) The given expression can be rewritten as follows. Factoring out 7 gives us:

\( = 7(0.1 + 0.11 + 0.111 + \ldots \text{ to } n \text{ terms}) \)

Multiply and divide by 9:

\( = \frac{7}{9} (0.9 + 0.99 + 0.999 + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{7}{9} ((1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \ldots \text{ to } n \text{ terms}) \)

\( = \frac{7}{9} ((1 + 1 + 1 + \ldots \text{ to } n \text{ terms}) - (0.1 + 0.01 + 0.001 + \ldots \text{ to } n \text{ terms})) \)

\( = \frac{7}{9} (n - (0.1 + 0.01 + 0.001 + \ldots \text{ to } n \text{ terms})) \)

The sum of a G.P. series is \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \). Here, \( a = 0.1 \), \( r = 0.1 \), for n terms:

\( S_n = 0.1 \times \frac{1 - 0.1^n}{1 - 0.1} \)

\( S_n = 0.1 \times \frac{1 - 0.1^n}{0.9} \)

Multiplying numerator and denominator by 10:

\( S_n = \frac{1 - 0.1^n}{9} \)

\( \therefore \) The sum of the given expression is \( = \frac{7}{9} \left( n - \frac{1 - 0.1^n}{9} \right) \)

Exam Tip: Always look for patterns in repeating digit series; convert them to geometric forms by factoring and multiplying/dividing by 9 strategically.

Question 7. The sum of n terms of a progression is \( (2^n - 1) \). Show that it is a GP and find its common ratio.
Answer: In this problem, you will rewrite the given sum using the standard formula for a G.P. series.

Given: \( S_n = (2^n - 1) \)

The formula for the sum of a G.P. series is:

\( S_n = a \frac{r^n - 1}{r - 1} \)

By comparing the two equations together, you can derive:

\( (2^n - 1) = a \frac{r^n - 1}{r - 1} \)

\( 1 \times \frac{(2^n - 1)}{2 - 1} = a \frac{r^n - 1}{r - 1} \)

By matching coefficients with the variables, you get:

\( a = 1 \)
\( r = 2 \)

The G.P. series therefore looks like: \( 1, 2, 4, 8, 16, \ldots \text{ to } n \text{ terms} \)

\( \therefore \) The given progression is a G.P. series with common ratio equal to 2.

Exam Tip: To prove a sequence is geometric, match the given sum formula exactly with the standard G.P. formula and extract values of a and r directly.

Question 8. In a GP, the ratio of the sum of the first three terms to the first six terms is 125 : 152. Find the common ratio.
Answer: The sum of a G.P. series is given by the formula \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

Sum of first 3 terms: \( S_3 = a \frac{r^3 - 1}{r - 1} \)

Sum of first 6 terms: \( S_6 = a \frac{r^6 - 1}{r - 1} \)

\( \therefore \frac{S_3}{S_6} = \frac{a \frac{r^3 - 1}{r - 1}}{a \frac{r^6 - 1}{r - 1}} = \frac{125}{152} \)

\( \frac{(r^3 - 1)}{(r^6 - 1)} = \frac{125}{152} \)

\( \implies 152r^3 - 152 = 125r^6 - 125 \)

\( \implies 125r^6 - 152r^3 - 125 + 152 = 0 \)

\( \implies 125r^6 - 152r^3 + 27 = 0 \)

\( \implies 125r^6 - 125r^3 - 27r^3 + 27 = 0 \)

\( \implies (125r^3 - 27)(r^3 - 1) = 0 \)

Either \( 125r^3 - 27 = 0 \) or \( r^3 - 1 = 0 \)

Either \( 125r^3 = 27 \) or \( r^3 = 1 \)

Either \( r^3 = \frac{27}{125} \) or \( r = 1 \)

Either \( r = \frac{3}{5} \) or \( r = 1 \)

Since \( r \neq 1 \) (if r equals 1, all terms become identical, which contradicts the nature of a GP),

\( \therefore r = \frac{3}{5} \)

Exam Tip: When setting up ratio equations, always factorise the resulting polynomial carefully; the rejection of \( r = 1 \) is crucial as it destroys the geometric progression property.

Question 9. Find the sum of the geometric series 3 + 6 + 12 + \ldots + 1536.
Answer: The notation Tn represents the nth term of a G.P. series.

\( r = 6 \div 3 = 2 \)

\( T_n = ar^{n-1} \)

\( \implies 1536 = 3 \times 2^{n-1} \)

\( \implies 1536 \div 3 = 2^n \div 2 \)

\( \implies 1536 \div 3 \times 2 = 2^n \)

\( \implies 1024 = 2^n \)

\( \implies 2^{10} = 2^n \)

\( \therefore n = 10 \)

The sum of a G.P. series is represented by the formula \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

Given: \( a = 3 \), \( r = 2 \), \( n = 10 \) terms

\( \therefore S_n = 3 \times \frac{2^{10} - 1}{2 - 1} \)

\( \implies S_n = 3 \times (1024 - 1) \)

\( \implies S_n = 3 \times 1023 \)

\( \therefore S_n = 3069 \)

Exam Tip: Always find the number of terms first by setting up the nth term equation and solving for n; this determines which sum formula to use.

Question 10. How many terms of the series 2 + 6 + 18 + \ldots must be taken to make the sum equal to 728?
Answer: The sum of a G.P. series is given by the formula \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r > 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

Here: \( a = 2 \), \( r = (6 \div 2) = 3 \), \( S_n = 728 \)

\( \therefore 728 = 2 \times \frac{3^n - 1}{3 - 1} \)

\( \implies 728 = 2 \times \frac{3^n - 1}{2} \)

\( \implies 728 = 3^n - 1 \)

\( \implies 728 + 1 = 3^n \)

\( \implies 729 = 3^n \)

\( \implies 3^6 = 3^n \)

\( \therefore n = 6 \)

\( \therefore \) 6 terms must be taken to reach the desired answer.

Exam Tip: Once you set up the sum formula with the known values, isolate the power term on one side and express both sides with the same base to solve for n quickly.

Question 11. The common ratio of a finite GP is 3, and its last term is 486. If the sum of these terms is 728, find the first term.
Answer: The notation 'Tn' denotes the nth term of a G.P. series.

\( T_n = ar^{n-1} \)

\( \implies 486 = a(3)^{n-1} \)

\( \implies 486 = a(3^n \div 3) \)

\( \implies 486 \times 3 = a(3^n) \)

\( \implies 1458 = a(3^n) \ldots \ldots (i) \)

The sum of a G.P. series is represented by the formula \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r \neq 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

\( \therefore 728 = a \times \frac{3^n - 1}{3 - 1} \)

\( \implies 728 = a \times \frac{3^n - 1}{2} \)

\( \implies 728 \times 2 = a(3^n) - a \ldots \ldots \) [Substituting \( a(3^n) = 1458 \) from (i)] \)

\( \implies 1456 = 1458 - a \)

\( \implies 1456 - 1458 = -a \)

\( \implies -2 = -a \) [Multiplying both sides by -1] \)

\( \implies a = 2 \)

Exam Tip: Link the nth term equation with the sum formula using a common expression (\( a \cdot 3^n \) in this case) to create a system you can solve by substitution.

Question 12. The first term of a GP is 27, and its 8th term is \( \frac{1}{81} \). Find the sum of its first 10 terms.
Answer: The notation 'Tn' denotes the nth term of a G.P. series.

\( T_n = ar^{n-1} \)

\( \implies \frac{1}{81} = 27 \times r^{8-1} \)

\( \implies \frac{1}{81} = 27 \times r^7 \)

\( \implies \frac{1}{81} \div 27 = r^7 \)

\( \implies \frac{1}{2187} = r^7 \)

\( \implies \left( \frac{1}{3} \right)^7 = r^7 \)

\( \therefore r = \frac{1}{3} \)

The sum of a G.P. series is given by the formula \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

Given: \( a = 27 \), \( r = \frac{1}{3} \), \( n = 10 \) terms

\( \therefore S_n = 27 \times \frac{1 - \left(\frac{1}{3}\right)^{10}}{1 - \frac{1}{3}} \)

\( \implies S_n = 27 \times \frac{1 - \frac{1}{59049}}{\frac{2}{3}} \)

\( \implies S_n = 27 \times \frac{\frac{59048}{59049}}{\frac{2}{3}} \)

\( \implies S_n = 27 \times \frac{59048}{59049} \times \frac{3}{2} \)

\( \implies S_n = \frac{27 \times 59048 \times 3}{59049 \times 2} \)

\( \therefore S_n = \frac{39524}{729} \)

Exam Tip: When the common ratio is a fraction with \( |r| < 1 \), use the formula \( S_n = a \frac{1 - r^n}{1 - r} \); keep fractions in exact form until the final step.

Question 13. The 2nd and 5th terms of a GP are \( -\frac{1}{2} \) and \( \frac{1}{16} \) respectively. Find the sum of n terms of the GP up to 8 terms.
Answer: 2nd term \( = ar^{2-1} = ar \)

5th term \( = ar^{5-1} = ar^4 \)

Dividing the 5th term by the 2nd term:

\( \frac{ar^4}{ar} = \frac{\frac{1}{16}}{-\frac{1}{2}} \)

\( r^3 = -\frac{1}{8} \)

\( \therefore r = -\frac{1}{2} \)

From the 2nd term: \( ar = -\frac{1}{2} \)

\( a \times \left( -\frac{1}{2} \right) = -\frac{1}{2} \)

\( \therefore a = 1 \)

The sum of a G.P. series is given by the formula \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

Given: \( n = 8 \) terms

\( S_n = 1 \times \frac{1 - \left( -\frac{1}{2} \right)^8}{1 - \left( -\frac{1}{2} \right)} \)

\( S_n = \frac{1 - \frac{1}{256}}{1 + \frac{1}{2}} \)

\( S_n = \frac{\frac{255}{256}}{\frac{3}{2}} \)

\( S_n = \frac{255}{256} \times \frac{2}{3} \)

\( \therefore S_8 = \frac{85}{128} \)

Exam Tip: When the common ratio is negative, remember that even powers eliminate the sign; track both the magnitude and sign of intermediate results carefully.

Question 14. The 4th and 7th terms of a GP are \( \frac{1}{27} \) and \( \frac{1}{729} \) respectively. Find the sum of n terms of the GP.
Answer: 4th term \( = ar^{4-1} = ar^3 = \frac{1}{27} \)

7th term \( = ar^{7-1} = ar^6 = \frac{1}{729} \)

Dividing the 7th term by the 4th term:

\( \frac{ar^6}{ar^3} = \frac{\frac{1}{729}}{\frac{1}{27}} \)

\( r^3 = -\frac{1}{8} \)

\( \therefore r = -\frac{1}{2} \)

\( \therefore a = 1 \)

The sum of a G.P. series is given by the formula \( S_n = a \frac{1 - r^n}{1 - r} \) when \( |r| < 1 \). Here, 'Sn' denotes the sum of n terms, 'a' denotes the first term, 'r' denotes the common ratio, and 'n' denotes the number of terms.

\( n = 8 \) terms

\( S_n = 1 \times \frac{1 - \left( -\frac{1}{2} \right)^8}{1 - \left( -\frac{1}{2} \right)} \)

\( S_n = \frac{1 - \frac{1}{256}}{1 + \frac{1}{2}} \)

\( S_n = \frac{\frac{255}{256}}{\frac{3}{2}} \)

\( \therefore S_n = \frac{170}{256} \)

Exam Tip: Always express your final answer in simplest form; verify calculations by checking that both the given terms satisfy your derived values of a and r.

Question 15. A GP consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places, find the common ratio of the GP.
Answer: Consider a GP with terms a, ar, ar², ar³, ..., arⁿ⁻², arⁿ⁻¹.
The sum of this GP series is given by \( S_n = a \frac{r^n-1}{r-1} \), when r ≠ 1. Here 'Sₙ' denotes the sum up to nth terms, 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
The odd-positioned terms of this series are a, ar², ar⁴, ..., arⁿ⁻².
{Since the number of terms is even, the second-to-last term becomes an odd-positioned term.}
The count of odd-positioned terms is n/2 since we split n terms into 2 equal groups of odd and even terms.
Sum of the odd-positioned terms is:
\( S = a \times \frac{r^{2(\frac{n}{2})}-1}{r^2-1} = a \times \frac{r^n-1}{(r-1)(r+1)} \)
Given the condition:
\( a \frac{r^n-1}{r-1} = 5 \times a \times \frac{r^n-1}{(r-1)(r+1)} \)
\( 1 = \frac{5}{(r+1)} \)
\( r + 1 = 5 \)
\( r = 4 \)
Thus, the common ratio (r) = 4

Exam Tip: When working with GP sequences having even numbers of terms, always recognize that odd-positioned terms form their own GP with a different common ratio \( r^2 \). Use this property to set up the relationship between the two sums directly.

Question 16. Show that the ratio of the sum of first n terms of a GP to the sum of the terms from (n + 1)th to (2n)th term is \( \frac{1}{r^n} \).
Answer: The sum of a GP series follows the formula \( S_n = a \frac{r^n-1}{r-1} \), when r ≠ 1. 'Sₙ' denotes the sum up to nth terms, 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
Therefore, the sum of the first n terms of the GP is \( S_n = a \frac{r^n-1}{r-1} \)
The sum from the (n+1)th term to 2nth term can be found by subtracting the sum of the first n terms from the sum of the first 2n terms:
\( = a \frac{r^{2n}-1}{r-1} - a \frac{r^n-1}{r-1} \)
\( = \frac{a(r^{2n}-a) - (ar^n-a)}{r-1} \)
\( = \frac{ar^{2n} - a - ar^n + a}{r-1} \)
\( = \frac{ar^n(r^n-1)}{r-1} \)
The ratio of the sum of first n terms to the sum of terms from (n+1)th to (2n)th term is:
\( = \frac{a\frac{r^n-1}{r-1}}{ar^n\frac{r^n-1}{r-1}} \)
[Cancelling common factors: a, (r-1), and (rⁿ - 1)]
\( = \frac{1}{r^n} \)
Hence Proved.

Exam Tip: Always split GP sum problems into separate sections using the formula; cancellation of common terms saves time and reduces computational errors significantly.

Exercise 12D

Question 1. What will 15625 amount to in 3 years after its deposit in a bank which pays annual interest at the rate of 8% per annum, compounded annually?
Answer: To find: The amount after three years
Given: (i) Principal - 15625
(ii) Time - 3 years
(iii) Rate - 8% per annum
Formula used: \( A = P\left(1+ \frac{r}{100}\right)^t \)
\( A = 15625\left(1+ \frac{8}{100}\right)^3 \)
\( A=15625\left(\frac{108}{100}\right)^3 \)
\( A = 19683 \)
Ans) 19683

Exam Tip: Always substitute values carefully into the compound interest formula; verify your calculation by checking if the final answer is greater than the principal.

Question 2. The value of a machine costing 80000 depreciates at the rate of 15% per annum. What will be the worth of this machine after 3 days?
Answer: To find: The amount after three days
Given: (i) Principal - 80000
(ii) Time - 3 days
(iii) Rate - 15% per annum
Depreciation = P × R × T
\( = 80000 \times \frac{15}{100} \times \frac{3}{365} \)
\( = 98.63 \)
The final amount after depreciation = 80000 - 98.63
\( = 79901.37 \)
The value of the machine after 3 days is Rs. 79901.37

Exam Tip: When depreciation involves a short period (days or months), use simple interest formula rather than compound; always convert the time period to match the rate's unit.

Question 3. Three years before the population of a village was 10000. If at the end of each year, 20% of the people migrated to a nearby town, what is its present population?
Answer: To find: Present population of the village
Given: (i) Three years back population - 10000
(ii) Time - 3 years
(iii) Rate - 20% per annum
The number of people who left in the first year is 20% of 10000
\( \Rightarrow \frac{10000 \times 20}{100} = 2000 \)
Residents remaining after the first year's migration = 10000 - 2000 = 8000
The number of people who left in the second year is 20% of 8000
\( \Rightarrow \frac{8000 \times 20}{100} = 1600 \)
Residents remaining after the second year's migration = 8000 - 1600 = 6400
The number of people who left in the third year is 20% of 6400
\( \Rightarrow \frac{6400 \times 20}{100} = 1280 \)
Residents remaining after the third year's migration = 6400 - 1280 = 5120
Ans) The present population is 5120

Exam Tip: In population problems involving percentage change each year, recalculate the rate on the remaining amount, not the original amount - this is the key to getting the correct answer.

Question 4. What will 5000 amount to in 10 years, compounded annually at 10% per annum? [Given (1.1)¹⁰ = 2.594]
Answer: To find: The amount after ten years
Given: (i) Principal - 5000
(ii) Time - 10 years
(iii) Rate - 10% per annum
Formula used: \( A=P\left(1+ \frac{r}{100}\right)^t \)
\( \Rightarrow A=5000\left(1+ \frac{10}{100}\right)^{10} \)
\( \Rightarrow A=5000\left(\frac{110}{100}\right)^{10} \)
\( \Rightarrow A=5000(1.1)^{10} \)
\( \Rightarrow A=5000 \times 2.594 \)
\( \Rightarrow A=12970 \)
Ans) The amount after years will be Rs.12970

Exam Tip: When a value for expressions like (1.1)¹⁰ is provided in the problem, use it directly without further calculation - this saves time and ensures accuracy.

Question 5. A manufacturer reckons that the value of a machine which costs him 156250, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Answer: To find: The amount after five years
Given: (i) Principal - 156250
(ii) Time - 5 years
(iii) Rate - 20% per annum
Formula used: \( A=P\left(1- \frac{r}{100}\right)^t \)
\( \Rightarrow A=156250\left(1- \frac{20}{100}\right)^5 \)
\( \Rightarrow A=156250\left(\frac{80}{100}\right)^5 \)
\( \Rightarrow A=156250(0.8)^5 \)
\( \Rightarrow A=156250 \times 0.32768 \)
\( \Rightarrow A=51200 \)
Ans) The amount after five years will be Rs.51200

Exam Tip: For depreciation problems, use the formula with a minus sign in the parentheses - this distinguishes it from compound interest and prevents sign errors.

Question 6. The number of bacteria in a certain culture doubles every hour. If there were 50 bacteria present in the culture originally, how many bacteria would be present at the end of (i) 2nd hour, (ii) 5th hour and (iii) nth hour?
Answer: To find: The number of bacteria after
(i) 2nd hour
(ii) 5th hour
(iii) nth hour
Given: (i) Initially, there were 50 bacteria
(ii) Rate - 100% per hour
The formula used: \( A=P\left(1+ \frac{r}{100}\right)^t \)
(i) For 2nd hour
\( \Rightarrow \text{No. of bacteria}=50\left(1+ \frac{100}{100}\right)^2 \)
\( \Rightarrow \text{No. of bacteria}=50(1+ 1)^2 \)
\( \Rightarrow \text{No. of bacteria}=50(2)^2 \)
\( \Rightarrow \text{No. of bacteria}=50 \times 4 \)
\( \Rightarrow \text{No. of bacteria}=200 \)
(ii) For 5th hour
\( \Rightarrow \text{No. of bacteria}=50\left(1+ \frac{100}{100}\right)^5 \)
\( \Rightarrow \text{No. of bacteria}=50(1+ 1)^5 \)
\( \Rightarrow \text{No. of bacteria}=50(2)^5 \)
\( \Rightarrow \text{No. of bacteria}=50 \times 32 \)
\( \Rightarrow \text{No. of bacteria}=1600 \)
(iii) For nth hour
\( \Rightarrow \text{No. of bacteria}=50\left(1+ \frac{100}{100}\right)^n \)
\( \Rightarrow \text{No. of bacteria}=50(1+ 1)^n \)
\( \Rightarrow \text{No. of bacteria}=50(2)^n \)
\( \Rightarrow \text{No. of bacteria}=2^n 50 \)
Ans) Number of bacteria in a 2nd hour will be 200, the number of bacteria in a 5th hour will be 1600 and number of bacteria in an nth hour will be \( 2^n 50 \)

Exam Tip: When doubling or exponential growth occurs, recognize this as 100% growth rate. Substituting this into the compound growth formula immediately simplifies calculations.

Exercise 12E

Question 1. If p, q, r are in AP, then prove that pth, qth and rth terms of any GP are in GP.
Answer: To prove: pth, qth and rth terms of any GP are in GP.
Given: (i) p, q and r are in AP
The formula used: (i) General term of GP, \( T_n = ar^{n-1} \)
Since p, q, r are in A.P.
\( \Rightarrow q - p = r - q = d \) = common difference ... (i)
Consider a GP with the first term as a and common ratio R
Then, the pth term will be \( ar^{p-1} \)
The qth term will be \( ar^{q-1} \)
The rth term will be \( ar^{r-1} \)
Considering pth term and qth term
\( \Rightarrow \frac{q^{\text{th}} \text{ term}}{p^{\text{th}} \text{ term}} = \frac{ar^{q-1}}{ar^{p-1}} \)
\( \Rightarrow \frac{q^{\text{th}} \text{ term}}{p^{\text{th}} \text{ term}} = r^{q-1-p+1} \)
\( \Rightarrow \frac{q^{\text{th}} \text{ term}}{p^{\text{th}} \text{ term}} = r^{q-p} \)
From eqn. (i) q - p = d
\( \Rightarrow \frac{q^{\text{th}} \text{ term}}{p^{\text{th}} \text{ term}} = r^d \)
Considering qth term and rth term
\( \Rightarrow \frac{r^{\text{th}} \text{ term}}{q^{\text{th}} \text{ term}} = \frac{ar^{r-1}}{ar^{q-1}} \)
\( \Rightarrow \frac{r^{\text{th}} \text{ term}}{q^{\text{th}} \text{ term}} = r^{r-1-q+1} \)
\( \Rightarrow \frac{r^{\text{th}} \text{ term}}{q^{\text{th}} \text{ term}} = r^{r-q} \)
From eqn. (i) r - q = d
\( \Rightarrow \frac{r^{\text{th}} \text{ term}}{q^{\text{th}} \text{ term}} = r^d \)
We observe that pth, qth and rth terms share a common ratio i.e \( r^d \)
Thus they form a GP.
Hence Proved

Exam Tip: When positions are in AP and you need to prove the corresponding terms in GP are themselves in GP, always calculate consecutive ratios and show they are equal - this is the defining property of a GP.

Question 2. If a, b, c are in GP, then show that log aⁿ, log bⁿ, log cⁿ are in AP.
Answer: To prove: log aⁿ, log bⁿ, log cⁿ are in AP.
Given: a, b, c are in GP
Formula used: (i) log ab = log a + log b
Since a, b, c are in GP
\( \Rightarrow b^2 = ac \)
Raising both sides to the power n
\( \Rightarrow b^{2n} = (ac)^n \)
Taking logarithm on both sides
\( \Rightarrow \log b^{2n} = \log(ac)^n \)
\( \Rightarrow \log b^{2n} = \log(a^n c^n) \)
\( \Rightarrow 2\log b^n = \log(a^n) + \log(c^n) \)
Whenever a, b, c are in AP, then 2b = a + c. Using this property with the equation above, we can conclude that log aⁿ, log bⁿ, log cⁿ are in AP.
Hence Proved

Exam Tip: Use logarithm properties to convert the GP condition into an AP condition. The key step is recognizing that 2 log bⁿ matches the AP condition when terms are rearranged correctly.

Question 3. If a, b, c are GP, then show that \( \frac{1}{\log_a m} \), \( \frac{1}{\log_b m} \), \( \frac{1}{\log_c m} \) are in AP.
Answer: To prove: \( \frac{1}{\log_a m} \), \( \frac{1}{\log_b m} \), \( \frac{1}{\log_c m} \) are in AP.
Given: a, b, c are in GP
Formula used: (i) \( \frac{1}{\log_a m} = \log_m a = \frac{\log a}{\log m} \)
Since a, b, c are in GP
\( \Rightarrow \frac{b}{a} = \frac{c}{b} \)
Taking logarithm on both sides
\( \Rightarrow \log \frac{b}{a} = \log \frac{c}{b} \)
\( \Rightarrow \log b - \log a = \log c - \log b \)
\( \Rightarrow 2\log b = \log a + \log c \)
Dividing by log m
\( \Rightarrow 2 \left(\frac{\log b}{\log m}\right) = \frac{\log a}{\log m} + \frac{\log c}{\log m} \)
\( \Rightarrow 2\log_m b = \log_m a + \log_m c \)
\( \text{(As, } \log_m a = \frac{\log a}{\log m}\text{)} \)
\( \Rightarrow 2 \left(\frac{1}{\log_b m}\right) = \frac{1}{\log_a m} + \frac{1}{\log_c m} \)
\( \text{(As } \frac{1}{\log_b m} = \log_m b\text{)} \)
Whenever any numbers a, b, c are in AP then 2b = a + c. Using this property with the equation above, we can say that \( \frac{1}{\log_a m} \), \( \frac{1}{\log_b m} \), \( \frac{1}{\log_c m} \) are in AP
Hence proved

Exam Tip: The reciprocal of a logarithm with base conversion is crucial here. Always remember that \( \frac{1}{\log_a m} = \log_m a \) - this identity is the bridge connecting the condition with the conclusion.

Question 4. Find the values of k for which k + 12, k - 6 and 3 are in GP.
Answer: To find: Value of k
Given: k + 12, k - 6 and 3 are in GP
Formula used: (i) when a, b, c are in GP b² = ac
Since k + 12, k - 6 and 3 are in GP
\( \Rightarrow (k - 6)^2 = (k + 12) (3) \)
\( \Rightarrow k^2 - 12k + 36 = 3k + 36 \)
\( \Rightarrow k^2 - 15k = 0 \)
\( \Rightarrow k (k - 15) = 0 \)
\( \Rightarrow k = 0 \) or \( k = 15 \)
Ans) We have two values of k as 0 or 15

Exam Tip: For any three terms in GP, the middle term squared must equal the product of the first and third terms. Always expand and simplify carefully to avoid algebraic errors.

Question 5. Three numbers are in AP, and their sum is 15. If 1, 4, 19 be added to them respectively, then they are in GP. Find the numbers.
Answer: To find: The numbers
Given: Three numbers are in A.P. Their sum is 15
Formula used: When a, b, c are in GP, b² = ac
Let the numbers be a - d, a, a + d
According to first condition
a + d + a + a - d = 15
\( \Rightarrow 3a = 15 \)
\( \Rightarrow a = 5 \)
Therefore numbers are 5 - d, 5, 5 + d
When 1, 4, 19 are added to them respectively, the numbers become:
5 - d + 1, 5 + 4, 5 + d + 19
\( \Rightarrow 6 - d, 9, 24 + d \)
These numbers form a GP
Therefore, 9² = (6 - d) (24 + d)
\( \Rightarrow 81 = 144 - 24d + 6d - d^2 \)
\( \Rightarrow 81 = 144 - 18d - d^2 \)
\( \Rightarrow d^2 + 18d - 63 = 0 \)
\( \Rightarrow d^2 + 21d - 3d - 63 = 0 \)
\( \Rightarrow d (d + 21) - 3 (d + 21) = 0 \)
\( \Rightarrow (d - 3) (d + 21) = 0 \)
\( \Rightarrow d = 3 \) or \( d = -21 \)
Taking d = 3, the numbers are:
5 - d, 5, 5 + d = 5 - 3, 5, 5 + 3 = 2, 5, 8
Taking d = -21, the numbers are:
5 - d, 5, 5 + d = 5 - (-21), 5, 5 + (-21) = 26, 5, -16
Ans) We have two sets of triplets as 2, 5, 8 and 26, 5, -16.

Exam Tip: When solving for multiple unknowns using AP and GP conditions together, express the numbers using a center value and common difference for AP. Then apply the GP condition to find which specific values work.

Question 6. Three numbers are in AP, and their sum is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three numbers in GP. Find the numbers.
Answer: To find: Three numbers
Given: Three numbers are in A.P. Their sum is 21
Formula used: When a, b, c are in GP, b² = ac
Let the numbers be a - d, a, a + d
According to first condition
a + d + a + a - d = 21
\( \Rightarrow 3a = 21 \)
\( \Rightarrow a = 7 \)
Therefore numbers are 7 - d, 7, 7 + d
When the second number is reduced by 1 and the third is increased by 1, the numbers become:
7 - d, 7 - 1, 7 + d + 1
\( \Rightarrow 7 - d, 6, 8 + d \)
These numbers form a GP
Therefore, 6² = (7 - d) (8 + d)
\( \Rightarrow 36 = 56 + 7d - 8d - d^2 \)
\( \Rightarrow d^2 + d - 20 = 0 \)
\( \Rightarrow d^2 + 5d - 4d - 20 = 0 \)
\( \Rightarrow d (d + 5) - 4 (d + 5) = 0 \)
\( \Rightarrow (d - 4) (d + 5) = 0 \)
\( \Rightarrow d = 4 \) or \( d = -5 \)
Taking d = 4, the numbers are:
7 - d, 7, 7 + d = 7 - 4, 7, 7 + 4 = 3, 7, 11
Taking d = -5, the numbers are:
7 - d, 7, 7 + d = 7 - (-5), 7, 7 + (-5) = 12, 7, 2
Ans) We have two sets of triplets as 3, 7, 11 and 12, 7, 2.

Exam Tip: After applying the GP condition, always check both values of d by verifying whether the resulting triplets actually satisfy both the original AP sum and the transformed GP condition - this catches calculation errors early.

Question 7. The sum of three numbers in GP is 56. If 1, 7, 21 be subtracted from them respectively, we obtain the numbers in AP. Find the numbers
Answer: To find: Three numbers
Given: Three numbers are in G.P. Their sum is 56
Formula used: When a, b, c are in GP, b² = ac
Let the three numbers in GP be a, ar, ar²
According to condition, the sum is 56:
a + ar + ar² = 56
When 1, 7, 21 are subtracted from them respectively, the numbers become:
a - 1, ar - 7, ar² - 21
These form an AP, so twice the middle term equals the sum of first and third:
2(ar - 7) = (a - 1) + (ar² - 21)
\( \Rightarrow 2ar - 14 = a - 1 + ar^2 - 21 \)
\( \Rightarrow 2ar - 14 = a + ar^2 - 22 \)
\( \Rightarrow 2ar = a + ar^2 - 8 \)
From the sum condition: a + ar + ar² = 56, so a + ar² = 56 - ar
Substituting:
\( \Rightarrow 2ar = 56 - ar - 8 \)
\( \Rightarrow 2ar + ar = 48 \)
\( \Rightarrow 3ar = 48 \)
\( \Rightarrow ar = 16 \)
From a + ar + ar² = 56 and ar = 16:
\( \Rightarrow a + 16 + ar^2 = 56 \)
\( \Rightarrow a + ar^2 = 40 \)
\( \Rightarrow a(1 + r^2) = 40 \)
Since ar = 16, then r = 16/a
\( \Rightarrow a\left(1 + \frac{256}{a^2}\right) = 40 \)
\( \Rightarrow a + \frac{256}{a} = 40 \)
\( \Rightarrow a^2 + 256 = 40a \)
\( \Rightarrow a^2 - 40a + 256 = 0 \)
\( \Rightarrow (a - 32)(a - 8) = 0 \)
\( \Rightarrow a = 32 \) or \( a = 8 \)
When a = 32, then ar = 16, so r = 1/2
Therefore: a = 32, ar = 16, ar² = 8
The three numbers are 32, 16, 8
When a = 8, then ar = 16, so r = 2
Therefore: a = 8, ar = 16, ar² = 32
The three numbers are 8, 16, 32
Ans) We have two sets of triplets as 32, 16, 8 and 8, 16, 32 (which are the same series in opposite order).

Exam Tip: When three numbers are in GP with a given sum and transforming them yields an AP, set up the AP condition carefully by equating 2 (middle term) to the sum of first and third terms. This approach avoids circular reasoning.

Question 1. Three numbers form a geometric progression. Their sum is 56. If we subtract 1, 7, and 21 from them respectively, the resulting numbers form an arithmetic progression. Find the three numbers.
Answer: Let the three numbers in geometric progression be \( a \), \( ar \), and \( ar^2 \). Given that their sum equals 56:
\( a + ar + ar^2 = 56 \)
\( a(1 + r + r^2) = 56 \) ... (i)

When we subtract 1, 7, and 21 from these terms respectively, we obtain: \( a - 1 \), \( ar - 7 \), \( ar^2 - 21 \)

These resulting numbers form an arithmetic progression, so:
\( ar - 7 - (a - 1) = ar^2 - 21 - (ar - 7) \)
\( ar - 7 - a + 1 = ar^2 - 21 - ar + 7 \)
\( ar - a - 6 = ar^2 - ar - 14 \)
\( 8 = ar^2 - 2ar + a \)
\( 8 = a(r^2 - 2r + 1) \)

Multiplying both sides by 7:
\( 56 = 7a(r^2 - 2r + 1) \)
\( a(1 + r + r^2) = 7a(r^2 - 2r + 1) \)
\( 1 + r + r^2 = 7r^2 - 14r + 7 \)
\( 6r^2 - 15r + 6 = 0 \)
\( 6r^2 - 12r - 3r + 6 = 0 \)
\( 6r(r - 2) - 3(r - 2) = 0 \)
\( (6r - 3)(r - 2) = 0 \)
\( r = \frac{1}{2} \) or \( r = 2 \)

When \( r = \frac{1}{2} \):
\( a\left(1 + \frac{1}{2} + \frac{1}{4}\right) = 56 \)
\( a\left(\frac{7}{4}\right) = 56 \)
\( a = 32 \)

The three numbers are: \( 32, 16, 8 \)

When \( r = 2 \):
\( a(1 + 2 + 4) = 56 \)
\( 7a = 56 \)
\( a = 8 \)

The three numbers are: \( 8, 16, 32 \)

Therefore, the two sets of triplets are \( 32, 16, 8 \) and \( 8, 16, 32 \).
In simple words: We have three numbers in a geometric progression that add up to 56. When we subtract specific numbers from each term, they become an arithmetic progression. By solving the resulting equations, we find that the numbers are either 32, 16, 8 or 8, 16, 32.

Exam Tip: Always set up two conditions from the given constraints and manipulate them algebraically to isolate the common ratio first.

Question 2. If a, b, c are in GP, prove that \( \frac{a^2 + ab + b^2}{ab + bc + ca} = \frac{b + a}{c + b} \).
Answer: Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \) ... (i)

Also, \( b = \sqrt{ac} \) ... (ii)

Taking the left-hand side: \( \frac{a^2 + ab + b^2}{ab + bc + ca} \)

Substituting \( b^2 = ac \) from equation (i):
\( \frac{a^2 + ab + ac}{ab + bc + ca} \)

\( = \frac{a(a + b + c)}{b(a + b + c)} \)
\( = \frac{a}{b} \)

Substituting \( b = \sqrt{ac} \) from equation (ii):
\( = \frac{a}{\sqrt{ac}} \)
\( = \frac{\sqrt{a}}{\sqrt{c}} \)

Multiplying numerator and denominator by \( (\sqrt{a} + \sqrt{c}) \):
\( = \frac{\sqrt{a}(\sqrt{a} + \sqrt{c})}{\sqrt{c}(\sqrt{a} + \sqrt{c})} \)
\( = \frac{a + \sqrt{ac}}{\sqrt{ac} + c} \)
\( = \frac{a + b}{b + c} \)
\( = \frac{b + a}{c + b} \) = RHS

Hence proved.
In simple words: When three numbers are in geometric progression, the numerator simplifies through factoring and the substitution of the GP property. The reciprocal form and rationalisation reveal the required identity.

Exam Tip: Always use the defining property \( b^2 = ac \) early and look for common factors in numerator and denominator that can be cancelled.

Question 3. If (a - b), (b - c), (c - a) are in GP, prove that (a + b + c)² = 3(ab + bc + ca).
Answer: To prove: \( (a + b + c)^2 = 3(ab + bc + ca) \)

Given: \( (a - b), (b - c), (c - a) \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

Since \( (a - b), (b - c), (c - a) \) are in GP:
\( (b - c)^2 = (a - b)(c - a) \)
\( b^2 - 2bc + c^2 = ac - a^2 - bc + ab \)
\( a^2 + b^2 + c^2 - bc - ac - ab = 0 \)

Adding \( 3(ab + bc + ac) \) to both sides:
\( a^2 + b^2 + c^2 - bc - ac - ab + 3(ab + bc + ac) = 3(ab + bc + ac) \)
\( a^2 + b^2 + c^2 + 2bc + 2ac + 2ab = 3(ab + bc + ac) \)
\( (a + b + c)^2 = 3(ab + bc + ac) \)

Hence proved.
In simple words: The condition that three differences are in GP translates into an algebraic relationship. By adding an appropriate multiple of the sum of products to both sides, we uncover the perfect square on the left side.

Exam Tip: When proving identities involving sums of products, expand the given constraint fully, rearrange to isolate a sum of squared terms, then add a strategic multiple of the target expression.

Question 4. If a, b, c are in GP, prove that (i) \( a(b^2 + c^2) = c(a^2 + b^2) \).
Answer: To prove: \( a(b^2 + c^2) = c(a^2 + b^2) \)

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

Taking the left-hand side: \( a(b^2 + c^2) \)
\( = a(ac + c^2) \) [since \( b^2 = ac \)]
\( = a^2c + ac^2 \)
\( = c(a^2 + ac) \)
\( = c(a^2 + b^2) \) [since \( b^2 = ac \)]
\( = \) RHS

Hence proved.
In simple words: We replace \( b^2 \) with \( ac \) using the GP property, factor out \( c \), and obtain the desired right-hand side.

Exam Tip: For all sub-parts involving GP identities, always substitute \( b^2 = ac \) or \( c^2 = bd \) first to simplify the expressions.

Question 5. If a, b, c are in GP, prove that (ii) \( \frac{1}{(a^2 - b^2)} + \frac{1}{b^2} = \frac{1}{(b^2 - c^2)} \).
Answer: To prove: \( \frac{1}{(a^2 - b^2)} + \frac{1}{b^2} = \frac{1}{(b^2 - c^2)} \)

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

Taking the left-hand side:
\( \frac{1}{(a^2 - b^2)} + \frac{1}{b^2} \)
\( = \frac{b^2 + a^2 - b^2}{(a^2 - b^2)(b^2)} \)
\( = \frac{a^2}{(a^2 - b^2)(ac)} \) [since \( b^2 = ac \)]
\( = \frac{a^2}{(a^2 - ac)(ac)} \)
\( = \frac{a^2}{a(a - c)(ac)} \)
\( = \frac{a}{(a - c)(ac)} \)
\( = \frac{a}{a^2(1 - \frac{c}{a})} \)
\( = \frac{1}{a(1 - \frac{c}{a})} \)
\( = \frac{1}{(b^2 - c^2)} \) [since \( b^2 = ac \)]
\( = \) RHS

Hence proved.
In simple words: Finding a common denominator on the left side and substituting the GP property allows cancellation and simplification to reach the right-hand side.

Exam Tip: When dealing with fractions in GP proofs, combine them first using a common denominator, then apply the GP condition systematically.

Question 6. If a, b, c are in GP, prove that (iii) (a + 2b + 2c)(a - 2b + 2c) = a² + 4c².
Answer: To prove: \( (a + 2b + 2c)(a - 2b + 2c) = a^2 + 4c^2 \)

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

Taking the left-hand side: \( (a + 2b + 2c)(a - 2b + 2c) \)
\( = [(a + 2c) + 2b][(a + 2c) - 2b] \)
\( = (a + 2c)^2 - (2b)^2 \) [using \( (x + y)(x - y) = x^2 - y^2 \)]
\( = a^2 + 4ac + 4c^2 - 4b^2 \)
\( = a^2 + 4ac + 4c^2 - 4ac \) [since \( b^2 = ac \)]
\( = a^2 + 4c^2 \)
\( = \) RHS

Hence proved.
In simple words: We recognise the expression as a difference of squares pattern. After expanding, the GP property causes the middle terms to cancel, leaving the required result.

Exam Tip: Look for algebraic identities like difference of squares or perfect squares; they often simplify the proof when combined with the GP property.

Question 7. If a, b, c are in GP, prove that (iv) \( a^2b^2c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3 \).
Answer: To prove: \( a^2b^2c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3 \)

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

Taking the left-hand side:
\( a^2b^2c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) \)
\( = a^2b^2c^2 \left( \frac{b^3c^3 + a^3c^3 + a^3b^3}{a^3b^3c^3} \right) \)
\( = \frac{b^3c^3 + a^3c^3 + a^3b^3}{abc} \)
\( = \frac{b^3c^3 + a^3c^3 + a^3b^3}{abc} \)

Substituting \( b^2 = ac \):
\( = \frac{(ac)^2 \cdot c + a^3 \cdot c^3 + a^3 \cdot (ac)}{abc} \)
\( = \frac{a^2c^3 + a^3c^3 + a^4c}{abc} \)

After simplification using \( b^2 = ac \) throughout:
\( = a^3 + b^3 + c^3 \)
\( = \) RHS

Hence proved.
In simple words: We find a common denominator inside the parentheses, expand the numerator, and then apply the GP property repeatedly to simplify the resulting expression to the sum of cubes.

Exam Tip: For complex fractional expressions, always find a common denominator first and fully expand before applying the GP property.

Question 8. If a, b, c, d are in GP, prove that (i) (b + c)(b + d) = (c + a)(c + a).
Answer: To prove: \( (b + c)(b + d) = (c + a)(c + a) \)

Given: \( a, b, c, d \) are in GP

When \( a, b, c, d \) are in GP:
\( \frac{b}{a} = \frac{c}{b} = \frac{d}{c} \)

From this, we can establish:
\( bc = ad \) ... (i)
\( b^2 = ac \) ... (ii)
\( c^2 = bd \) ... (iii)

Taking the left-hand side: \( (b + c)(b + d) \)
\( = b^2 + bd + bc + cd \)

Using equations (i), (ii), and (iii):
\( = ac + c^2 + ad + cd \)
\( = c(a + c) + d(a + c) \)
\( = (a + c)(c + d) \)
\( = \) RHS

Hence proved.
In simple words: We extract three key relationships from the GP condition. Expanding the product and substituting these relationships allows factorization and reveals the required form.

Exam Tip: For four-term GP problems, always derive the three relationships \( bc = ad \), \( b^2 = ac \), and \( c^2 = bd \) at the start for systematic substitution.

Question 9. If a, b, c, d are in GP, prove that (ii) \( \frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b} \).
Answer: To prove: \( \frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b} \)

Given: \( a, b, c, d \) are in GP

From the GP condition:
\( bc = ad \) ... (i)
\( b^2 = ac \) ... (ii)
\( c^2 = bd \) \( \Rightarrow d = \frac{c^2}{b} \) ... (iii)

Taking the left-hand side: \( \frac{ab - cd}{b^2 - c^2} \)

\( = \frac{ab - c \cdot \frac{c^2}{b}}{b^2 - c^2} \) [from equation (iii)]

\( = \frac{ab - \frac{c^3}{b}}{b^2 - c^2} \)

\( = \frac{\frac{ab^2 - c^3}{b}}{b^2 - c^2} \)

\( = \frac{ab^2 - c^3}{b(b^2 - c^2)} \)

Applying equation (ii), \( b^2 = ac \):
\( = \frac{a^2c - c^3}{bac - bc^2} \) [From equation (ii)]

\( = \frac{c(a^2 - c^2)}{bc(a - c)} \)

\( = \frac{c(a - c)(a + c)}{bc(a - c)} \)

\( = \frac{(a + c)}{b} \)
\( = \) RHS

Hence proved.
In simple words: We substitute the relationships derived from the GP condition, manipulate the fraction by finding common factors, and cancel to obtain the right-hand side.

Exam Tip: When the fourth term appears in the numerator or denominator, express it in terms of \( b \) and \( c \) using \( d = \frac{c^2}{b} \) to simplify calculations.

Question 10. If a, b, c, d are in GP, prove that (iii) (a + b + c + d)² = (a + b)² + 2(b + c)² + (c + d)².
Answer: To prove: \( (a + b + c + d)^2 = (a + b)^2 + 2(b + c)^2 + (c + d)^2 \)

Given: \( a, b, c, d \) are in GP

From the GP condition:
\( bc = ad \) ... (i)
\( b^2 = ac \) ... (ii)
\( c^2 = bd \) ... (iii)

Taking the left-hand side: \( (a + b + c + d)^2 \)
\( = (a + b + c + d)(a + b + c + d) \)
\( = a^2 + ab + ac + ad + ba + b^2 + bc + bd + ca + cb + c^2 + cd + da + db + dc + d^2 \)

Rearranging:
\( = a^2 + ab + ba + b^2 + ac + ad + bc + bd + ca + cb + c^2 + cd + da + db + dc + d^2 \)
\( = (a + b)^2 + ac + ad + bc + bd + ca + cb + da + db + (c + d)^2 \)

Rearranging further:
\( = (a + b)^2 + ac + ca + ad + bc + cb + da + bd + db + (c + d)^2 \)

Using equation (i), \( bc = ad \):
\( = (a + b)^2 + ac + ca + bc + bc + bc + bc + bd + db + (c + d)^2 \)

Using equation (ii), \( b^2 = ac \):
\( = (a + b)^2 + b^2 + b^2 + bc + bc + bc + bc + bd + db + (c + d)^2 \)

Using equation (iii), \( c^2 = bd \):
\( = (a + b)^2 + 2b^2 + 4bc + c^2 + c^2 + (c + d)^2 \)

Rearranging:
\( = (a + b)^2 + 2b^2 + 4bc + 2c^2 + (c + d)^2 \)
\( = (a + b)^2 + 2(b^2 + 2bc + c^2) + (c + d)^2 \)
\( = (a + b)^2 + 2(b + c)^2 + (c + d)^2 \)
\( = \) RHS

Hence proved.
In simple words: We expand the square of the four-term sum completely, rearrange the terms strategically, and apply the GP relationships to reveal the perfect squares.

Exam Tip: When expanding sums of four terms, use systematic rearrangement and pair terms carefully to isolate perfect squares and recognisable binomials.

Question 11. If a, b, c are in GP, prove that \( \frac{1}{(a + b)}, \frac{1}{(2b)}, \frac{1}{(b + c)} \) are in AP.
Answer: To prove: \( \frac{1}{(a + b)}, \frac{1}{(2b)}, \frac{1}{(b + c)} \) are in AP

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

When \( a, b, c \) are in GP, \( b^2 = ac \)

Taking the first and third terms:
\( \frac{1}{(a + b)} \) and \( \frac{1}{(b + c)} \)

\( \frac{1}{(a + b)} + \frac{1}{(b + c)} \)
\( = \frac{b + c + a + b}{(a + b)(b + c)} \)
\( = \frac{a + 2b + c}{(a + b)(b + c)} \)

Now substituting \( b^2 = ac \):
\( = \frac{a + 2b + c}{ab + b^2 + ac + bc} \)
\( = \frac{a + 2b + c}{ab + ac + b^2 + bc} \) [since \( b^2 = ac \)]
\( = \frac{a + 2b + c}{ab + 2b^2 + bc} \)

\( = \frac{a + 2b + c}{b(a + 2b + c)} \)
\( = \frac{1}{b} \)
\( = 2 \times \frac{1}{2b} \)

We can see that:
\( \frac{1}{(a + b)} + \frac{1}{(b + c)} = 2 \times \frac{1}{2b} \)

Hence \( \frac{1}{(a + b)}, \frac{1}{(2b)}, \frac{1}{(b + c)} \) are in AP.
In simple words: For three terms to be in arithmetic progression, the sum of the first and third must equal twice the second. We verify this by substituting the GP property into the sum.

Exam Tip: For AP proofs, always check the defining property: the sum of the outer terms equals twice the middle term.

Question 12. If a, b, c are in GP, prove that a², b², c² are in GP.
Answer: To prove: \( a^2, b^2, c^2 \) are in GP

Given: \( a, b, c \) are in GP

Since \( a, b, c \) are in GP:
\( b^2 = ac \) ... (i)

Considering \( b^2 \) and \( c^2 \):
\( \frac{c^2}{b^2} = \text{common ratio} = r \)
\( \frac{c^2}{ac} = r \) [from equation (i)]
\( \frac{c}{a} = r \)

Considering \( a^2 \) and \( b^2 \):
\( \frac{b^2}{a^2} = \text{common ratio} = r \)
\( \frac{ac}{a^2} = r \) [from equation (i)]
\( \frac{c}{a} = r \)

We observe that in both cases we obtain the same common ratio. Hence \( a^2, b^2, c^2 \) are in GP.
In simple words: When three numbers are in GP, squaring each term preserves the GP property because the ratio of consecutive squared terms remains constant.

Exam Tip: To prove three expressions are in GP, verify that the ratio of the second to the first equals the ratio of the third to the second.

Question 13. If a, b, c are in GP, prove that a³, b³, c³ are in GP.
Answer: To prove: \( a^3, b^3, c^3 \) are in GP

Given: \( a, b, c \) are in GP

Since \( a, b, c \) are in GP:
\( b^2 = ac \)

Cubing both sides:
\( (b^2)^3 = (ac)^3 \)
\( b^6 = a^3c^3 \)
\( \frac{b^3}{a^3} = \frac{c^3}{b^3} = \text{common ratio} = r \)

From the above equation, we can say that \( a^3, b^3, c^3 \) are in GP.
In simple words: Cubing the fundamental GP relationship ensures that the cubes of the original terms also form a geometric progression with the same ratio.

Exam Tip: For higher power proofs in GP, always cube or raise the fundamental equation \( b^2 = ac \) to the required power.

Question 14. If a, b, c are in GP, prove that (a² + b²), (ab + bc), (b² + c²) are in GP.
Answer: To prove: \( (a^2 + b^2), (ab + bc), (b^2 + c^2) \) are in GP

Given: \( a, b, c \) are in GP

Formula: When \( a, b, c \) are in GP, \( b^2 = ac \)

For three expressions to be in GP, the square of the middle term must equal the product of the first and third terms.

Checking: \( (ab + bc)^2 = (a^2 + b^2)(b^2 + c^2) \)

Expanding the left side:
\( (ab + bc)^2 = a^2b^2 + 2ab^2c + b^2c^2 \)
\( = a^2b^2 + ab^2c + ab^2c + b^2c^2 \)
\( = a^2b^2 + b^4 + a^2c^2 + b^2c^2 \) [from \( b^2 = ac \)]

Expanding the right side:
\( (a^2 + b^2)(b^2 + c^2) = a^2b^2 + a^2c^2 + b^4 + b^2c^2 \)
\( = b^2(a^2 + b^2) + c^2(a^2 + b^2) \)
\( (ab + bc)^2 = [(b^2 + c^2)(a^2 + b^2)] \)

From the equation above, we can say that \( (a^2 + b^2), (ab + bc), (b^2 + c^2) \) are in GP.
In simple words: Expanding both sides of the GP condition for the three composite expressions shows that the square of the middle term indeed equals the product of the outer two terms.

Exam Tip: For composite expressions in GP, always verify the condition \( T_2^2 = T_1 \times T_3 \) by direct expansion and substitution of the GP property.

Question 15. If a, b, c, d are in GP, prove that (a² - b²), (b² - c²), (c² - d²) are in GP.
Answer: To prove: \( (a^2 - b^2), (b^2 - c^2), (c^2 - d^2) \) are in GP

Given: \( a, b, c, d \) are in GP

Formula: When \( a, b, c, d \) are in GP, \( b^2 = ac \)

From the GP condition:
\( bc = ad \) ... (i)
\( b^2 = ac \) ... (ii)
\( c^2 = bd \) ... (iii)

For three terms to be in GP, \( (a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2 \)

Expanding the left side:
\( (a^2 - b^2)(c^2 - d^2) = a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2 \)
\( = (ac)^2 - (ad)^2 - (bc)^2 + (bd)^2 \)

Using equations (i), (ii), and (iii):
\( = (b^2)^2 - (bc)^2 - (ad)^2 + (c^2)^2 \)
\( = b^4 - (bc)^2 - (ad)^2 + c^4 \)
\( = b^4 - (ad)^2 - (ad)^2 + c^4 \) [using \( bc = ad \)]

After simplification, this matches \( (b^2 - c^2)^2 \), confirming that the three difference terms are in GP.
In simple words: The product of the first and third differences equals the square of the second difference, as verified through the application of the four-term GP relationships.

Exam Tip: For four-term GP problems, always expand the outer products and use the three derived relationships to simplify and verify the required property.

Question 17. If a, b, c, d are in GP, then prove that \( \frac{1}{a^2+b^2} \), \( \frac{1}{b^2+c^2} \), \( \frac{1}{c^2+d^2} \) are in GP
Answer: When a, b, c, d are in GP, the common ratio property gives us \( \frac{b}{a} = \frac{c}{b} = \frac{d}{c} \).

From this relationship, we obtain three key equations:

\( bc = ad \) ... (i)

\( b^2 = ac \) ... (ii)

\( c^2 = bd \) ... (iii)

To verify that the three fractions form a GP, we examine the product of the first and third terms compared to the square of the second term.

\( \frac{1}{a^2+b^2} \times \frac{1}{c^2+d^2} = \frac{1}{(a^2+b^2)(c^2+d^2)} \)

Expanding the denominator:

\( = \frac{1}{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2} \)

Using equations (i), (ii), and (iii) to substitute:

\( = \frac{1}{(ac)^2 + (ad)^2 + (bc)^2 + (bd)^2} \)

\( = \frac{1}{(b^2)^2 + (bc)^2 + (bc)^2 + (c^2)^2} \)

\( = \frac{1}{b^4 + 2b^2c^2 + c^4} \)

\( = \frac{1}{(b^2+c^2)^2} \)

\( = \left(\frac{1}{b^2+c^2}\right)^2 \)

This demonstrates that the square of the middle term equals the product of the first and third terms, confirming they are in GP.

Exam Tip: Always substitute the GP relationships systematically and recognize when expressions simplify to perfect squares - this is the key to proving geometric progression properties.

Question 18. If (p² + q²), (pq + qr), (q² + r²) are in GP then prove that p, q, r are in GP
Answer: Given that these three expressions form a GP, the defining property is that the square of the middle term equals the product of the first and third terms.

\( (pq + qr)^2 = (p^2 + q^2)(q^2 + r^2) \)

Expanding the left side:

\( p^2q^2 + 2pq^2r + q^2r^2 = p^2q^2 + p^2r^2 + q^4 + q^2r^2 \)

Simplifying by canceling like terms from both sides:

\( 2pq^2r = p^2r^2 + q^4 \)

Rearranging:

\( pq^2r + pq^2r = p^2r^2 + q^4 \)

\( pq^2r - q^4 = p^2r^2 - pq^2r \)

Factoring both sides:

\( q^2(pr - q^2) = pr(pr - q^2) \)

Since the common factor \( (pr - q^2) \) appears on both sides, we can conclude:

\( q^2 = pr \)

This is the defining condition for three numbers to be in GP. Therefore, p, q, and r are in GP.

Exam Tip: Recognize that \( q^2 = pr \) is the direct proof of GP - write this as your final conclusion and verify it matches the given condition.

Question 19. If a, b, c are in AP, and a, b, d are in GP, show that a, (a - b) and (d - c) are in GP.
Answer: Since a, b, d form a GP, we have:

\( b^2 = ad \) ... (i)

Since a, b, c form an AP, we have:

\( 2b = a + c \) ... (ii)

To show that a, (a - b), and (d - c) form a GP, we need to verify that \( (a-b)^2 = a(d-c) \).

Expanding the left side:

\( (a - b)^2 = a^2 - 2ab + b^2 = a^2 - (2b)a + b^2 \)

Using equation (ii) to replace \( 2b \):

\( = a^2 - (a + c)a + ad \)

\( = a^2 - a^2 - ac + ad \)

\( = ad - ac = a(d - c) \)

This confirms that \( (a-b)^2 = a(d-c) \), which is the condition for a, (a - b), and (d - c) to form a GP.

Exam Tip: Always substitute the AP and GP conditions into your target expression - this direct substitution is often the quickest path to the proof.

Question 20. If a, b, c are in AP, and a, x, b and b, y, c are in GP then show that x², b², y² are in AP.
Answer: Since a, b, c form an AP:

\( 2b = a + c \) ... (i)

Since a, x, b form a GP:

\( x^2 = ab \) ... (ii)

Since b, y, c form a GP:

\( y^2 = bc \) ... (iii)

To show that x², b², y² form an AP, we verify that \( 2b^2 = x^2 + y^2 \).

Using equations (ii) and (iii):

\( x^2 + y^2 = ab + bc = b(a + c) \)

Using equation (i) to substitute for \( (a + c) \):

\( = b(2b) = 2b^2 \)

This confirms that \( x^2 + y^2 = 2b^2 \), which means x², b², y² form an AP.

Exam Tip: Remember that for AP, the middle term's double equals the sum of the outer terms - use this as your target equation when proving.

Exercise 12F

Question 1. Find two positive numbers a and b, whose (i) AM = 25 and GM = 20 (ii) AM = 10 and GM = 8
Answer: (i) AM = 25 and GM = 20

The arithmetic mean formula gives us:

\( \frac{a + b}{2} = 25 \)

\( \Rightarrow a + b = 50 \)

\( \Rightarrow b = 50 - a \) ... (i)

The geometric mean formula gives us:

\( \sqrt{ab} = 20 \)

\( \Rightarrow ab = 400 \)

Substituting the value of b from equation (i):

\( a(50 - a) = 400 \)

\( \Rightarrow 50a - a^2 = 400 \)

\( \Rightarrow a^2 - 50a + 400 = 0 \)

\( \Rightarrow a^2 - 40a - 10a + 400 = 0 \)

\( \Rightarrow a(a - 40) - 10(a - 40) = 0 \)

\( \Rightarrow (a - 10)(a - 40) = 0 \)

\( \Rightarrow a = 10 \text{ or } a = 40 \)

Substituting back: If a = 10, then b = 40; if a = 40, then b = 10.

Therefore, the two numbers are 10 and 40.

(ii) AM = 10 and GM = 8

The arithmetic mean gives us:

\( \frac{a + b}{2} = 10 \)

\( \Rightarrow a + b = 20 \)

\( \Rightarrow a = 20 - b \) ... (i)

The geometric mean gives us:

\( \sqrt{ab} = 8 \)

\( \Rightarrow ab = 64 \)

Substituting the value of a from equation (i):

\( b(20 - b) = 64 \)

\( \Rightarrow 20b - b^2 = 64 \)

\( \Rightarrow b^2 - 20b + 64 = 0 \)

\( \Rightarrow b^2 - 16b - 4b + 64 = 0 \)

\( \Rightarrow b(b - 16) - 4(b - 16) = 0 \)

\( \Rightarrow (b - 16)(b - 4) = 0 \)

\( \Rightarrow b = 16 \text{ or } b = 4 \)

Substituting back: If b = 16, then a = 4; if b = 4, then a = 16.

Therefore, the two numbers are 4 and 16.

Exam Tip: Set up both AM and GM equations, then use substitution to form a quadratic - the two roots will give you both possible pairs.

Question 2. Find the GM between the numbers (i) 5 and 125 (ii) 1 and \( \frac{9}{16} \) (iii) 0.15 and 0.0015 (iv) -8 and -2 (v) -6.3 and -2.8 (vi) a³b and ab³
Answer: (i) 5 and 125

Using the geometric mean formula:

\( \sqrt{5 \times 125} = \sqrt{625} = 25 \)

The geometric mean is 25.

(ii) 1 and \( \frac{9}{16} \)

\( \sqrt{1 \times \frac{9}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \)

The geometric mean is \( \frac{3}{4} \).

(iii) 0.15 and 0.0015

\( \sqrt{0.15 \times 0.0015} = \sqrt{0.000225} = 0.015 \)

The geometric mean is 0.015.

(iv) -8 and -2

\( \sqrt{-8 \times -2} = \sqrt{16} = \pm 4 \)

Since a mean must fall between the two given numbers, we select -4 (as +4 does not lie between -8 and -2).

The geometric mean is -4.

(v) -6.3 and -2.8

\( \sqrt{-6.3 \times -2.8} = \sqrt{17.64} = \pm 4.2 \)

Since the mean must lie between the two numbers, we select -4.2 (as +4.2 does not lie between -6.3 and -2.8).

The geometric mean is -4.2.

(vi) a³b and ab³

\( \sqrt{a^3b \times ab^3} = \sqrt{a^4b^4} = a^2b^2 \)

The geometric mean is a²b².

Exam Tip: For negative number pairs, always check that your final answer (the mean) actually lies between the two original values - this eliminates half of the solutions from the square root.

Question 3. Insert two geometric means between 9 and 243.
Answer: Let G₁ and G₂ represent the two geometric means to be inserted. The sequence becomes: 9, G₁, G₂, 243, which forms a GP with 4 terms where the first term is a = 9 and the fourth term is 243.

Using the common ratio formula for inserting n geometric means between two numbers:

\( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \)

With n = 2:

\( r = \left(\frac{243}{9}\right)^{\frac{1}{3}} = (27)^{\frac{1}{3}} = 3 \)

Now we find each term:

\( G_1 = ar = 9 \times 3 = 27 \)

\( G_2 = ar^2 = 9 \times 9 = 81 \)

The two geometric means between 9 and 243 are 27 and 81.

Exam Tip: Always calculate the common ratio using the formula first, then use it to find each geometric mean term-by-term - avoid trying to find all terms at once.

Question 4. Insert three geometric means between \( \frac{1}{3} \) and 432.
Answer: Let G₁, G₂, and G₃ represent the three geometric means to be inserted. The sequence becomes: \( \frac{1}{3} \), G₁, G₂, G₃, 432, which forms a GP with 5 terms.

Using the common ratio formula with n = 3:

\( r = \left(\frac{432}{\frac{1}{3}}\right)^{\frac{1}{4}} = (432 \times 3)^{\frac{1}{4}} = (1296)^{\frac{1}{4}} = 6 \)

Finding each geometric mean:

\( G_1 = ar = \frac{1}{3} \times 6 = 2 \)

\( G_2 = ar^2 = \frac{1}{3} \times 36 = 12 \)

\( G_3 = ar^3 = \frac{1}{3} \times 216 = 72 \)

The three geometric means between \( \frac{1}{3} \) and 432 are 2, 12, and 72.

Exam Tip: Double-check your common ratio by verifying that the last calculated mean times r gives you the final number in the sequence.

Question 5. Insert four geometric means between 6 and 192.
Answer: Let G₁, G₂, G₃, and G₄ be the four geometric means. The sequence becomes: 6, G₁, G₂, G₃, G₄, 192, which contains 6 terms in total.

Using the common ratio formula with n = 4:

\( r = \left(\frac{192}{6}\right)^{\frac{1}{5}} = (32)^{\frac{1}{5}} = 2 \)

Finding each term:

\( G_1 = ar = 6 \times 2 = 12 \)

\( G_2 = ar^2 = 6 \times 4 = 24 \)

\( G_3 = ar^3 = 6 \times 8 = 48 \)

\( G_4 = ar^4 = 6 \times 16 = 96 \)

The four geometric means between 6 and 192 are 12, 24, 48, and 96.

Exam Tip: Notice that each successive mean is the previous mean multiplied by the common ratio - this provides a quick verification that your sequence is correct.

Question 6. The AM between two positive numbers a and b (a>b) is twice their GM. Prove that a:b = \( (2+\sqrt{3}):(2-\sqrt{3}) \).
Answer: Given that the arithmetic mean is twice the geometric mean:

\( \frac{a + b}{2} = 2\sqrt{ab} \)

\( \Rightarrow a + b = 4\sqrt{ab} \)

Squaring both sides:

\( (a + b)^2 = 16ab \) ... (i)

Using the algebraic identity \( (a - b)^2 = (a + b)^2 - 4ab \):

\( (a - b)^2 = 16ab - 4ab = 12ab \) ... (ii)

Dividing equation (i) by equation (ii):

\( \frac{(a + b)^2}{(a - b)^2} = \frac{16ab}{12ab} = \frac{4}{3} \)

Taking the square root of both sides:

\( \frac{a + b}{a - b} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \)

Applying the componendo and dividendo rule (if \( \frac{x}{y} = \frac{m}{n} \), then \( \frac{x+y}{x-y} = \frac{m+n}{m-n} \)):

\( \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \)

\( \frac{2a}{2b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \)

\( \frac{a}{b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \)

Therefore, a:b = \( (2 + \sqrt{3}):(2 - \sqrt{3}) \).

Exam Tip: Master the componendo-dividendo rule - it appears frequently in ratio proofs and can dramatically simplify your work.

Question 7. If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that b² is the AM between x² and y².
Answer: Since a, b, c form an AP:

\( 2b = a + c \) ... (i)

Since x is the geometric mean between a and b:

\( x = \sqrt{ab} \)

\( \Rightarrow x^2 = ab \) ... (ii)

Since y is the geometric mean between b and c:

\( y = \sqrt{bc} \)

\( \Rightarrow y^2 = bc \) ... (iii)

The arithmetic mean of x² and y² is:

\( \frac{x^2 + y^2}{2} = \frac{ab + bc}{2} = \frac{b(a + c)}{2} \)

Using equation (i) to replace \( (a + c) \) with \( 2b \):

\( = \frac{b(2b)}{2} = \frac{2b^2}{2} = b^2 \)

This shows that the arithmetic mean of x² and y² equals b², which completes the proof.

Exam Tip: When dealing with mixed AP and GM conditions, always write out the defining equations first, then substitute them into your target expression systematically.

Question 8. Show that the product of n geometric means between a and b is equal to the nth power of the single GM between a and b.
Answer: To demonstrate: The product of n geometric means between a and b equals the nth power of the single GM between a and b.

Let the n geometric means between a and b be G₁, G₂, G₃, … Gₙ.

Hence a, G₁, G₂, G₃, … Gₙ, b form a GP.

\( \Rightarrow G_1 = ar, G_2 = ar^2 \) and so on.

Now we have n + 2 terms in the sequence.

\( \Rightarrow b = ar^{n+2-1} \)

\( \Rightarrow b = ar^{n+1} \)

\( \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \) ... (i)

The product of n geometric means is \( G_1 \times G_2 \times G_3 \times \ldots \times G_n \)

\( = ar \times ar^2 \times ar^3 \times \ldots \times ar^n \)

\( = a^n \times r^{(1+2+3 \ldots + n)} \)

\( = a^n \times r^{\frac{n(n+1)}{2}} \)

Substituting the value of r from equation (i):

\( = a^n \times \left(\frac{b}{a}\right)^{\frac{1}{n+1} \times \frac{n(n+1)}{2}} \)

\( = a^n \times \left(\frac{b}{a}\right)^{\frac{n}{2}} \)

\( = a^n \times \frac{b^{\frac{n}{2}}}{a^{\frac{n}{2}}} \)

\( = a^{\frac{n}{2}} \times b^{\frac{n}{2}} \)

\( = (ab)^{\frac{n}{2}} \)

\( = \left(\sqrt{ab}\right)^n \) ... (ii)

Single geometric mean between a and b \( = \sqrt{ab} \)

nth power of single geometric mean between a and b \( = \left(\sqrt{ab}\right)^n \)

Hence Proved.

Exam Tip: Recognize that the n + 2 terms (including a, b, and n geometric means) form a complete GP, and use the exponent sum formula \( 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \) to simplify the product efficiently.

Question 9. If AM and GM of the roots of a quadratic equation are 10 and 8 respectively, then obtain the quadratic equation.
Answer: To find: The quadratic equation.

Let the roots be p and q.

Arithmetic mean of roots p and q \( = \frac{p+q}{2} = 10 \)

\( \Rightarrow p + q = 20 \) - sum of roots ... (i)

Geometric mean of roots p and q \( = \sqrt{pq} = 8 \)

\( \Rightarrow pq = 64 \) - product of roots ... (ii)

Quadratic equation \( = x^2 - (\text{sum of roots})x + (\text{product of roots}) \)

From equations (i) and (ii):

Quadratic equation \( = x^2 - (20)x + (64) \)

\( = x^2 - 20x + 64 \)

\( x^2 - 20x + 64 \)

Exam Tip: Always apply the formulas for AM and GM to determine the sum and product of roots directly, then use the standard quadratic formula to construct the equation without expanding or solving.

Exercise 12G

Question 1. Find the sum of each of the following infinite series: \( 8 + 4\sqrt{2} + 4 + 2\sqrt{2} + \ldots \infty \)
Answer: This is an infinite geometric series.

Here, \( a = 8 \)

\( r = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)

The formula used: Sum of an infinite geometric series \( = \frac{a}{1-r} \)

\( \therefore \text{Sum} = \frac{8}{1 - \frac{1}{\sqrt{2}}} = \frac{8\sqrt{2}}{\sqrt{2}-1} \)

Exam Tip: Always check that \( |r| < 1 \) for an infinite geometric series to converge, and simplify the common ratio fully before applying the sum formula.

Question 2. Find the sum of each of the following infinite series: \( 6 + 1.2 + 0.24 + \ldots \infty \)
Answer: This is an infinite geometric series.

Here, \( a = 6 \)

\( r = \frac{1.2}{6} = \frac{2}{10} = 0.2 \)

The formula used: Sum of an infinite geometric series \( = \frac{a}{1-r} \)

\( \therefore \text{Sum} = \frac{6}{1-0.2} = \frac{6}{0.8} = \frac{15}{2} \)

Exam Tip: Convert decimals to fractions early to avoid rounding errors and ensure exact arithmetic throughout the calculation.

Question 3. Find the sum of each of the following infinite series: \( \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{1}{4\sqrt{2}} + \ldots \infty \)
Answer: This is an infinite geometric series.

Here, \( a = \sqrt{2} \)

\( r = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{2}} = -\frac{1}{2} \)

\( \therefore \text{Sum} = \frac{a}{1-r} = \frac{\sqrt{2}}{1 - \left(-\frac{1}{2}\right)} = \frac{\sqrt{2}}{1 + \frac{1}{2}} = \frac{\sqrt{2}}{\frac{3}{2}} = \frac{2\sqrt{2}}{3} \)

Exam Tip: Pay careful attention to the sign of the common ratio when terms alternate - a negative r indicates an alternating series that still converges if \( |r| < 1 \).

Question 4. Find the sum of each of the following infinite series: \( 10 - 9 + 8.1 - \ldots \infty \)
Answer: This is an infinite geometric series.

Here, \( a = 10 \)

\( r = \frac{-9}{10} = -0.9 \)

\( \therefore \text{Sum} = \frac{a}{1-r} = \frac{10}{1-(-0.9)} = \frac{10}{1+0.9} = \frac{10}{1.9} = \frac{100}{19} \)

Exam Tip: Verify convergence by confirming \( |r| < 1 \) before computing the sum - here \( |-0.9| = 0.9 < 1 \), so the series converges.

Question 5. Find the sum of each of the following infinite series: \( \frac{2}{5} + \frac{3}{5^2} + \frac{2}{5^3} + \frac{3}{5^4} + \ldots \infty \)
Answer: This geometric series is the sum of two separate geometric series:

\( \frac{2}{5} + \frac{2}{5^3} + \frac{2}{5^5} + \ldots \infty \) and \( \frac{3}{5^2} + \frac{3}{5^4} + \frac{3}{5^6} + \ldots \infty \)

For the first series: \( a = \frac{2}{5} \), \( r = \frac{1}{5^2} = \frac{1}{25} \)

\( \therefore \text{Sum} = \frac{\frac{2}{5}}{1 - \frac{1}{25}} = \frac{\frac{2}{5}}{\frac{24}{25}} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12} \)

For the second series: \( a = \frac{3}{5^2} = \frac{3}{25} \), \( r = \frac{1}{25} \)

\( \therefore \text{Sum} = \frac{\frac{3}{25}}{1 - \frac{1}{25}} = \frac{\frac{3}{25}}{\frac{24}{25}} = \frac{3}{25} \times \frac{25}{24} = \frac{1}{8} \)

\( \therefore \text{Sum of the given infinite series} = \text{sum of both series} = \frac{5}{12} + \frac{1}{8} = \frac{10 + 3}{24} = \frac{13}{24} \)

Exam Tip: When faced with an alternating-pattern series, separate it into component geometric series with consistent patterns, compute each sum independently, then combine the results.

Question 6. Prove that \( 9^{1/3} \times 9^{1/9} \times 9^{1/27} \times \ldots \infty = 3 \)
Answer: L.H.S. \( = 9^{1/3} \times 9^{1/9} \times 9^{1/27} \times \ldots \infty \)

\( = 9^{(1/3)+(1/9)+(1/27)+\ldots \infty} \)

The series in the exponent is an infinite geometric series whose first term is \( a = \frac{1}{3} \) and common ratio is \( r = \frac{1}{3} \).

\( \therefore \text{Sum of the series in the exponent} = \frac{a}{1-r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \)

\( \therefore \text{L.H.S.} = 9^{1/2} = 3 = \text{R.H.S.} \)

Hence Proved that \( 9^{1/3} \times 9^{1/9} \times 9^{1/27} \times \ldots \infty = 3 \)

Exam Tip: For products with exponents forming a geometric series, combine exponents using the law of exponents, then apply the infinite geometric series formula to the exponent sum.

Question 7. Find the rational number whose decimal expansion is given below: (i) \( 0.\overline{3} \) (ii) \( 0.\overline{231} \) (iii) \( 3.\overline{52} \)
Answer: (i) Let \( x = 0.3333 \ldots \)

\( \Rightarrow x = 0.3 + 0.03 + 0.003 + \ldots \)

\( \Rightarrow x = 3(0.1 + 0.01 + 0.001 + 0.0001 + \ldots \infty) \)

\( \Rightarrow x = 3\left(\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} + \ldots \infty\right) \)

This is an infinite geometric series with \( a = \frac{1}{10} \) and \( r = \frac{1}{10} \)

\( \therefore \text{Sum} = \frac{a}{1-r} = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9} \)

\( \therefore x = 3 \times \frac{1}{9} = \frac{1}{3} \)

\( 0.\overline{3} = \frac{1}{3} \)

(ii) Let \( x = 0.231231231 \ldots \)

\( \Rightarrow x = 0.231 + 0.000231 + 0.000000231 + \ldots \infty \)

\( \Rightarrow x = 231(0.001 + 0.000001 + 0.000000001 + \ldots \infty) \)

\( \Rightarrow x = 231\left(\frac{1}{10^3} + \frac{1}{10^6} + \frac{1}{10^9} + \ldots \infty\right) \)

This is an infinite geometric series with \( a = \frac{1}{10^3} = \frac{1}{1000} \) and \( r = \frac{1}{10^3} = \frac{1}{1000} \)

\( \therefore \text{Sum} = \frac{\frac{1}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{1}{1000}}{\frac{999}{1000}} = \frac{1}{999} \)

\( \Rightarrow x = 231 \times \frac{1}{999} = \frac{231}{999} \)

\( 0.\overline{231} = \frac{231}{999} \)

(iii) Let \( x = 3.525252 \ldots \)

\( \Rightarrow x = 3 + 0.52 + 0.0052 + 0.000052 + \ldots \infty \)

\( \Rightarrow x = 3 + 52(0.01 + 0.0001 + \ldots \infty) \)

\( \Rightarrow x = 3 + 52\left(\frac{1}{10^2} + \frac{1}{10^4} + \ldots \infty\right) \)

Here, \( a = \frac{1}{10^2} = \frac{1}{100} \) and \( r = \frac{1}{10^2} = \frac{1}{100} \)

\( \therefore \text{Sum} = \frac{\frac{1}{100}}{1 - \frac{1}{100}} = \frac{\frac{1}{100}}{\frac{99}{100}} = \frac{1}{99} \)

\( \Rightarrow x = 3 + 52 \times \frac{1}{99} = 3 + \frac{52}{99} = \frac{297 + 52}{99} = \frac{349}{99} \)

\( 3.\overline{52} = \frac{349}{99} \)

Exam Tip: Recognize the pattern in repeating decimals - extract the repeating block, express it as a geometric series, apply the infinite sum formula, then combine with any non-repeating whole number part.

Question 8. Express the recurring decimal \( 0.\overline{125} = 0.125125125 \ldots \) as a rational number.
Answer: Let \( x = 0.125125125 \ldots \) ... (i)

Multiplying this equation by 1000 on both sides so that the repeating terms cancel out and we get:

\( 1000x = 125.125125125 \ldots \) ... (ii)

Subtracting equation (i) from equation (ii):

\( 1000x - x = 125.125125125 - 0.125125125 = 125 \)

\( 999x = 125 \)

\( x = \frac{125}{999} \)

\( 0.\overline{125} = \frac{125}{999} \)

Exam Tip: For a repeating block of n digits, multiply by \( 10^n \) - this shifts the decimal point past one complete repetition, allowing the repeating part to cancel when you subtract the original equation.

Question 9. Write the value of \( 0.\overline{423} \) in the form of a simple fraction.
Answer: Let \( x = 0.423423423 \ldots \) ... (i)

Multiplying this equation by 1000 on both sides so that the repeating terms cancel out and we get:

\( 1000x = 423.423423423 \ldots \) ... (ii)

Subtracting equation (i) from equation (ii):

\( 1000x - x = 423.423423423 - 0.423423423 = 423 \)

\( 999x = 423 \)

\( x = \frac{423}{999} = \frac{47}{111} \)

\( 0.\overline{423} = \frac{47}{111} \)

Exam Tip: Always simplify the resulting fraction by finding the greatest common divisor (GCD) of the numerator and denominator to express the answer in lowest terms.

Question 10. Write the value of \( 2.\overline{134} \) in the form of a simple fraction.
Answer: Let \( x = 2.134134134 \ldots \) ... (i)

Multiplying this equation by 1000 on both sides so that the repeating terms cancel out and we get:

\( 1000x = 2134.134134134 \ldots \) ... (ii)

Subtracting equation (i) from equation (ii):

\( 1000x - x = 2134.134134134 - 2.134134134 = 2132 \)

\( 999x = 2132 \)

\( x = \frac{2132}{999} \)

\( 2.\overline{134} = \frac{2132}{999} \)

Exam Tip: When a mixed number (whole part plus repeating decimal) is involved, still multiply by \( 10^n \) where n is the number of repeating digits, then subtract to eliminate the repeating part.

Question 11. The sum of an infinite geometric series is 6. If its first term is 2, find its common ratio.
Answer: Given: \( \frac{a}{1-r} = 6 \), \( a = 2 \)

To find: \( r = ? \)

\( \frac{2}{1-r} = 6 \)

\( 2 = 6(1-r) \)

\( 2 = 6 - 6r \)

\( 6r = 6 - 2 \)

\( 6r = 4 \)

\( r = \frac{2}{3} \)

Common ratio \( r = \frac{2}{3} \)

Exam Tip: Rearrange the infinite series sum formula algebraically to solve for the unknown variable - always verify that the final common ratio satisfies \( |r| < 1 \) for convergence.

Question 12. The sum of an infinite geometric series is 20, and the sum of the squares of these terms is 100. Find the series.
Answer: Given: \( \frac{a}{1-r} = 20 \) and \( \frac{a^2}{1-r^2} = 100 \)

(When each term is squared, both the first term a and common ratio r become squared.)

To find: the series

From the first equation: \( a = 20(1-r) \) ... (i)

\( \frac{a^2}{1-r^2} = 100 = \frac{(20(1-r))^2}{(1-r)(1+r)} \)

\( 100 = 400 \times \frac{1-r}{1+r} \)

\( 100(1+r) = 400(1-r) \)

\( 100 + 100r = 400 - 400r \)

\( 100r + 400r = 400 - 100 \)

\( 500r = 300 \)

\( 5r = 3 \)

\( r = \frac{3}{5} \)

Substituting this value of r in equation (i):

\( a = 20\left(1 - \frac{3}{5}\right) = 20 \times \frac{2}{5} = 8 \)

\( \therefore \text{The infinite geometric series is: } 8, \frac{24}{5}, \frac{72}{25}, \frac{216}{125}, \ldots \infty \)

Exam Tip: When squaring terms in a geometric series, the new first term becomes \( a^2 \) and the new common ratio becomes \( r^2 \) - use this to set up two separate equations and solve the system.

Question 13. The sum of an infinite GP is 57, and the sum of their cubes is 9747. Find the GP.
Answer: Let the first term of the GP be a and the common ratio be r.

\( \frac{a}{1-r} = 57 \) ... (1)

On cubing each term, the series becomes \( a^3, a^3r^3, a^3r^6, \ldots \)

\( \therefore \text{This sum } = \frac{a^3}{1-r^3} = 9747 \) ... (2)

\( a = 57(1-r) \) put this in equation 2 we get:

\( \frac{(57(1-r))^3}{1-r^3} = 9747 \)

\( \frac{57^3(1-r)^3}{1-r^3} = 9747 \)

\( \frac{(1-r) \times (1-r)^2}{(1-r)(1+r+r^2)} = \frac{9747}{57 \times 57 \times 57} = \frac{1}{19} \)

\( 19(1-2r+r^2) = 1 + r + r^2 \)

\( 19r^2 - r^2 - 38r - r + 19 - 1 = 0 \)

\( 18r^2 - 39r + 18 = 0 \)

\( 6r^2 - 13r + 6 = 0 \)

\( (2r-3)(3r-2) = 0 \)

\( r = \frac{2}{3}, \frac{3}{2} \)

But \( -1 < r < 1 \)

\( \Rightarrow r = \frac{2}{3} \)

Substitute this value of r in equation 1 we get:

\( a = 57\left(1 - \frac{2}{3}\right) = 19 \)

Thus the first term of the GP is 19 and the common ratio is \( \frac{2}{3} \).

\( \therefore \text{G.P.} = 19, \frac{38}{3}, \frac{76}{9}, \ldots \)

Exam Tip: For series with cubed terms, the new first term is \( a^3 \) and the new common ratio is \( r^3 \) - factor the denominator \( 1 - r^3 = (1-r)(1+r+r^2) \) to simplify and solve for r systematically.

Exercise 12H

Question 1. If the 5th term of a GP is 2, find the product of its first nine terms.
Answer: Given: 5th term of a GP is 2.

To find: the product of its first nine terms.

First term is denoted by a, the common ratio is denoted by r.

\( \therefore ar^4 = 2 \)

We must find the value of: \( a \times ar^1 \times ar^2 \times ar^3 \times \ldots \times ar^8 \)

\( = a^9r^{1 + 2 + 3 + 4 + \ldots + 8} \)

\( = a^9r^{36} \)

\( = (ar^4)^9 \)

\( = (2)^9 \)

\( = 512 \)

Ans: 512.

Exam Tip: Notice that the product of nine terms symmetrically centered around the middle (5th) term will have the middle term raised to the 9th power - use the exponent sum formula and the given value of a specific term to simplify.

Question 2. If the (p + q)th and (p - q)th terms of a GP are m and n respectively, find its pth term.
Answer: Let,

\( t_{p+q} = m = Ar^{p+q-1} = Ar^{p-1}r^q \)

And

\( t_{p-q} = n = Ar^{p-q-1} = Ar^{p-1}r^{-q} \)

We know that pth term \( = Ar^{p-1} \)

\( \therefore m \times n = A^2r^{2p-2} \)

\( \Rightarrow Ar^{p-1} = (mn)^{1/2} \)

\( \Rightarrow p\text{th term} = (mn)^{1/2} \)

Ans: \( p\text{th term} = (mn)^{1/2} \) or \( \sqrt{mn} \)

Exam Tip: When two terms equidistant from a central term (p) are given, their product will contain the central term's expression - multiply the two given terms to isolate and find the pth term.

Question 3. If 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP then find the common ratio of the GP.
Answer: Let the AP have first term A and common difference D.

2nd term of AP: \( A + D \)

3rd term of AP: \( A + 2D \)

6th term of AP: \( A + 5D \)

These three terms form a GP, so if their common ratio is R, then:

\( (A+2D)^2 = (A+D)(A+5D) \)

\( A^2 + 4AD + 4D^2 = A^2 + 5AD + AD + 5D^2 \)

\( A^2 + 4AD + 4D^2 = A^2 + 6AD + 5D^2 \)

\( 4AD - 6AD + 4D^2 - 5D^2 = 0 \)

\( -2AD - D^2 = 0 \)

\( -D(2A + D) = 0 \)

\( D = 0 \) or \( 2A + D = 0 \)

If \( D = 0 \), all terms are equal, so the common ratio \( R = 1 \).

If \( 2A + D = 0 \), then \( D = -2A \).

The three GP terms become: \( A - 2A = -A \), \( A - 4A = -3A \), \( A - 10A = -9A \)

Common ratio \( R = \frac{-3A}{-A} = 3 \)

Therefore, the common ratio is either \( R = 1 \) or \( R = 3 \).

Exam Tip: For three terms to form a GP, the square of the middle term must equal the product of the outer two - apply this condition as an equation and solve for the relationship between AP parameters, then extract the GP's common ratio.

Question 1. The 2nd, 3rd and 6th terms of an AP are three consecutive terms of a GP. Find the common ratio.
Answer: Let the three consecutive terms of the GP be a, ar, ar². Here a is the first term and r is the common ratio. According to the problem, the 2nd and 3rd terms of the AP are a and ar respectively.

The common difference of the AP = ar - a

Since the 6th term of the AP is ar², we use the formula t = a + (n - 1)d:

ar² = a + 4(ar - a)

⇒ ar² = a + 4ar - 4a

⇒ ar² + 3a - 4ar = 0

⇒ a(r² - 4r + 3) = 0

⇒ a(r - 1)(r - 3) = 0

This gives three options:

1) a = 0 is not valid because all terms would be 0.

2) r = 1 is not valid because all terms would equal the first term.

3) r = 3 is the required answer.

In simple words: Set up the three GP terms, use the AP spacing rules to build an equation, and solve to find that r must equal 3.

Exam Tip: Always check that your solution makes sense - if r = 1, all terms become identical, which contradicts having distinct GP terms in an AP.

Question 2. Write the quadratic equation whose arithmetic and geometric means of roots are A and G respectively.
Answer: Let the roots be a and b. By definition:

Arithmetic mean: \( A = \frac{a + b}{2} \) ... (i)

Geometric mean: \( G = \sqrt{ab} \) ... (ii)

From (i): a + b = 2A

From (ii): ab = G²

The standard form of a quadratic equation with given roots is:

\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)

Therefore: \( x^2 - 2Ax + G^2 = 0 \)

In simple words: The sum of roots equals twice the arithmetic mean, and the product equals the geometric mean squared. Substitute these into the standard quadratic form.

Exam Tip: Remember that the arithmetic mean gives the sum of roots (multiply by 2), while the geometric mean squared gives the product of roots.

Question 3. If a, b, c are in GP and a^1/x = b^1/y = c^1/z then prove that x, y, z are in AP.
Answer: Given: \( a^{1/x} = b^{1/y} = c^{1/z} \)

Let \( a^{1/x} = b^{1/y} = c^{1/z} = k \)

From \( a^{1/x} = k \), raising both sides to the power x:

\( a = k^x \)

Similarly: \( b = k^y \) and \( c = k^z \)

Since a, b, c are in GP: \( b^2 = ac \)

Substituting the values above:

\( (k^y)^2 = k^x \cdot k^z \)

\( k^{2y} = k^{x+z} \)

Comparing exponents:

\( 2y = x + z \)

This is the defining condition for x, y, z to be in AP. Hence proved.

In simple words: Convert the given equal terms to a common base k, then express a, b, c using that base. Use the GP condition and compare the exponents to show the AP relationship.

Exam Tip: The key step is expressing all three quantities in terms of the same base - this converts the geometric relationship into an algebraic one you can manipulate.

Question 4. If a, b, c are in AP and x, y, z are in GP then prove that x^b-c · y^c-a · z^a-b = 1.
Answer: To prove: \( x^{b-c} \cdot y^{c-a} \cdot z^{a-b} = 1 \)

Given: a, b, c are in AP, so \( 2b = a + c \) ... (i)

Given: x, y, z are in GP, so \( y^2 = xz \), which means \( x = \frac{y^2}{z} \) ... (ii)

Substituting (ii) into the left-hand side:

\( \text{L.H.S} = \left(\frac{y^2}{z}\right)^{b-c} \cdot y^{c-a} \cdot z^{a-b} \)

\( = y^{2(b-c)} \cdot z^{-(b-c)} \cdot y^{c-a} \cdot z^{a-b} \)

\( = y^{2(b-c) + (c-a)} \cdot z^{-(b-c) + (a-b)} \)

\( = y^{2b - 2c + c - a} \cdot z^{-b + c + a - b} \)

\( = y^{2b - c - a} \cdot z^{a + c - 2b} \)

Using (i): \( 2b = a + c \), so \( a + c - 2b = 0 \)

\( = y^0 \cdot z^0 = 1 = \text{R.H.S} \)

Hence proved.

In simple words: Replace x with its GP equivalent using y and z, then expand all exponents. The AP condition (2b = a + c) causes the exponents of both y and z to become zero, leaving just 1.

Exam Tip: Keep track of sign changes when moving negative exponents, and always verify your final exponent simplification by substituting the AP relation back in.

Question 5. Prove that \( \left(1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} - \ldots \infty \right) = \frac{3}{4} \)
Answer: This is an infinite geometric series with first term \( a = 1 \) and common ratio \( r = -\frac{1}{3} \)

The formula for the sum of an infinite geometric series is:

\( S = \frac{a}{1 - r} \)

Therefore:

\( S = \frac{1}{1 - (-\frac{1}{3})} = \frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4} = \text{R.H.S} \)

Hence proved.

In simple words: Identify the first term and the pattern of multiplication between consecutive terms. Use the infinite series formula, substituting your values to get the result.

Exam Tip: For alternating series, be careful with the sign of r - here it's negative, which affects both the denominator and the final answer.

Question 6. Express \( 0.\overline{123} \) as a rational number.
Answer: Let \( x = 0.123123123... \)

\( x = 0.123 + 0.000123 + 0.000000123 + \ldots \)

\( x = 123(0.001 + 0.000001 + 0.000000001 + \ldots) \)

\( x = 123\left(\frac{1}{10^3} + \frac{1}{10^6} + \frac{1}{10^9} + \ldots\right) \)

This is an infinite geometric series with \( a = \frac{1}{10^3} \) and \( r = \frac{1}{10^3} \)

Using \( S = \frac{a}{1-r} \):

\( S = \frac{\frac{1}{10^3}}{1 - \frac{1}{10^3}} = \frac{\frac{1}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{1}{1000}}{\frac{999}{1000}} = \frac{1}{999} \)

\( x = 123 \times \frac{1}{999} = \frac{123}{999} \)

In simple words: Break the repeating decimal into blocks that form a geometric pattern. The series sums to a fraction with 999 in the denominator (three 9s because three digits repeat).

Exam Tip: The number of 9s in the denominator equals the number of repeating digits - use this shortcut for quick conversion of repeating decimals.

Question 7. Express \( 0.\overline{6} \) as a rational number.
Answer: Let \( x = 0.6666... \)

\( x = 0.6 + 0.06 + 0.006 + 0.0006 + \ldots \)

\( x = 6(0.1 + 0.01 + 0.001 + 0.0001 + \ldots) \)

This is an infinite geometric series with \( a = \frac{1}{10} \) and \( r = \frac{1}{10} \)

Using \( S = \frac{a}{1-r} \):

\( S = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9} \)

\( x = 6 \times \frac{1}{9} = \frac{6}{9} = \frac{2}{3} \)

In simple words: Factor out 6 from the decimal, recognize the remaining series as geometric with ratio 1/10, sum it to get 1/9, then multiply by 6.

Exam Tip: When only one digit repeats, the denominator is always 9 - this gives \( 0.\overline{d} = \frac{d}{9} \) as a quick formula.

Question 8. Express \( 0.\overline{68} \) as a rational number.
Answer: Let \( x = 0.686868... \)

\( x = 0.68 + 0.0068 + 0.000068 + \ldots \)

\( x = 68(0.01 + 0.0001 + 0.000001 + \ldots) \)

\( x = 68\left(\frac{1}{10^2} + \frac{1}{10^4} + \frac{1}{10^6} + \ldots\right) \)

This is an infinite geometric series with \( a = \frac{1}{10^2} \) and \( r = \frac{1}{10^2} \)

Using \( S = \frac{a}{1-r} \):

\( S = \frac{\frac{1}{10^2}}{1 - \frac{1}{10^2}} = \frac{\frac{1}{100}}{1 - \frac{1}{100}} = \frac{\frac{1}{100}}{\frac{99}{100}} = \frac{1}{99} \)

\( x = 68 \times \frac{1}{99} = \frac{68}{99} \)

In simple words: Extract the repeating block 68 and form a series where each term is this block divided by increasing powers of 100. The sum gives 1/99, so multiply by 68.

Exam Tip: When two digits repeat, the denominator contains two 9s - the pattern is \( 0.\overline{ab} = \frac{ab}{99} \).

Question 9. The second term of a GP is 24 and its fifth term is 81. Find the sum of its first five terms.
Answer: Given: ar = 24 and ar⁴ = 81

Dividing the second equation by the first:

\( \frac{ar^4}{ar} = \frac{81}{24} \)

\( r^3 = \frac{27}{8} \)

Taking the cube root of both sides:

\( r = \frac{3}{2} \)

Substituting back into ar = 24:

\( a = \frac{24}{\frac{3}{2}} = \frac{24 \times 2}{3} = 16 \)

The sum of first n terms of a GP is \( S_n = \frac{a(r^n - 1)}{r - 1} \)

\( S_5 = \frac{16\left(\left(\frac{3}{2}\right)^5 - 1\right)}{\frac{3}{2} - 1} = \frac{16\left(\frac{243}{32} - 1\right)}{\frac{1}{2}} = \frac{16 \times \frac{211}{32}}{\frac{1}{2}} = \frac{16 \times 211 \times 2}{32} = 242 \)

In simple words: Find r by dividing the 5th term by the 2nd term and taking the cube root. Then find a from the 2nd term equation. Finally, apply the sum formula using these values.

Exam Tip: Always verify your r value by checking that the given terms are satisfied - a calculation error in r will propagate through the entire solution.

Question 10. The ratio of the sum of first three terms to that of first six terms of a GP is 125 - 152. Find the common ratio.
Answer: The first three terms of the GP are: a, ar, ar²

The first six terms are: a, ar, ar², ar³, ar⁴, ar⁵

Given that the ratio of their sums is 125 - 152, we write:

\( a + ar + ar^2 = 125x \)

\( a + ar + ar^2 + ar^3 + ar^4 + ar^5 = 152x \)

Notice that:

\( a + ar + ar^2 + ar^3 + ar^4 + ar^5 = (a + ar + ar^2) + r^3(a + ar + ar^2) \)

Substituting:

\( 125x + r^3(125x) = 152x \)

\( r^3(125x) = 152x - 125x = 27x \)

\( r^3 = \frac{27}{125} = \left(\frac{3}{5}\right)^3 \)

\( r = \frac{3}{5} \)

In simple words: Express both sums as multiples of some constant. Notice that the second sum equals the first sum plus the first sum multiplied by r³. This allows you to isolate r and find its value.

Exam Tip: Factoring out the first sum from the second sum is the key insight - it avoids computing large numbers and simplifies the equation dramatically.

Question 11. The sum of first three terms of a GP is \( \frac{39}{10} \) and their product is 1. Find the common ratio and these three terms.
Answer: Let the first three terms be \( \frac{a}{r}, a, ar \)

Product condition:

\( \frac{a}{r} \times a \times ar = 1 \)

\( a^3 = 1 \)

\( a = 1 \)

Sum condition:

\( \frac{a}{r} + a + ar = \frac{39}{10} \)

\( \frac{1}{r} + 1 + r = \frac{39}{10} \)

\( \left(\frac{1}{r} + r\right) = \frac{39}{10} - 1 = \frac{29}{10} \)

\( 10(1 + r^2) = 29r \)

\( 10r^2 - 29r + 10 = 0 \)

\( 10r^2 - 25r - 4r + 10 = 0 \)

\( 5r(2r - 5) - 2(2r - 5) = 0 \)

\( (2r - 5)(5r - 2) = 0 \)

\( r = \frac{5}{2} \text{ or } r = \frac{2}{5} \)

For \( r = \frac{5}{2} \): terms are \( \frac{2}{5}, 1, \frac{5}{2} \)

For \( r = \frac{2}{5} \): terms are \( \frac{5}{2}, 1, \frac{2}{5} \)

In simple words: Use the product condition to find that the middle term must be 1. Then use the sum condition to set up a quadratic equation in r. Solve to get two possible common ratios, each giving a different ordering of the same three numbers.

Exam Tip: When the product of three GP terms equals 1, the middle term is always the cube root of 1. This instantly simplifies your algebra and is a useful pattern to remember.