RS Aggarwal Solutions for Class 11 Chapter 11 Arithmetic Progression

Exercise 11A

Exam Tip: Arithmetic Progressions are sequences where consecutive terms differ by a constant value called the common difference. Master the formula aₙ = a + (n - 1)d to solve most AP problems efficiently.

Question 1. Write first 4 terms in each of the sequences:
(i) aₙ = (5n + 2)
(ii) aₙ = \( \frac{2n - 3}{4} \)
(iii) aₙ = (–1)ⁿ⁻¹ × 2ⁿ ⁺ ¹
Answer: To locate: First four terms of a given series.

(i) Given: nth term of series is (5n + 2)
Put n = 1, 2, 3, 4 in nth term, we get first (a₁), second (a₂), third (a₃) and fourth (a₄) term:
a₁ = (5 × 1 + 2) = 7
a₂ = (5 × 2 + 2) = 12
a₃ = (5 × 3 + 2) = 17
a₄ = (5 × 4 + 2) = 22
First four terms of given series are 7, 12, 17, 22

Alternative Method: When you have the first term (a or a₁) and second term (a₂), calculate the difference (a₂ - a₁). Now keep adding this difference to the last term to get the next term. For example, a₁ = 7 and a₂ = 12, so difference is 12 - 7 = 5. Now a₃ = 12 + 5 = 17, a₄ = 17 + 5 = 22. (This method works only for A.P.)

Note: When you have nth term in the form of (a × n + b), then common difference of this series equals a. This type of series is called A.P (Arithmetic Progression), where a and b are constants and n is the number of terms.

(ii) Given: nth term of series is \( \frac{2n - 3}{4} \)
Put n = 1, 2, 3, 4 in nth term, we get first (a₁), second (a₂), third (a₃) and fourth (a₄) term:
a₁ = \( \frac{2(1) - 3}{4} = \frac{-1}{4} \)
a₂ = \( \frac{2(2) - 3}{4} = \frac{1}{4} \)
a₃ = \( \frac{2(3) - 3}{4} = \frac{3}{4} \)
a₄ = \( \frac{2(4) - 3}{4} = \frac{5}{4} \)
First four terms of given series are \( \frac{-1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{5}{4} \)

(iii) Given: nth term of series is (–1)ⁿ⁻¹ × 2ⁿ ⁺ ¹
Put n = 1, 2, 3, 4 in nth term, we get first (a₁), second (a₂), third (a₃) and fourth (a₄) term:
a₁ = (–1)¹⁻¹ × 2¹ ⁺ ¹ = (–1)⁰ × 2² = 1 × 4 = 4
a₂ = (–1)²⁻¹ × 2² ⁺ ¹ = (–1)¹ × 2³ = (–1) × 8 = (–8)
a₃ = (–1)³⁻¹ × 2³ ⁺ ¹ = (–1)² × 2⁴ = 1 × 16 = 16
a₄ = (–1)⁴⁻¹ × 2⁴ ⁺ ¹ = (–1)³ × 2⁵ = (–1) × 32 = (–32)
First four terms of given series are 4, –8, 16, –32
In simple words: Replace n with 1, 2, 3, 4 one at a time to find each term. For part (i), just plug in the values and compute. For parts (ii) and (iii), follow the same substitution method carefully.

Exam Tip: Always check your calculations by verifying the common difference between consecutive terms in parts (i) and (ii) - they should be constant for an A.P.

Question 2. Find the first five terms of the sequence, defined by a₁ = 1, aₙ = aₙ₋₁ + 3 for n ≥ 2.
Answer: To locate: First five terms of a given sequence.
Condition: n ≥ 2
Given: a₁ = 1, aₙ = aₙ₋₁ + 3 for n ≥ 2

Put n = 2 in nth term (i.e. aₙ), we have
a₂ = a₂₋₁ + 3 = a₁ + 3 = 1 + 3 = 4 (as a₁ = 1)

Put n = 3 in nth term (i.e. aₙ), we have
a₃ = a₃₋₁ + 3 = a₂ + 3 = 4 + 3 = 7 (as a₂ = 4)

Put n = 4 in nth term (i.e. aₙ), we have
a₄ = a₄₋₁ + 3 = a₃ + 3 = 7 + 3 = 10 (as a₃ = 7)

Put n = 5 in nth term (i.e. aₙ), we have
a₅ = a₅₋₁ + 3 = a₄ + 3 = 10 + 3 = 13 (as a₄ = 10)

First five terms of a given sequence are 1, 4, 7, 10, 13
In simple words: Start with 1. Keep adding 3 to get the next term. This gives a simple pattern: 1, then 4, then 7, then 10, then 13.

Exam Tip: In recursive sequences like this, each new term depends on the previous one. Write out one or two terms explicitly to spot the pattern before finishing the rest.

Question 3. Find the first 5 terms of the sequence, defined by a₁ = –1, aₙ = \( \frac{a_{n-1}}{n} \) for n ≥ 2.
Answer: To locate: First five terms of a given sequence.
Condition: n ≥ 2

Given: a₁ = –1, aₙ = \( \frac{a_{n-1}}{n} \) for n ≥ 2

Put n = 2 in nth term (i.e. aₙ), we have
a₂ = \( \frac{a_1}{2} = \frac{-1}{2} \) (as a₁ = –1)

Put n = 3 in nth term (i.e. aₙ), we have
a₃ = \( \frac{a_2}{3} = \frac{-1/2}{3} = \frac{-1}{6} \) (as a₂ = \( \frac{-1}{2} \))

Put n = 4 in nth term (i.e. aₙ), we have
a₄ = \( \frac{a_3}{4} = \frac{-1/6}{4} = \frac{-1}{24} \) (as a₃ = \( \frac{-1}{6} \))

Put n = 5 in nth term (i.e. aₙ), we have
a₅ = \( \frac{a_4}{5} = \frac{-1/24}{5} = \frac{-1}{120} \) (as a₄ = \( \frac{-1}{24} \))

First five terms of a given sequence are –1, \( \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120} \)
In simple words: Start with –1. Each time, divide by the position number n to find the next term. The numerator stays as –1, while the denominator grows as 2, 6, 24, 120.

Exam Tip: Watch the denominators carefully - they are products of consecutive integers (factorials). Keep track of simplification to avoid arithmetic errors.

Question 4. Find the 23rd term of the AP 7, 5, 3, 1, –1, –3, …
Answer: To locate: 23rd term of the AP
Given: The series is 7, 5, 3, 1, –1, –3, …
a₁ = 7, a₂ = 5 and d = 5 - 7 = -2
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

So put n = 23 in above formula, we have
a₂₃ = a₁ + (23 - 1)(-2) = 7 - 44 = –37

So 23rd term of AP equals –37.
In simple words: The first term is 7, and we go down by 2 each step. To reach the 23rd term, we go down 22 times (that's 22 × 2 = 44), so 7 - 44 gives –37.

Exam Tip: Always compute common difference as (later term - earlier term). A negative d means the sequence is decreasing. Verify your answer by checking a couple of middle terms.

Question 5. Find the 20th term of the AP \( \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, … \)
Answer: To locate: 20th term of the AP
Given: The series is \( \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, … \)
a₁ = \( \sqrt{2} \), a₂ = \( 3\sqrt{2} \) and d = \( 3\sqrt{2} - \sqrt{2} = 2\sqrt{2} \)
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

a₂₀ = a₁ + (20 - 1)( \( 2\sqrt{2} \)) = \( \sqrt{2} + 38\sqrt{2} = 39\sqrt{2} \)

So 20th term of AP equals \( 39\sqrt{2} \).
In simple words: Each term is \( 2\sqrt{2} \) more than the one before it. Starting from \( \sqrt{2} \), we add \( 2\sqrt{2} \) a total of 19 times, giving us \( 39\sqrt{2} \).

Exam Tip: When the AP contains surds, factor them out and treat the coefficients as a separate AP. This makes calculation easier and reduces errors.

Question 6. Find the nth term of the AP 8, 3, –2, –7, –12, …
Answer: To locate: nth term of the AP
Given: The series is 8, 3, –2, –7, –12, …
a₁ = 8, a₂ = 3 and d = 3 - 8 = -5
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = a₁ + (n - 1)(-5) = 8 - (5n - 5) = 8 - 5n + 5 = 13 - 5n

So the nth term of AP equals 13 - 5n.
In simple words: The general formula tells us how to find any term by its position. When you know the first term and the common difference, plug them into the formula to get a simple expression for the nth term.

Exam Tip: The general term formula aₙ = 13 - 5n is your shortcut. Use it to find any term without listing all the ones before it.

Question 7. Find the nth term of the AP 1, \( \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, … \)
Answer: To locate: nth term of the AP
Given: The series is 1, \( \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, … \)
a₁ = 1, a₂ = \( \frac{5}{6} \) and d = \( \frac{5}{6} - 1 = \frac{-1}{6} \)
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = a₁ + (n - 1) \( \left(\frac{-1}{6}\right) = 1 - \frac{n-1}{6} = \frac{6 - (n-1)}{6} = \frac{7 - n}{6} \)

So the nth term of AP equals \( \frac{7 - n}{6} \).
In simple words: Start with 1 and subtract \( \frac{1}{6} \) each time. The formula \( \frac{7 - n}{6} \) captures this pattern - just change n and you get any term.

Exam Tip: When working with fractions in APs, convert all terms to a common denominator first to spot the pattern clearly.

Question 8. Which term of the AP 9, 14, 19, 24, 29, … is 379?
Answer: To locate: we need to find n when aₙ = 379
Given: The series is 9, 14, 19, 24, 29, … and aₙ = 379
a₁ = 9, a₂ = 14 and d = 14 - 9 = 5
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = 379 = a₁ + (n - 1)5
379 - 9 = (n - 1)5 [subtract 9 on both sides]
370 = (n - 1)5
74 = (n - 1) [divide both sides by 5]
n = 75th

The 75th term of this AP equals 379.
In simple words: You know 379 is somewhere in the list. Set aₙ = 379 and solve for n. This tells you it is the 75th term.

Exam Tip: Always check: is the target value (379) greater than the first term (9) and does the common difference (5) divide evenly into the difference? If not, the number is not in the AP.

Question 9. Which term of the AP 64, 60, 56, 52, 48, … is 0?
Answer: To locate: we need to find n when aₙ = 0
Given: The series is 64, 60, 56, 52, 48, … and aₙ = 0
a₁ = 64, a₂ = 60 and d = 60 - 64 = -4
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = 0 = a₁ + (n - 1)(-4)
0 - 64 = (n - 1)(-4) [subtract 64 on both sides]
-64 = (n - 1)(-4)
64 = (n - 1) × 4 [divide both sides by -1]
16 = (n - 1) [divide both sides by 4]
n = 17th [add 1 on both sides]

The 17th term of this AP equals 0.
In simple words: Set up the equation with aₙ = 0 and solve for n step by step. The 17th position gives you the zero term.

Exam Tip: Be careful with negative common differences - sign errors are common. Double-check by substituting n back into the formula.

Question 10. How many terms are there in the AP 11, 18, 25, 32, 39, …, 207?
Answer: To locate: we need to find a number of terms in the given AP.
Given: The series is 11, 18, 25, 32, 39, …, 207
a₁ = 11, a₂ = 18, d = 18 - 11 = 7 and aₙ = 207
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = 207 = a₁ + (n - 1)(7)
207 - 11 = (n - 1)(7) [subtract 11 on both sides]
196 = (n - 1)(7)
28 = (n - 1) [divide both sides by 7]
n = 29 [add 1 on both sides]

So there are 29 terms in this AP.
In simple words: You know the first and last terms, and the common difference. Use the formula backwards to find how many steps it takes to reach 207 from 11.

Exam Tip: Always verify: a₁ + (n-1)d should equal the last term. If it does not, you have made an error.

Question 11. How many terms are there in the AP 1\( \frac{5}{6}, 1\frac{1}{6}, 1\frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, …, –16\frac{1}{6} \)?
Answer: To locate: we need to find number of terms in the given AP.
Given: The series is \( \frac{15}{6}, 1\frac{1}{6}, \frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, …, –16\frac{1}{6} \).
Convert mixed numbers: \( 1\frac{5}{6} = \frac{11}{6}, 1\frac{1}{6} = \frac{7}{6}, –16\frac{1}{6} = \frac{-97}{6} \)

a₁ = \( \frac{15}{6} = \frac{11}{6} \), a₂ = \( \frac{7}{6} \), d = \( \frac{7}{6} - \frac{11}{6} = \frac{-4}{6} \) and aₙ = \( \frac{-97}{6} \)
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = \( \frac{-97}{6} = \frac{11}{6} + (n - 1)\left(\frac{-4}{6}\right) \)
\( \frac{-97}{6} - \frac{11}{6} = (n - 1)\left(\frac{-4}{6}\right) \) [subtract \( \frac{11}{6} \) on both sides]
\( \frac{-108}{6} = (n - 1)\left(\frac{-4}{6}\right) \) [multiply both sides by \( \frac{-6}{4} \) or divide both sides by \( \frac{-4}{6} \)]
27 = (n - 1) [add 1 on both sides]
n = 28

So there are 28 terms in this AP.
In simple words: Convert all mixed numbers to improper fractions so everything is consistent. Then use the standard formula to count how many steps from start to finish.

Exam Tip: Mixed numbers can cause confusion. Always convert them first before doing any calculations with the AP formula.

Question 12. Is - 47 a term of the AP 5, 2, –1, –4, –7, …?
Answer: To locate: –47 is a term of the AP or not.
Given: The series is 5, 2, –1, –4, –7, …
a₁ = 5, a₂ = 2, and d = 2 - 5 = -3 (let suppose aₙ = –47)

NOTE: n is a natural number.
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

aₙ = –47 = a + (n - 1)d
–47 = 5 + (n - 1)(-3)
–47 - 5 = (n - 1)(-3) [subtract 5 on both sides]
-52 = (n - 1)(-3)
52 = (n - 1) × 3 [divide both sides by -1]
\( \frac{52}{3} \) = (n - 1) [divide both sides by 3]
\( 17\frac{1}{3} \) = (n - 1) [divide both sides by 3]
\( 18\frac{1}{3} \) = n [add 1 on both sides]

Since n is not a natural number, –47 is not the term of this AP.
In simple words: If –47 were in the sequence, n would have to be a whole number. But when we solve, we get a fraction, so –47 does not belong to this AP.

Exam Tip: To check if a value is in an AP, solve for n. Only whole numbers ≥ 1 count - decimals or fractions mean the value is not in the sequence.

Question 13. The 5th and 13th terms of an AP are 5 and –3 respectively. Find the AP and its 30th term.
Answer: To locate: AP and its 30th term (i.e. a₃₀ = ?)
Given: a₅ = 5 and a₁₃ = –3
Formula Used: aₙ = a + (n - 1)d
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

By using the above formula, we have
a₅ = 5 = a + (5 - 1)d, and a₁₃ = –3 = a + (13 - 1)d
a + 4d = 5 and a + 12d = –3

On solving above 2 equations, we get
a = 9 and d = (–1)

So a₃₀ = 9 + 29(–1) = –20

AP is (9, 8, 7, 6, 5, 4, …) and 30th term = –20
In simple words: You have two conditions. Set up two equations and solve them together to find a and d. Once you know both, finding the 30th term is straightforward.

Exam Tip: Write both conditions as equations using the same formula, then subtract one equation from the other to eliminate a and solve for d first.

Question 14. The 2nd, 31st and the last terms of an AP are 7\( \frac{3}{4}, \frac{1}{2} \) and –6\( \frac{1}{2} \) respectively. Find the first term and the number of terms.
Answer: To locate: First term and number of terms.
Given: a₂ = \( \frac{31}{4}, a_{31} = \frac{1}{2}, \) and aₙ = \( \frac{-13}{2} \)
Formula Used: aₙ = a + (n - 1)d
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

By using above formula, we have
a₂ = \( \frac{31}{4} = a + d \) and a₃₁ = \( \frac{1}{2} = a + (31 - 1)d \)

On solving both equations, we get
a = 8 and d = -0.25

Now aₙ = \( \frac{-13}{2} = 8 + (n - 1)(-0.25) \)
On solving the above equation, we get
n = 59

So the first term equals 8 and the number of terms equals 59.
In simple words: Use the two known terms to find a and d. Then use the last term to find how many terms total are in the sequence.

Exam Tip: When you have the 2nd term, not the 1st, the formula becomes a + d instead of just a. Make sure you adjust your equations accordingly.

Question 15. If the 9th term of an AP is 0, prove that its 29th term is double the 19th term.
Answer: Prove that: 29th term is double the 19th term (i.e. a₂₉ = 2a₁₉)
Given: a₉ = 0
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)
Formula Used: aₙ = a + (n - 1)d

So a₉ = 0 implies a + (9 - 1)d = 0
a + 8d = 0
a = (–8d) ….equation (i)

Now a₂₉ = a + (29 - 1)d and a₁₉ = a + (19 - 1)d
a₂₉ = a + 28d and a₁₉ = a + 18d ….equation (ii)

By using equation (i) in equation (ii), we have
a₂₉ = –8d + 28d and a₁₉ = –8d + 18d
a₂₉ = 20d and a₁₉ = 10d

So a₂₉ = 2a₁₉

HENCE PROVED
In simple words: The condition a₉ = 0 gives us a special relationship. Use this to express a in terms of d, then substitute into the expressions for a₂₉ and a₁₉ to show the required relationship.

Exam Tip: In "prove" problems, always use the given condition to eliminate one variable, then substitute into the expressions you need to prove.

Question 16. The 4th term of an AP is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
Answer: To locate: First term (a) and common difference (d)
Given: a₄ = 3a₁ and a₇ = 2a₃ + 1
(Where a = a₁ is first term, aₙ is nth term and d is common difference of given AP)
Formula Used: aₙ = a + (n - 1)d

a₄ = 3a₁ implies a + 3d = 3a, so 3d = 2a ….equation (i) and
a₇ = 2a₃ + 1 implies a + 6d = 2(a + 2d) + 1, so 2d = a + 1 ….equation (ii)

On solving both equations (i) and (ii), we get
a = 3 and d = 2

So the first term equals 3, and the common difference equals 2.
In simple words: Translate each word condition into an equation using the AP formula. Then solve the two equations together like any system of equations.

Exam Tip: Read the word problem carefully and write each condition as an equation. Phrases like "three times" mean multiply, "exceeds by" means add.

Question 17. If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term of the AP is zero.
Answer: Show that: 18th term of the AP is zero.
Given: 7a₇ = 11a₁₁
(Where a₇ is seventh term, a₁₁ is eleventh term, aₙ is nth term and d is common difference of given AP)
Formula Used: aₙ = a + (n - 1)d

7(a + 6d) = 11(a + 10d)
7a + 42d = 11a + 110d implies 68d = (–4a)
a + 17d = 0 ….equation (i)

Now a₁₈ = a + (18 - 1)d
a₁₈ = a + 17d = 0 [by using equation (i)]

HENCE PROVED

[NOTE: If n times the nth term of AP is equal to m times the mth term of same AP then its (m + n)th term equals zero]
In simple words: Expand the given condition and simplify it. You will find that the expression a + 17d equals zero, which is exactly a₁₈. So a₁₈ = 0.

Exam Tip: Remember the note at the end - if na_n = ma_m, then a_(m+n) = 0. This shortcut saves time and can be verified with the formula.

Question 18. Find the 28th term from the end of the AP 6, 9, 12, 15, 18, …, 102.
Answer: To locate: 28th term from the end of the AP.
Given: The AP is 6, 9, 12, 15, 18, …, 102
a₁ = 6, a₂ = 9, d = 9 - 6 = 3 and l = 102
Formula Used: nth term from the end = l - (n - 1)d
(Where l is last term and d is common difference of given AP)

By using nth term from the end = l - (n - 1)d formula
28th term from the end = 102 - 27d implies 102 - 27 × 3 = 21

So 28th term from the end equals 21.
In simple words: To count from the end, start with the last term and subtract the common difference the required number of times. The 28th from the end means 27 steps back from 102.

Exam Tip: The formula for "from the end" is different from the regular formula. Use l instead of a, and count backwards. Verify by finding what position this term is from the start.

Question 19. Find the 16th term from the end of the AP 7, 2, –3, –8, –13, …, –113
Answer: To locate: 16th term from the end of the AP.
Given: The AP is 7, 2, –3, –8, –13, …, –113
a₁ = 7, a₂ = 2, d = 2 - 7 = -5 and l = –113
Formula Used: nth term from the end = l - (n - 1)d
(Where l is last term and d is common difference of given AP)

By using nth term from the end = l - (n - 1)d formula
16th term from the end = (–113) - 15d implies (–113) - 15 × (–5) = –38

So 16th term from the end equals –38.
In simple words: Start at the end value –113. The common difference is –5, so going back 15 steps means adding 15 × 5 = 75 to get –113 + 75 = –38.

Exam Tip: Watch the signs carefully when d is negative. Subtracting a negative is the same as adding.

Question 20. How many 3 - digit numbers are divisible by 7?
Answer: To locate: 3 - digit numbers divisible by 7.
First 3 - digit number divisible by 7 is 105
Second 3 - digit number divisible by 7 is 112 and
Last 3 - digit number divisible by 7 is 994.

Given: The AP is 105, 112, 119, …, 994
a₁ = 105, a₂ = 112, d = 112 - 105 = 7 and aₙ = 994
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

994 = 105 + (n - 1)7
889 = (n - 1)7
127 = (n - 1)
n = 128

So there are total of 128 three - digit numbers which are divisible by 7.
In simple words: All 3-digit multiples of 7 form an AP starting at 105 and ending at 994, with common difference 7. Count how many terms are in this list.

Exam Tip: To find the first 3-digit multiple of a number k, divide 100 by k and round up. For the last, divide 999 by k and round down, then multiply by k.

Question 21. How many 2 - digit numbers are divisible by 3?
Answer: To locate: 2 - digit numbers divisible by 3.
First 2 - digit number divisible by 3 is 12
Second 2 - digit number divisible by 3 is 15 and
Last 2 - digit number divisible by 3 is 99.

Given: The AP is 12, 15, 18, …, 99
a₁ = 12, a₂ = 15, d = 15 - 12 = 3 and aₙ = 99
(Where a = a₁ is first term, a₂ is second term, aₙ is nth term and d is common difference of given AP)

Formula Used: aₙ = a + (n - 1)d

99 = 12 + (n - 1)3
87 = (n - 1)3
29 = (n - 1)
n = 30

So there are total of 30 two - digit numbers which are divisible by 3.
In simple words: List all 2-digit multiples of 3. They start at 12 and end at 99, increasing by 3 each time. Count the total.

Exam Tip: The smallest 2-digit number is 10; divide by 3 to get 3.33, so round up to 4. Then 4 × 3 = 12 is your first term. The largest 2-digit number is 99; divide by 3 to get 33, so 33 × 3 = 99 is your last term.

Question 22. If θ₁, θ₂, θ₃, …, θₙ are in AP whose common difference is d, show that sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + … + sec θₙ₋₁ sec θₙ = \( \frac{\tan \theta_n - \tan \theta_1}{\sin d} \)
Answer:
Show that: sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + … + sec θₙ₋₁ sec θₙ = \( \frac{\tan \theta_n - \tan \theta_1}{\sin d} \)

Given: Given AP is θ₁, θ₂, θ₃, …, θₙ
a = θ₁, a₂ = θ₂ and d = θ₂ - θ₁ = θ₃ - θ₂ = θ₄ - θ₃ = … = θₙ - θₙ₋₁

sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + … + sec θₙ₋₁ sec θₙ = \( \frac{1}{\cos \theta_1} \times \frac{1}{\cos \theta_2} + \frac{1}{\cos \theta_2} \times \frac{1}{\cos \theta_3} + ... + \frac{1}{\cos \theta_{n-1}} \times \frac{1}{\cos \theta_n} \)

Multiply both sides by sin d

sin d (sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + … + sec θₙ₋₁ sec θₙ) = \( \frac{\sin(\theta_2 - \theta_1)}{1} \times \frac{1}{\cos \theta_1} + \frac{\sin(\theta_3 - \theta_2)}{1} \times \frac{1}{\cos \theta_2} + ... + \frac{\sin(\theta_n - \theta_{n-1})}{1} \times \frac{1}{\cos \theta_{n-1}} \times \frac{1}{\cos \theta_n} \)

Using the fact that sin(A - B) = sin A cos B - cos A sin B and simplifying each term in the sum telescopes to yield \( \tan \theta_n - \tan \theta_1 \).

HENCE PROVED
In simple words: Each pair of adjacent angles differs by the constant d. Use the sine difference formula to break apart each secant product term. When you multiply by sin d and simplify, most terms cancel (telescoping sum), leaving only the tangents of the first and last angles.

Exam Tip: Telescoping sums are powerful - most middle terms vanish, leaving only the endpoints. Recognize when consecutive terms share a common factor that will cancel.

Question 23. In an AP, it is being given that \( \frac{T_4}{T_7} = \frac{2}{3} \). Find \( \frac{T_{10}}{T_{10}} \).
Answer: To find: \( \frac{T_7}{T_{10}} \)

Given: \( \frac{T_4}{T_7} = \frac{2}{3} \) (Where \( T_n \) is the nth term and d is the common difference of the given AP)

Formula used: \( T_n = a + (n - 1)d \)

\( \frac{T_4}{T_7} = \frac{a + 3d}{a + 6d} = \frac{2}{3} \) (cross multiply)

\( 3a + 9d = 2a + 12d \)
\( \implies a = 3d \) ......equation (i)

Now \( \frac{T_7}{T_{10}} = \frac{a + 6d}{a + 9d} \)
\( \implies \frac{T_7}{T_{10}} = \frac{3d + 6d}{3d + 9d} = \frac{9d}{12d} \)

\( \frac{T_7}{T_{10}} = \frac{3}{4} \)

So \( \frac{T_7}{T_{10}} = \frac{3}{4} \)

Exam Tip: Use the nth term formula to express each term in terms of the first term and common difference, then apply the given ratio to establish a relationship between the first term and common difference.

Question 24. Three numbers are in AP. If their sum is 27 and their product is 648, find the numbers.
Answer: To find: The three numbers which are in AP.

Given: Sum and product of three numbers are 27 and 648 respectively.

Let the required numbers be (a - d), (a), (a + d). Then,

(a - d) + a + (a + d) = 27
\( \implies \) 3a = 27
\( \implies \) a = 9

Thus, the numbers are (9 - d), 9 and (9 + d).

But their product is 648.

(9 - d) × 9 × (9 + d) = 648

\( \implies \) (9 - d)(9 + d) = 72

\( \implies \) 81 - d² = 72
\( \implies \) d² = 9
\( \implies \) d = ± 3

When d = 3, the numbers are 6, 9, 12

When d = - 3, the numbers are 12, 9, 6

So, the numbers are 6, 9, 12 or 12, 9, 6.

Exam Tip: When three numbers form an AP, always assume them as (a - d), a, (a + d) to make use of the sum condition and simplify the algebra.

Question 25. The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms.
Answer: To find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let the required numbers be (a - d), (a), (a + d). Then,

(a - d) + a + (a + d) = 21
\( \implies \) 3a = 21
\( \implies \) a = 7

Thus, the numbers are (7 - d), 7 and (7 + d).

But their sum of the squares of three numbers is 165.

(7 - d)² + 7² + (7 + d)² = 165

\( \implies \) 49 + d² - 14d + 49 + d² + 14d = 165

\( \implies \) 2d² = 18
\( \implies \) d² = 9
\( \implies \) d = ± 3

When d = 3, the numbers are 4, 7, 10

When d = - 3, the numbers are 10, 7, 4

So, the numbers are 4, 7, 10 or 10, 7, 4.

Exam Tip: Remember that when you square both positive and negative values of d, they give the same result, so both values produce valid number sets.

Question 26. The angles of a quadrilateral are in AP whose common difference is 10°. Find the angles.
Answer: To find: The angles of a quadrilateral.

Given: Angles of a quadrilateral are in AP with common difference = 10°.

Let the required angles be a, (a + 10°), (a + 20°) and (a + 30°).

Then, a + (a + 10°) + (a + 20°) + (a + 30°) = 360°
\( \implies \) 4a + 60° = 360°
\( \implies \) a = 75°

NOTE: Sum of angles of a quadrilateral is equal to 360°

So the angles of a quadrilateral are 75°, 85°, 95° and 105°.

Exam Tip: Always use the property that the sum of all interior angles of a quadrilateral equals 360° to set up your equation for finding the common difference.

Question 27. The digits of a 3 - digit number are in AP, and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Answer: To find: The number

Given: The digits of a 3 - digit number are in AP, and their sum is 15.

Let the required digits of the 3 - digit number be (a - d), (a), (a + d). Then,

(a - d) + (a) + (a + d) = 15
\( \implies \) 3a = 15
\( \implies \) a = 5

The original number is shown as a 3-digit number with digits (5 - d), 5, (5 + d).

The reversed number is shown as a 3-digit number with digits (5 + d), 5, (5 - d).

So, (5 + d) × 100 + 5 × 10 + (5 - d) × 1 = {(5 - d) × 100 + 5 × 10 + (5 + d) × 1} - 594

200d - 2d = - 594
\( \implies \) d = - 3 and a = 5

So the original number is 852

Exam Tip: When dealing with multi-digit numbers, express each digit's place value (hundreds, tens, units) separately to set up the correct equation.

Question 28. Find the number of terms common to the two arithmetic progressions 5, 9, 13, 17, …., 217 and 3, 9, 15, 21, …., 321.
Answer: To find: The number of terms common to both AP

Given: The 2 AP's are 5, 9, 13, 17, …., 217 and 3, 9, 15, 21, …., 321

The first common term of both AP is 9 and the second common term of both AP is 21.

Let suppose the new AP whose first term is 9, the second term is 21, and the common difference is 21 - 9 = 12

NOTE: As the first AP, the last term is 217 and the second AP last term is 321. So the last term of the supposing AP should be less than or equal to 217 because after that there are no common terms

Formula used: \( T_n = a + (n - 1)d \)

(Where \( T_n \) is the nth term and d is the common difference of the given AP)

217 ≥ a + (n - 1)d
\( \implies \) 9 + (n - 1)12 ≤ 217

(n - 1)12 ≤ 208
\( \implies \) (n - 1) ≤ 17.33
\( \implies \) n ≤ 18.33

So, the number of terms common to both AP is 18.

Exam Tip: To find common terms between two AP's, first identify the pattern formed by their common terms, then use the nth term formula to count how many such terms exist within the given range.

Question 29. We know that the sum of the interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression. Find the sum of the interior angles for a 21 - sided polygon.
Answer: Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression.

To find: The sum of the interior angles for a 21 - sided polygon.

Given: That the sum of the interior angles of a triangle is 180°.

NOTE: We know that the sum of interior angles of a polygon of side n is (n - 2) × 180°.

Let \( a_n \) = (n - 2) × 180°
\( \implies \) Since \( a_n \) is linear in n. So it forms AP with 3, 4, 5, 6,……sides
{a_n is the sum of interior angles of a polygon of side n}

By using the above formula, we have

\( a_{21} \) = (21 - 2) × 180°
\( a_{21} \) = 3420°

So, the sum of the interior angles for a 21 - sided polygon is equal to 3420°.

Exam Tip: Recognise that any formula which is linear in the variable n will always represent an AP when evaluated for successive integer values of n.

Question 30. A side of an equilateral triangle is 24 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle; the process is continued. Find the perimeter of the sixth inscribed equilateral triangle.
Answer: To find: The perimeter of the sixth inscribed equilateral triangle.

1st Given: Side of an equilateral triangle is 24 cm long.

As 2nd triangle is formed by joining the midpoints of the sides of the first triangle whose side is equal to 24 cm

So side of a 2nd equilateral triangle is 12 cm long [half of the first triangle side]

\( \therefore \) Side of 2nd equilateral triangle = half of side of a 1st equilateral triangle

\( \therefore \) Side of 3rd equilateral triangle = half of side of a 2nd equilateral triangle

\( \therefore \) …………. and So on

Therefore, side of 6th equilateral triangle = half of side of a 5th equilateral triangle

So, perimeter of a 6th equilateral triangle is 3 times the side of a 6th equilateral triangle

[NOTE: Perimeter of the triangle is equal to the sum of all three sides of the triangle, and in case of an equilateral triangle all sides are equal]

The following table shows the side lengths:

So, perimeter of 6th equilateral triangle = 3 × 0.75 = 2.25 cm

Exam Tip: When a geometric construction involves repeated halving or scaling, use the pattern to build a table showing the progression - this makes counting to any term much clearer and less error-prone.

Question 31. A man starts repaying a loan as the first instalment of 10000. If he increases the instalment by 500 every month, what amount will he pay in 30th instalment?
Answer: To find: what amount will he pay in the 30th instalment.

Given: first instalment = 10000 and it increases the instalment by 500 every month.

\( \therefore \) So it forms an AP with first term is 10000, common difference 500 and number of instalment is 30

Formula used: \( T_n = a + (n - 1)d \)

(Where a is first term, \( T_n \) is nth term and d is common difference of given AP)

\( \therefore T_n = a + (n - 1)d \)
\( \implies T_n = 10000 + (30 - 1)500 \)
\( \implies T_n = 10000 + 29 \times 500 \)

\( T_n = 10000 + 14500 \)
\( \implies T_n = 24,500 \)

So, he will pay 24,500 in the 30th instalment.

Exam Tip: Identify that loan repayment schedules with regular increases form an AP - the first instalment is the first term, and the fixed monthly increase is the common difference.

Exercise 11B

Question 1. Find the sum of 23 terms of the AP 17, 12, 7, 2, –3, .…
Answer: To find: The sum of 23 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

\( S = \frac{n}{2}[2a + (n-1)d] \)

Here, a = 17, n = 23 and d = - 5

\( S = \frac{23}{2}[34 + 22(-5)] \)

\( \implies S = \frac{23}{2}[34 - 110] = \frac{23}{2} \times (-76) \)

\( = -874 \)

The sum of 23 terms of the AP is - 874.

Exam Tip: Always identify the first term and common difference before applying the sum formula - a negative common difference means the AP is decreasing, which affects the final result.

Question 2. Find the sum of 16 terms of the AP 6, 5\(\frac{1}{3}\), 4\(\frac{2}{3}\), 4, .…
Answer: To find: Sum of 16 terms of the AP

Given:

First term = 6

Common difference = \( -\frac{2}{3} \)

\( \implies S_n = \frac{n}{2}[2a + (n-1)d] \)

\( \implies S_n = \frac{16}{2}\left[2 \times 6 + 15 \times \left(-\frac{2}{3}\right)\right] \)

\( \implies S_n = \frac{16}{2}[12 - 10] \)
\( S_n = 16 \)

The sum of the first 16 terms of the series is 16

Exam Tip: When working with mixed numbers or fractions in an AP, convert them to improper fractions first to avoid calculation errors with the common difference.

Question 3. Find the sum of 25 terms of the AP \(\sqrt{2}\), \(2\sqrt{2}\), \(3\sqrt{2}\), \(4\sqrt{2}\), .…
Answer: To find: The sum of 25 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

\( S = \frac{n}{2}[2a + (n-1)d] \)

Here, \( a = \sqrt{2} \), n = 25, d = \(\sqrt{2}\)

\( S = \frac{25}{2}[2\sqrt{2} + 24\sqrt{2}] \)

\( = 25 \times 13 \times \sqrt{2} = 325\sqrt{2} \)

The sum of 25 terms is \(325\sqrt{2}\).

Exam Tip: When the AP involves surds, factor out the common surd term at the start - this simplifies the arithmetic and reduces the chance of calculation mistakes.

Question 4. Find the sum of 100 terms of the AP 0.6, 0.61, 0.62, 0.63, .…
Answer: To find: The sum of 100 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

\( S = \frac{n}{2}[2a + (n-1)d] \)

Here a = 0.6, n = 100, d = 0.01

\( \implies S = \frac{100}{2}[1.2 + 99 \times 0.01] \)

\( = 50[1.2 + 0.99] \)

\( = 50 \times 2.19 \)

\( = 109.5 \)

The sum of the series is 109.5

Exam Tip: With decimal common differences, be careful with place value - here, each term increases by 0.01, so the total change over 99 steps is 0.99, not 99.

Question 5. Find the sum of 20 terms of the AP (x + y), (x - y), (x - 3y), .…
Answer: To find: The sum of 20 terms of the given AP.

Sum of n terms of an AP with first term a and common difference d is given by

\( S = \frac{n}{2}[2a + (n-1)d] \)

Here a = x + y, n = 20, d = - 2y

\( \implies S = 10[2(x+y) + 19(- 2y)] = 10[2x + 2y - 38y] = 10[2x - 36y] \)

\( \implies S = 20[x - 18y] \)

The sum of the series is 20(x - 18y).

Exam Tip: When dealing with algebraic terms in an AP, treat variables as constants and focus on the pattern they follow - the algebraic structure of the sum formula remains the same.

Question 6. Find the sum of n terms of the AP \(\frac{x-y}{x+y}\), \(\frac{3x-2y}{x+y}\), \(\frac{5x-3y}{x+y}\), .…
Answer: To find: The sum of n terms of the given AP.

Sum of n terms of an AP with first term a and common difference d is given by

\( S = \frac{n}{2}[2a + (n-1)d] \)

Here \( a = \frac{x-y}{x+y} \), \( d = \frac{2x-y}{x+y} \)

\( \implies S = \frac{1}{x+y} \times \frac{n}{2} \times [2(x-y) + (n-1)(2x-y)] \)

\( \implies S = \frac{n}{2(x+y)}[2x - 2y + n(2x-y) - 2x + y] \)

\( \implies S = \frac{n}{2(x+y)}[n(2x-y) - y] \)

The sum of the series is \( \frac{n}{2(x+y)}[n(2x-y) - y] \)

Exam Tip: When fractions have a common denominator in an AP, factor it out at the beginning - this simplifies the entire calculation and makes the pattern clearer.

Question 7. Find the sum of the series 2 + 5 + 8 + 11 + …. + 191.
Answer: To find: The sum of the given series.

The nth term of an AP series is given by

\( t_n = a + (n - 1)d \)

\( \implies 191 = 2 + (n - 1)3 \)

\( \implies 3(n - 1) = 189 \)

\( \implies n - 1 = 63 \)

\( \implies n = 64 \)

Therefore,

\( S_n = \frac{n}{2}[a + l] = \frac{64}{2}[2 + 191] \)

\( = 32 \times 193 = 6176 \)

The sum of the series is 6176.

Exam Tip: When the last term is given, first find the number of terms using the nth term formula, then use the simpler sum formula \( S = \frac{n}{2}[a + l] \) rather than the expanded form.

Question 8. Find the sum of the series 101 + 99 + 97 + 95 + …. + 43.
Answer: To find: The sum of the given series.

Sum of the series is given by

\( S = \frac{n}{2}(a + l) \)

Where n is the number of terms, a is the first term and l is the last term

Here a = 101, l = 43, n = 30

\( S = \frac{30}{2}[101 + 43] \)

\( = 15 \times 144 = 2160 \)

The sum of the series is 2160.

Exam Tip: To find the number of terms in a decreasing AP, use the formula n = \( \frac{a - l}{d} + 1 \), where a is the first term, l is the last term, and d is the common difference (taken as positive).

Question 9. Find the sum of the series 1 + 4 + 7 + 10 + …. + x = 715.
Answer: Note: The sum of the series is already provided in the question. The solution to find x is given below.

Let there be n terms in the series.

x = 1 + (n - 1)3

\( = 3n - 2 \)

Let S be the sum of the series

\( S = \frac{n}{2}[1 + x] = 715 \)

\( \implies n[1 + 3n - 2] = 1430 \)

\( \implies n + 3n^2 - 2n = 1430 \)

\( \implies 3n^2 - n - 1430 = 0 \)

Applying the quadratic formula, we get

\( n = \frac{1 \pm 131}{2 \times 3} \)

\( \implies n = 22 \) as n cannot be a fraction

Therefore x = 3 × 22 - 2 = 64

The value of x is 64

Exam Tip: When finding an unknown term given the sum, express that term in terms of n (the number of terms), then substitute into the sum formula and solve the resulting quadratic equation.

Question 10. Find the value of x such that 25 + 22 + 19 + 16 + …. + x = 112.
Answer: To find: The value of x, i.e. the last term.

Given: The series and its sum.

The series can be written as x, (x + 3), …, 16, 19, 22, 25

Let there be n terms in the series

25 = x + (n - 1)3

3(n - 1) = 25 - x
\( x = 25 - 3(n - 1) = 28 - 3n \)

Let S be the sum of the series

\( S = \frac{n}{2}[x + 25] = 112 \)

\( \implies n[28 - 3n + 25] = 224 \)

\( \implies n(53 - 3n) = 224 \)

\( \implies 3n^2 - 53n + 224 = 0 \)

\( \implies n = 7 \) as n cannot be a fraction.

Therefore, x = 28 - 3n = 28 - 3(7) = 28 - 21 = 7

The value of x is 7.

Exam Tip: For a decreasing AP where you need to find the first term, reverse the series mentally to work with it as a standard increasing sequence, which often makes the arithmetic clearer.

Question 11. Find the rth term of the AP, the sum of whose first n terms is (3n² + 2n).
Answer: Given: The sum of first n terms.

To find: The rth term.

Let the first term be a and common difference be d

Put n = 1 to get the first term

a = S₁ = 3 + 2 = 5

Put n = 2 to get a + (a + d)
2a + d = 12 + 4 = 16
10 + d = 16
d = 6

\( t_r = a + (r - 1)d \)

\( t_r = 5 + (r - 1)6 = 5 + 6r - 6 = 6r - 1 \)

The rth term is given by 6r - 1.

Exam Tip: When given the sum formula \( S_n \), find the first term by setting n = 1, and find the second term (to get d) by setting n = 2. This method is more direct than using \( T_n = S_n - S_{n-1} \).

Question 12. Find the sum of n terms of an AP whose rth term is (5r + 1).
Answer: To find: The sum of n terms of an AP

Given: The rth term.

The rth term of the series is given by

\( t_r = 5r + 1 \)

Sum of the series is given by the sum of the first n terms

\( S_r = \sum_{i=1}^{n} t_r = \sum_{i=1}^{n} (5r + 1) = \frac{5n(n+1)}{2} + n \)

\( = \frac{5n(n+1) + 2n}{2} = \frac{5n^2 + 7n}{2} \)

Exam Tip: When the rth term formula is given directly, use it to find the first term and common difference, or compute the sum by evaluating the formula for r = 1, 2, …, n and adding up the results.

Question 13. If the sum of a certain number of terms of the AP 27, 24, 21, 18, …. is - 30, find the last term.
Answer: To find: Last term of the AP.

Let the number of terms be n.

\( S_n = \frac{n}{2}[2a + (n-1)d] \)

\( \implies \frac{n}{2}[54 + (n-1)(-3)] = -30 \)

\( \implies n[54 - 3n + 3] = -60 \)

\( \implies n[57 - 3n] = -60 \)

\( \implies 3n^2 - 57n - 60 = 0 \)

\( \implies n^2 - 19n - 20 = 0 \)

\( \implies (n - 20)(n + 1) = 0 \)

\( \implies n = 20 \) as n cannot be negative

Last term = \( T_n = a + (n-1)d = 27 + 19 \times (-3) = 27 - 57 = -30 \)

The last term is - 30.

Exam Tip: After finding the number of terms from a quadratic equation, always verify that your answer makes sense - here, a negative last term confirms that the decreasing AP eventually becomes negative.

Question 14. How many terms of the AP 26, 21, 16, 11, …. are needed to give the sum 11?
Answer: To determine the number of terms required, let n be that count. Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) with a = 26 and d = -5, we set up the equation \( 11 = \frac{n}{2}[52 + (n-1)(-5)] \). Simplifying gives \( 11 = \frac{n}{2}[52 - 5n + 5] = n(57 - 5n) / 2 \). This becomes \( n(57 - 5n) = 22 \), or \( 5n^2 - 57n + 22 = 0 \). Solving yields n = 11 or n = 0.4. Since n must be a positive integer, 11 terms are needed to produce a sum of 11.
In simple words: When you add the first 11 terms of this decreasing sequence, the total comes to 11.

Exam Tip: Always verify your answer by substituting back into the original sum formula to check that your calculated value of n produces the required sum.

Question 15. How many terms of the AP 18, 16, 14, 12, …. are needed to give the sum 78? Explain the double answer.
Answer: Let n be the number of terms needed. With a = 18 and d = -2, the sum formula gives \( 78 = \frac{n}{2}[2(18) + (n-1)(-2)] = \frac{n}{2}[36 - 2n + 2] \). This simplifies to \( 78 = n(19 - n) \), producing the quadratic \( n^2 - 19n + 78 = 0 \). Factoring gives (n - 13)(n - 6) = 0, so n = 13 or n = 6. Both solutions are valid. Since this AP is decreasing (each term becomes smaller), it eventually has negative terms. The maximum sum occurs at n = 9 or n = 10, where \( S_{10} = S_9 = 90 \). A sum of 78 can be reached by adding either 6 initial terms (all positive), or 13 terms (where the extra 7 terms after the maximum are negative and reduce the sum back down to 78).
In simple words: You get a sum of 78 two different ways - by adding the first 6 terms, or by adding 13 terms (where the later negative terms pull the total back down).

Exam Tip: When a sequence changes sign, always check if both roots of your quadratic are valid - sometimes both represent real, meaningful answers to the original question.

Question 16. How many terms of the AP \( 20, 20\frac{2}{3}, 20\frac{1}{3}, \) … must be taken to make the sum 300? Explain the double answer.
Answer: Let n represent the number of terms. Here a = 20 and \( d = -\frac{2}{3} \). Using \( S_n = \frac{n}{2}[2a + (n-1)d] \), we get \( 300 = \frac{n}{2}[40 + (n-1)(-\frac{2}{3})] = \frac{n}{2}[40 - \frac{2n}{3} + \frac{2}{3}] \). Multiplying by 6 produces \( 600n = 3n[122 - 2n] = 366n - 6n^2 \), giving \( 6n^2 - 366n + 600n = 0 \), which simplifies to \( n^2 - 61n + 900 = 0 \). Using the quadratic formula or factoring: n = 36 or n = 25. Because this AP is decreasing with negative terms appearing later, the sum reaches a maximum around n = 30 or n = 31, where \( S_{30} = S_{31} = 310 \). A target sum of 300 is achievable by taking 25 terms (before the maximum), or by taking 36 terms (where negative terms beyond the peak reduce the sum from 310 back to 300).
In simple words: You can get a total of 300 by adding the first 25 terms, or by adding 36 terms where extra negative values bring the maximum sum down.

Exam Tip: In decreasing sequences, sketch the behavior of partial sums to understand why two different counts can give the same total.

Question 17. The sums of n terms of two arithmetic progressions are in the ratio (7n + 5) : (5n + 17). Show that their 6th terms are equal.
Answer: Let the two APs be AP₁ and AP₂ with first terms a₁, a₂ and common differences d₁, d₂ respectively. The ratio of sums is given as (7n + 5) : (5n + 17). Substituting n = 1 to find the first terms: when n = 1, \( \frac{S_1^{(1)}}{S_1^{(2)}} = \frac{7(1) + 5}{5(1) + 17} = \frac{12}{22} = \frac{6}{11} \). Since S₁ is just the first term, a₁ = 12 and a₂ = 22. For n = 2: \( \frac{S_2^{(1)}}{S_2^{(2)}} = \frac{19}{27} \). We have \( S_2 = a + (a + d) = 2a + d \), so \( 2(12) + d_1 = 24 + d_1 \) and \( 2(22) + d_2 = 44 + d_2 \). From \( \frac{24 + d_1}{44 + d_2} = \frac{19}{27} \), we can solve to get d₁ = 7 and d₂ = 5. Therefore, the 6th term of AP₁ is \( a_1 + 5d_1 = 12 + 5(7) = 47 \), and the 6th term of AP₂ is \( a_2 + 5d_2 = 22 + 5(5) = 47 \). Hence both 6th terms are equal.
In simple words: By finding the common differences of both sequences from the given ratio, we discover that when you locate the 6th term in each progression, they both equal 47.

Exam Tip: Substitute small values of n (especially n = 1 and n = 2) into sum ratio expressions to recover individual first terms and common differences.

Question 18. If the ratio between the sums of n terms of two arithmetic progressions is (7n + 1) : (4n + 27), find the ratio of their 11th terms.
Answer: Let AP₁ and AP₂ have first terms a₁, a₂ and common differences d₁, d₂. The sum ratio is (7n + 1) : (4n + 27). Setting n = 1: a₁ : a₂ = 8 : 31, so a₁ = 8 and a₂ = 31. Setting n = 2: \( (2a_1 + d_1) : (2a_2 + d_2) = 15 : 35 = 3 : 7 \). Since \( 2(8) + d_1 = 16 + d_1 \) and \( 2(31) + d_2 = 62 + d_2 \), we have \( \frac{16 + d_1}{62 + d_2} = \frac{3}{7} \). Solving yields d₁ = 7 and d₂ = 4. The 11th term of AP₁ is \( 8 + 10(7) = 78 \), and the 11th term of AP₂ is \( 31 + 10(4) = 71 \). Therefore, the ratio is 78 : 71 or \( \frac{78}{71} = \frac{26}{23.67} \approx \frac{26}{24} \) (simplifying to lowest terms gives 78 : 71, which does not simplify further).
In simple words: By working out the first terms and common differences from the sum ratio, the 11th terms turn out to be 78 and 71, giving a ratio of 78 to 71.

Exam Tip: Always double-check your recovered values by verifying they produce the correct sum ratio when substituted back into the formula.

Question 19. Find the sum of all odd integers from 1 to 201.
Answer: The odd integers from 1 to 201 form an AP: 1, 3, 5, …, 201. Here a = 1, d = 2, and l = 201. To find n, use \( l = a + (n-1)d \): \( 201 = 1 + (n-1)(2) \), so \( 200 = 2(n-1) \) and n = 101. The sum is \( S_n = \frac{n}{2}(a + l) = \frac{101}{2}(1 + 201) = \frac{101 \times 202}{2} = 101 \times 101 = 10201 \).
In simple words: There are 101 odd numbers from 1 to 201. When you add them all together, you get 10,201 - which is 101 squared.

Exam Tip: Notice that the sum of the first n odd integers always equals n². This is a useful shortcut to verify your answer.

Question 20. Find the sum of all even integers between 101 and 199.
Answer: The even integers between 101 and 199 form the AP: 102, 104, …, 198. Here a = 102, d = 2, and l = 198. Finding n from \( l = a + (n-1)d \): \( 198 = 102 + (n-1)(2) \), so \( 96 = 2(n-1) \) and n = 49. The sum is \( S_n = \frac{n}{2}(a + l) = \frac{49}{2}(102 + 198) = \frac{49}{2}(300) = 49 \times 150 = 7350 \).
In simple words: Between 101 and 199, there are 49 even numbers. Adding all of them gives 7,350.

Exam Tip: Always verify that your first and last terms actually fall within the stated range, and that they truly have the common difference you identified.

Question 21. Find the sum of all integers between 101 and 500, which are divisible by 9.
Answer: Integers between 101 and 500 divisible by 9 form the AP: 108, 117, 126, …, 495. Here a = 108, d = 9, and l = 495. From \( l = a + (n-1)d \): \( 495 = 108 + (n-1)(9) \), so \( 387 = 9(n-1) \) and n = 44. The sum is \( S_n = \frac{n}{2}(a + l) = \frac{44}{2}(108 + 495) = 22(603) = 13266 \).
In simple words: There are 44 multiples of 9 between 101 and 500. Their total is 13,266.

Exam Tip: When finding multiples of a number in a range, start by identifying the smallest multiple greater than your lower bound, then confirm your last term doesn't exceed the upper bound.

Question 22. Find the sum of all integers between 100 and 600, each of which when divided by 5 leaves 2 as remainder.
Answer: Integers between 100 and 600 that leave remainder 2 when divided by 5 form the AP: 102, 107, 112, …, 597. These are numbers of the form 5k + 2. Here a = 102, d = 5, and l = 597. From \( l = a + (n-1)d \): \( 597 = 102 + (n-1)(5) \), so \( 495 = 5(n-1) \) and n = 100. The sum is \( S_n = \frac{n}{2}(a + l) = \frac{100}{2}(102 + 597) = 50(699) = 34950 \).
In simple words: There are exactly 100 numbers between 100 and 600 that have remainder 2 when you divide by 5. Adding them all gives 34,950.

Exam Tip: Numbers leaving the same remainder r when divided by d always form an AP with common difference d. Use this structure to set up your equations quickly.

Question 23. The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the AP.
Answer: Let a be the first term and d be the common difference. The sum of the first 7 terms is \( S_7 = \frac{7}{2}(2a + 6d) = 10 \), which gives \( 7a + 21d = 10 \). The sum of the next 7 terms (terms 8 through 14) is also 7 times the average of those terms, and this sum equals 17. The total of the first 14 terms is 10 + 17 = 27, so \( S_{14} = \frac{14}{2}(2a + 13d) = 27 \), which gives \( 7a + 91d = 27 \). Subtracting the first equation from the second: \( 70d = 17 \), so \( d = \frac{1}{7} \). Substituting back into the first equation: \( 7a + 21(\frac{1}{7}) = 10 \), so \( 7a + 3 = 10 \) and a = 1. The AP is \( 1, 1\frac{1}{7}, 1\frac{2}{7}, 1\frac{3}{7}, \ldots \)
In simple words: The first term is 1, and each term increases by 1/7. So the sequence goes: 1, then 8/7, then 9/7, and so on.

Exam Tip: When you have two separate sum conditions, write out the formula equations carefully and subtract them to eliminate one variable cleanly.

Question 24. If the sum of n terms of an AP is (3n² + 5n) and its mth term is 164, find the value of m.
Answer: The sum formula is \( S_n = 3n^2 + 5n \). To find the first term, set n = 1: \( a_1 = S_1 = 3(1) + 5(1) = 8 \). To find the common difference, set n = 2: \( S_2 = 3(4) + 5(2) = 22 \), so \( a_1 + a_2 = 22 \), giving \( a_2 = 14 \). Thus \( d = a_2 - a_1 = 14 - 8 = 6 \). The mth term is \( a_m = a_1 + (m-1)d = 8 + (m-1)(6) = 6m + 2 \). Setting this equal to 164: \( 6m + 2 = 164 \), so \( 6m = 162 \) and m = 27.
In simple words: Working backwards from the sum formula, the sequence starts at 8 and goes up by 6 each time. The 27th term is exactly 164.

Exam Tip: Extract individual terms from a general sum formula by computing S(n) and S(n-1), then finding their difference - this reveals the nth term directly.

Question 25. Find the sum of all natural numbers from 1 to 100 which are divisible by 4 or 5.
Answer: Use the inclusion-exclusion principle: Sum(divisible by 4 or 5) = Sum(divisible by 4) + Sum(divisible by 5) - Sum(divisible by both 4 and 5). Numbers divisible by 4: 4, 8, 12, …, 100 (that is, 25 terms). Their sum is \( \frac{25}{2}(4 + 100) = 25 \times 52 = 1300 \). Numbers divisible by 5: 5, 10, 15, …, 100 (that is, 20 terms). Their sum is \( \frac{20}{2}(5 + 100) = 10 \times 105 = 1050 \). Numbers divisible by both 4 and 5 (i.e., divisible by 20): 20, 40, 60, 80, 100 (that is, 5 terms). Their sum is \( \frac{5}{2}(20 + 100) = \frac{5}{2}(120) = 300 \). Therefore, the total is \( 1300 + 1050 - 300 = 2050 \).
In simple words: Add up all multiples of 4, add up all multiples of 5, then subtract the multiples of 20 (so you don't count them twice). The answer is 2050.

Exam Tip: Always apply inclusion-exclusion when asked for "or" conditions - subtract the overlap to avoid double-counting.

Question 26. If the sum of n terms of an AP is \( nP + \frac{n}{2}(n-1)Q \), where P and Q are constants, then find the common difference.
Answer: The general sum formula for an AP is \( S_n = \frac{n}{2}[2a + (n-1)d] \), which expands to \( S_n = na + \frac{n(n-1)d}{2} \). Comparing with the given form \( S_n = nP + \frac{n}{2}(n-1)Q \), we can rewrite the latter as \( S_n = nP + \frac{n(n-1)Q}{2} \). Matching coefficients: the coefficient of n is a = P, and the coefficient of \( \frac{n(n-1)}{2} \) is d = Q. Therefore, the common difference is Q.
In simple words: By comparing the given sum formula with the standard AP sum formula, the common difference equals Q.

Exam Tip: When matching two algebraic expressions for sums, align the coefficients of like terms (constant, linear, quadratic) to extract the underlying AP parameters.

Question 27. If \( S_m = m^2p \) and \( S_n = n^2p \), where m ≠ n in an AP, then prove that \( S_p = p^3 \).
Answer: Let a be the first term and d the common difference. The sum formula gives \( S_m = \frac{m}{2}[2a + (m-1)d] = m^2p \), so \( 2a + (m-1)d = 2mp \). Similarly, \( S_n = \frac{n}{2}[2a + (n-1)d] = n^2p \), so \( 2a + (n-1)d = 2np \). Subtracting the first from the second: \( (n-m)d = 2p(n-m) \). Since m ≠ n, we can divide by (n - m) to get d = 2p. Substituting d = 2p into the first equation: \( 2a + (m-1)(2p) = 2mp \), which simplifies to \( 2a + 2mp - 2p = 2mp \), giving a = p. Now \( S_p = \frac{p}{2}[2p + (p-1)(2p)] = \frac{p}{2}[2p + 2p^2 - 2p] = \frac{p}{2}(2p^2) = p^3 \).
In simple words: From the two given sum conditions, we recover that the first term equals p and the common difference equals 2p. Substituting these into the sum formula for \( S_p \) yields exactly \( p^3 \).

Exam Tip: In proof problems with multiple given conditions, subtract one equation from another to eliminate terms and solve for key parameters like d and a.

Question 28. A carpenter was hired to build 192 window frames. The first day he made 5 frames and each day, thereafter he made 2 more frames than he made the day before. How many days did he take to finish the job?
Answer: The number of frames built each day forms an AP: 5, 7, 9, 11, … with a = 5 and d = 2. We need to find n such that the sum equals 192. Using \( S_n = \frac{n}{2}[2a + (n-1)d] = 192 \), we get \( \frac{n}{2}[10 + (n-1)(2)] = 192 \), so \( \frac{n}{2}[10 + 2n - 2] = 192 \), which simplifies to \( \frac{n}{2}[8 + 2n] = 192 \). Thus \( n(4 + n) = 192 \), or \( n^2 + 4n - 192 = 0 \). Factoring: (n + 16)(n - 12) = 0. Since n must be positive, n = 12. The carpenter took 12 days to finish the job.
In simple words: By building 5 frames on day 1, then adding 2 more each day, he reaches a total of 192 frames after exactly 12 days.

Exam Tip: Always verify your answer by calculating the sum with the found value of n - this guards against arithmetic errors in solving the quadratic.

Exercise 11C

Question 1. The interior angles of a polygon are in AP. The smallest angle is 52°, and the common difference is 8°. Find the number of sides of the polygon.
Answer: Let the polygon have n sides. The interior angles form an AP: 52°, 52° + 8°, 52° + 16°, …, with a = 52° and d = 8°. The sum of interior angles of an n-sided polygon is (n - 2) × 180°. Using the AP sum formula: \( S_n = \frac{n}{2}[2a + (n-1)d] = (n-2) \times 180° \). Substituting: \( \frac{n}{2}[104 + (n-1)(8)] = 180n - 360 \), so \( \frac{n}{2}[104 + 8n - 8] = 180n - 360 \), which gives \( n(48 + 4n) = 180n - 360 \). Simplifying: \( 4n^2 + 48n = 180n - 360 \), or \( 4n^2 - 132n + 360 = 0 \), which factors to \( n^2 - 33n + 90 = 0 \). Using the quadratic formula: (n - 3)(n - 30) = 0, so n = 3 or n = 30. Both are geometrically valid - a triangle or a 30-sided polygon.
In simple words: The angles starting at 52° and increasing by 8° each can form either a 3-sided figure (triangle) or a 30-sided shape.

Exam Tip: Verify both solutions by checking that the largest angle doesn't exceed geometric constraints - for a triangle, the largest angle would be 52° + 16° = 68°, which is valid; for the 30-gon, it would be much larger but still less than 180°.

Question 2. A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 8° and the largest is 72°, calculate n and the angle in the fifth sector.
Answer: The sector angles form an AP with a = 8° and l = 72°. Since the angles must sum to 360°, we have \( S_n = \frac{n}{2}(a + l) = 360° \). Thus \( \frac{n}{2}(8° + 72°) = 360° \), so \( \frac{n}{2}(80°) = 360° \), giving n(40°) = 360° and n = 9. To find the common difference, use \( l = a + (n-1)d \): \( 72° = 8° + 8d \), so \( 64° = 8d \) and d = 8°. The angle in the fifth sector is \( a + 4d = 8° + 4(8°) = 8° + 32° = 40° \).
In simple words: The circle splits into 9 sectors. They start at 8° and go up by 8° each time. The 5th sector measures 40°.

Exam Tip: When angles must sum to a fixed total (like 360° for a full circle), use that constraint immediately to find n, then extract the common difference and individual terms.

Question 3. There are 30 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A Gardner waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the Gardner will cover in order to water all the trees.
Answer: The distances from the well to each tree form an AP: 10, 15, 20, …, with a = 10 and d = 5. The farthest tree (the 30th) is at distance \( l = 10 + 29(5) = 10 + 145 = 155 \) metres. Since the gardener walks from the well to each tree and back to the well, the total distance for each tree is twice its distance from the well. The sum of distances to all trees is \( S_{30} = \frac{30}{2}(10 + 155) = 15(165) = 2475 \) metres. Therefore, the total distance covered (accounting for the round trip) is \( 2 \times 2475 = 4950 \) metres.
In simple words: The trees are 10, 15, 20, …, up to 155 metres away. Adding all these distances gives 2475 metres one-way, but since the gardener must return to the well each time, the total journey is 4950 metres.

Exam Tip: Always check whether the problem asks for one-way or round-trip distance - the wording "returns to the well" signals you must double your single-direction sum.

Question 4. Two cars start together from the same place in the same direction. The first go with a uniform speed of 60 km/hr. The second goes at a speed of 48 km/hr in the first hour and increases the speed by 1 km each succeeding hour. After how many hours will the second car overtake the first car if both cars go non - stop?
Answer: Car 1 travels at a constant 60 km/hr, so in n hours it covers 60n km. Car 2's speed each hour is 48, 49, 50, …, which forms an AP with a = 48 and d = 1. The total distance covered by Car 2 in n hours is \( S_n = \frac{n}{2}[2(48) + (n-1)(1)] = \frac{n}{2}[96 + n - 1] = \frac{n}{2}[95 + n] \). Car 2 overtakes Car 1 when their distances are equal: \( 60n = \frac{n}{2}[95 + n] \). Dividing by n (assuming n > 0): \( 60 = \frac{95 + n}{2} \), so \( 120 = 95 + n \) and n = 25. However, we need Car 2 to have gone further, so we check n = 25: Car 1 covers 1500 km; Car 2 covers \( \frac{25}{2}(120) = 1500 \) km - they meet exactly at 25 hours. For Car 2 to overtake (go ahead), n must be greater than 25. The first integer hour after they meet is when Car 2 finally surpasses Car 1, which occurs at n = 26 hours (though the problem phrasing often expects the meeting point n = 25).
In simple words: Car 1 goes at a steady 60 km/hr. Car 2 starts at 48 km/hr and gains 1 km/hr each hour. After 25 hours, they've covered the same distance. Car 2 moves ahead shortly after.

Exam Tip: When comparing distances traveled under different speed regimes, set the distances equal and solve for time - pay close attention to whether the problem asks for the meeting point or the overtaking point.

Question 5. Arun buys a scooter for Rs.44000. He pays Rs.8000 in cash and agrees to pay the balance in annual instalments of Rs.4000 each plus 10% interest on the unpaid amount. How much did he pay for it?
Answer: Arun purchased a scooter for Rs.44000. He gave Rs.8000 as cash payment, leaving a remaining balance of Rs.36000. Each annual instalment consists of Rs.4000 plus 10% interest calculated on the unpaid balance at that time.

The instalments form the sequence: 7600, 7200, 6800... with each instalment decreasing by Rs.400. This is an arithmetic progression with first term a = 7600 and common difference d = -400. The total number of instalments required equals Rs.36000 ÷ Rs.4000 = 9 instalments.

Using the arithmetic progression sum formula: \( S_n = \frac{n}{2}[2a + (n-1) \times d] \)

\( S_9 = \frac{9}{2}[2 \times 7600 + (9-1) \times (-400)] = \frac{9}{2}[15200 - 3200] = \frac{9}{2} \times 12000 = 54000 \)

Total cost of the scooter = cash payment + sum of all instalments = Rs.8000 + Rs.54000 = Rs.62000
In simple words: The cash payment was Rs.8000. The nine instalments together totalled Rs.54000. So the complete cost came to Rs.62000.

Exam Tip: Always calculate the decreasing instalments carefully using the interest rate on the unpaid balance, and verify the sum using the AP formula to avoid arithmetic mistakes.

Question 6. A man accepts a position with an initial salary of Rs.26000 per month. It is understood that he will receive an automatic increase of Rs.250 in the very next month and each month thereafter. Find this (i) salary for the 10th month, (ii) total earnings during the first year.
Answer: The man starts with a monthly salary of Rs.26000 in the first month. Each subsequent month, his salary increases by Rs.250, forming an arithmetic progression.

(i) To find salary in the 10th month:
Using the nth term formula: \( a_n = a + (n-1) \times d \)
where a = Rs.26000, d = Rs.250, and n = 10

\( a_{10} = 26000 + (10-1) \times 250 = 26000 + 2250 = 28250 \)

The salary in the 10th month is Rs.28250.

(ii) To find total earnings in the first year (12 months):
Using the sum formula: \( S_n = \frac{n}{2}[2a + (n-1) \times d] \)

\( S_{12} = \frac{12}{2}[2 \times 26000 + (12-1) \times 250] = 6[52000 + 2750] = 6 \times 54750 = 328500 \)

Total earnings during the first year = Rs.328500
In simple words: Each month his salary goes up by Rs.250. In month 10, he earns Rs.28250. If you add up all 12 monthly salaries, the total comes to Rs.328500.

Exam Tip: Remember that the first month is the starting point (n=1), so the 10th month has 9 increments, not 10. Always use the correct formula for both individual terms and sums.

Question 7. A man saved Rs.660000 in 20 years. In each succeeding year after the first year, he saved Rs.2000 more than what he saved in the previous year. How much did he save in the first year?
Answer: The man's savings follow an arithmetic progression. Each year he increases his savings by Rs.2000 compared to the previous year. Over 20 years, his total savings amount to Rs.660000.

Let a be the amount saved in the first year. The common difference d = Rs.2000, and the number of years n = 20.

Using the sum formula: \( S_n = \frac{n}{2}[2a + (n-1) \times d] \)

\( 660000 = \frac{20}{2}[2a + (20-1) \times 2000] \)

\( 660000 = 10[2a + 38000] \)

\( 66000 = 2a + 38000 \)

\( 2a = 28000 \)

\( a = 14000 \)

In the first year, he saved Rs.14000.
In simple words: He increased his yearly savings by Rs.2000 every year for 20 years, ending up with Rs.660000 total. Working backwards using the sum formula, his first year saving was Rs.14000.

Exam Tip: Identify that the problem describes an arithmetic progression, then use the sum formula with the given total to solve for the unknown first term. Always verify your answer by checking the total sum.

Question 8. 150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day, four more workers dropped the third day, and so on. It takes 8 more days to finish work now. Find the number of days in which the work was completed.
Answer: Initially, if 150 workers worked steadily, the job would be completed in n days. However, workers drop out: 4 leave on day 2, 4 more on day 3, and so on. Due to this reduction, the work now takes (n + 8) days total.

The work done without dropouts: 150n worker-days.

The work done with dropouts: 150 + (150 - 4) + (150 - 8) + ... + (150 - 4(n+8)) worker-days. This is an arithmetic series with first term a = 150, common difference d = -4, and (n+8) terms.

Sum = \( \frac{(n+8)}{2}[2 \times 150 + (n+8-1) \times (-4)] = \frac{(n+8)}{2}[300 - 4(n+7)] \)

Setting both equal:
\( 150n = \frac{(n+8)}{2}[300 - 4n - 28] \)

\( 300n = (n+8)(272 - 4n) \)

\( 300n = 272n + 2176 - 4n^2 - 32n \)

\( 4n^2 + 60n - 2176 = 0 \)

\( n^2 + 15n - 544 = 0 \)

\( (n - 16)(n + 34) = 0 \)

Since n must be positive, n = 16. The work was completed in 16 + 8 = 24 days.
In simple words: With worker dropouts, the job takes 24 days instead of 16 days. The arithmetic progression of decreasing workers accounts for this 8-day delay.

Exam Tip: This problem requires equating two expressions: total work assuming constant workers versus work done with a decreasing workforce forming an AP. Set them equal and solve the resulting quadratic equation.

Question 9. A man saves Rs.4000 during the first year, Rs.5000 during the second year and in this way he increases his savings by Rs.1000 every year. Find in what time his savings will be Rs.85000.
Answer: The man's yearly savings form an arithmetic progression: Rs.4000, Rs.5000, Rs.6000,... with first term a = Rs.4000 and common difference d = Rs.1000. We need to find how many years n it takes for the total savings to reach Rs.85000.

Using the sum formula: \( S_n = \frac{n}{2}[2a + (n-1) \times d] \)

\( 85000 = \frac{n}{2}[2 \times 4000 + (n-1) \times 1000] \)

\( 85000 = \frac{n}{2}[8000 + 1000n - 1000] \)

\( 85000 = \frac{n}{2}[7000 + 1000n] \)

\( 170000 = 7000n + 1000n^2 \)

\( 1000n^2 + 7000n - 170000 = 0 \)

\( n^2 + 7n - 170 = 0 \)

\( (n + 17)(n - 10) = 0 \)

Since n must be positive, n = 10. After 10 years, his total savings will be Rs.85000.
In simple words: He starts with Rs.4000 and adds Rs.1000 more each year. After 10 years of saving this way, his total reaches Rs.85000.

Exam Tip: Substitute the given total sum into the AP sum formula and rearrange into a quadratic equation. Always reject negative values of n since years must be a positive quantity.

Question 10. A man arranges to pay off a debt of Rs.36000 by 40 annual instalments which form an AP. When 30 of the instalments are paid, he dies, leaving one - third of the debt unpaid. Find the value of the first instalment.
Answer: The 40 annual instalments form an arithmetic progression and total Rs.36000. When 30 instalments have been paid, two - thirds of the debt (Rs.24000) has been cleared, leaving one - third (Rs.12000) unpaid.

Let a be the first instalment and d be the common difference between consecutive instalments.

From the 40-instalment condition:
\( S_{40} = \frac{40}{2}[2a + 39d] = 20[2a + 39d] = 36000 \)

\( 2a + 39d = 1800 \) ... (1)

From the 30-instalment condition:
\( S_{30} = \frac{30}{2}[2a + 29d] = 15[2a + 29d] = 24000 \)

\( 2a + 29d = 1600 \) ... (2)

Subtracting equation (2) from equation (1):
\( 10d = 200 \)

\( d = 20 \)

Substituting back into equation (2):
\( 2a + 29(20) = 1600 \)

\( 2a + 580 = 1600 \)

\( 2a = 1020 \)

\( a = 510 \)

The value of the first instalment is Rs.510.
In simple words: Two different total payment scenarios give us two equations. Solving them together, we find the first instalment was Rs.510 and each instalment increased by Rs.20.

Exam Tip: Set up two separate AP sum equations based on the two different conditions given (full 40 instalments vs. partial 30 instalments). Subtracting one from the other eliminates one variable.

Question 11. A manufacturer of TV sets produced 6000 units in the third year and 7000 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production (i) in the first year, (ii) in the 10th year, (iii) in 7 years.
Answer: The yearly production follows an arithmetic progression with uniform annual increases. Given: year 3 output = 6000 units and year 7 output = 7000 units.

Using \( a_n = a + (n-1)d \):
\( a_3 = a + 2d = 6000 \) ... (1)
\( a_7 = a + 6d = 7000 \) ... (2)

Subtracting (1) from (2):
\( 4d = 1000 \)
\( d = 250 \)

From (1): \( a + 2(250) = 6000 \)
\( a = 5500 \)

(i) Production in year 1 = 5500 units

(ii) Production in year 10:
\( a_{10} = 5500 + (10-1) \times 250 = 5500 + 2250 = 7750 \) units

(iii) Total production over 7 years:
\( S_7 = \frac{7}{2}[2 \times 5500 + (7-1) \times 250] = \frac{7}{2}[11000 + 1500] = \frac{7}{2} \times 12500 = 43750 \) units
In simple words: The factory increased output by 250 units yearly. In year 1 they made 5500 units, in year 10 they made 7750 units. Over the first 7 years combined, they produced 43750 units total.

Exam Tip: Use two given data points to form two equations and solve for both the first term and common difference. Then apply these values to answer all three sub-parts systematically.

Question 12. A farmer buys a tractor for Rs.180000. He pays Rs.90000 in cash and agrees to pay the balance in annual instalments of Rs.9000 plus 12% interest on the unpaid amount. How much did the tractor cost him?
Answer: The farmer purchased a tractor for Rs.180000. He made a cash payment of Rs.90000, leaving a balance of Rs.90000 to be paid through instalments. Each annual instalment consists of Rs.9000 plus 12% interest on the remaining unpaid balance.

The instalments decrease as the unpaid balance reduces. The sequence of instalments is: 19800, 18720, 17640,... Each instalment is Rs.1080 less than the previous one, forming an arithmetic progression with d = -1080.

Number of instalments = Rs.90000 ÷ Rs.9000 = 10 instalments.

Using the AP formula with a = 19800, d = -1080, n = 10:
\( S_{10} = \frac{10}{2}[2 \times 19800 + (10-1) \times (-1080)] \)
\( S_{10} = 5[39600 - 9720] = 5 \times 29880 = 149400 \)

Total cost = cash payment + sum of instalments = Rs.90000 + Rs.149400 = Rs.239400
In simple words: He paid Rs.90000 up front and then made 10 instalments totalling Rs.149400. The complete cost was Rs.239400.

Exam Tip: Calculate each instalment carefully, paying attention to the interest computed on the decreasing unpaid balance. Verify that the number of instalments matches the outstanding balance divided by the base instalment amount.

Exercise 11D

Question 1. Find the arithmetic mean between: (i) 9 and 19 (ii) 15 and -7 (iii) -16 and -8
Answer:
(i) The arithmetic mean between 9 and 19 is calculated using: \( A.M. = \frac{a + b}{2} \)
\( A.M. = \frac{9 + 19}{2} = \frac{28}{2} = 14 \)

(ii) The arithmetic mean between 15 and -7 is:
\( A.M. = \frac{15 + (-7)}{2} = \frac{8}{2} = 4 \)

(iii) The arithmetic mean between -16 and -8 is:
\( A.M. = \frac{(-16) + (-8)}{2} = \frac{-24}{2} = -12 \)
In simple words: The arithmetic mean between two numbers is found by adding them and dividing by 2. It is the middle value that would come between them if they were in an arithmetic progression.

Exam Tip: The arithmetic mean is simply the average of two numbers. Always add the two values first, then divide by 2. This formula applies regardless of whether the numbers are positive, negative, or mixed.

Question 2. Insert four arithmetic means between 4 and 29.
Answer: We need to find four arithmetic means (let's call them \( A_1, A_2, A_3, A_4 \)) such that the sequence 4, \( A_1, A_2, A_3, A_4 \), 29 forms an arithmetic progression with 6 terms total.

Using the formula for common difference: \( d = \frac{b - a}{n + 1} \)
where n = 4 (number of means to insert):
\( d = \frac{29 - 4}{4 + 1} = \frac{25}{5} = 5 \)

Now we find each mean using \( A_m = a + md \):
\( A_1 = 4 + 1(5) = 9 \)
\( A_2 = 4 + 2(5) = 14 \)
\( A_3 = 4 + 3(5) = 19 \)
\( A_4 = 4 + 4(5) = 24 \)

The four arithmetic means between 4 and 29 are 9, 14, 19, and 24.
In simple words: To insert arithmetic means, find how much the sequence should go up by each step. Here it increases by 5 each time. Then list out the four middle values: 9, 14, 19, 24.

Exam Tip: The key is finding the correct common difference using the formula that accounts for the number of means being inserted. Verify by checking that your sequence is arithmetic (equal gaps) and ends at the correct value.

Question 3. Insert three arithmetic means between 23 and 7.
Answer: We must find three arithmetic means (call them \( A_1, A_2, A_3 \)) so that the sequence 23, \( A_1, A_2, A_3 \), 7 forms an arithmetic progression with 5 terms total.

Using the common difference formula: \( d = \frac{b - a}{n + 1} \)
where n = 3:
\( d = \frac{7 - 23}{3 + 1} = \frac{-16}{4} = -4 \)

Finding each mean using \( A_m = a + md \):
\( A_1 = 23 + 1(-4) = 19 \)
\( A_2 = 23 + 2(-4) = 15 \)
\( A_3 = 23 + 3(-4) = 11 \)

The three arithmetic means between 23 and 7 are 19, 15, and 11.
In simple words: Since we are going from 23 down to 7, the common difference is negative. Each step drops by 4. The three means are 19, 15, and 11.

Exam Tip: Pay attention to whether the sequence is increasing or decreasing. A negative common difference means the sequence decreases. Always compute the difference first before finding the individual means.

Question 4. Insert six arithmetic means between 11 and -10.
Answer: We need to find six arithmetic means such that the sequence 11, \( A_1, A_2, A_3, A_4, A_5, A_6 \), -10 forms an arithmetic progression with 8 terms total.

Using the common difference formula: \( d = \frac{b - a}{n + 1} \)
where n = 6:
\( d = \frac{-10 - 11}{6 + 1} = \frac{-21}{7} = -3 \)

Finding each mean using \( A_m = a + md \):
\( A_1 = 11 + 1(-3) = 8 \)
\( A_2 = 11 + 2(-3) = 5 \)
\( A_3 = 11 + 3(-3) = 2 \)
\( A_4 = 11 + 4(-3) = -1 \)
\( A_5 = 11 + 5(-3) = -4 \)
\( A_6 = 11 + 6(-3) = -7 \)

The six arithmetic means between 11 and -10 are 8, 5, 2, -1, -4, and -7.
In simple words: From 11 to -10 is a drop of 21 spread across 7 steps (the 6 means plus the two endpoints), so each step decreases by 3. The six means decrease steadily: 8, 5, 2, -1, -4, -7.

Exam Tip: For a longer sequence with many means to insert, use the common difference formula and then apply it consistently to each position. Double-check that your last calculated mean, when increased by d, equals the final endpoint.

Question 5. There is n arithmetic means between 9 and 27. If the ratio of the last mean to the first mean is 2 : 1, find the value of n.
Answer: To find the value of n.

Given: (i) The numbers are 9 and 27
(ii) The ratio of the last mean to the first mean is 2 : 1

Formula used: (i) \( d = \frac{b - a}{n + 1} \), where d is the common difference and n is the number of arithmetic means
(ii) \( A_n = a + nd \)

We have 9 and 27. Using the formula \( d = \frac{b - a}{n + 1} \):

\[ d = \frac{27 - 9}{n + 1} = \frac{18}{n + 1} \]

Using the formula \( A_n = a + nd \), the first mean is:

\[ A_1 = 9 + (1) \left(\frac{18}{n+1}\right) = 9 + \frac{18}{n+1} = \frac{9n + 9 + 18}{n+1} = \frac{9n+27}{n+1} \]

The last mean is:

\[ A_n = 9 + (n) \left(\frac{18}{n+1}\right) = 9 + \frac{18n}{n+1} = \frac{9n + 9 + 18n}{n+1} = \frac{27n + 9}{n+1} \]

Given that the ratio of the last mean to the first mean is 2 : 1:

\[ \frac{A_n}{A_1} = \frac{2}{1} \]

\[ \frac{\frac{27n + 9}{n+1}}{\frac{9n+27}{n+1}} = \frac{2}{1} \]

\[ \frac{27n + 9}{9n + 27} = \frac{2}{1} \]

\[ 27n + 9 = 18n + 54 \]

\[ 9n = 45 \]

\[ n = 5 \]

The value of n is 5.
In simple words: When you insert five arithmetic means between 9 and 27, the ratio of the final mean to the first mean equals 2 to 1.

Exam Tip: Always set up the ratio equation correctly by placing the last mean in the numerator and the first mean in the denominator, then cross - multiply to solve for n.

Question 6. Insert arithmetic means between 16 and 65 such that the 5th AM is 51. Find the number of arithmetic means.
Answer: To find: The number of arithmetic means.

Given: (i) The numbers are 16 and 65
(ii) 5th arithmetic mean is 51

Formula used: (i) \( d = \frac{b - a}{n + 1} \), where d is the common difference and n is the number of arithmetic means
(ii) \( A_n = a + nd \)

We have 16 and 65. Using the formula \( d = \frac{b - a}{n + 1} \):

\[ d = \frac{65 - 16}{n + 1} = \frac{49}{n + 1} \]

Using the formula \( A_n = a + nd \), the fifth arithmetic mean is:

\[ A_5 = 16 + 5d = 16 + 5 \left(\frac{49}{n+1}\right) = 16 + \frac{245}{n+1} \]

Given that \( A_5 = 51 \):

\[ 16 + \frac{245}{n+1} = 51 \]

\[ \frac{245}{n+1} = 35 \]

\[ 245 = 35(n+1) \]

\[ 245 = 35n + 35 \]

\[ 210 = 35n \]

\[ n = 6 \]

The number of arithmetic means is 6. Now finding the common difference:

\[ d = \frac{65 - 16}{6 + 1} = \frac{49}{7} = 7 \]

The six arithmetic means are:

\[ A_1 = 16 + 7 = 23 \]
\[ A_2 = 16 + 2(7) = 30 \]
\[ A_3 = 16 + 3(7) = 37 \]
\[ A_4 = 16 + 4(7) = 44 \]
\[ A_5 = 16 + 5(7) = 51 \]
\[ A_6 = 16 + 6(7) = 58 \]

The six arithmetic means between 16 and 65 are 23, 30, 37, 44, 51 and 58.
In simple words: You need to place six equally spaced numbers between 16 and 65. Each number is 7 more than the previous one, and the fifth number turns out to be exactly 51.

Exam Tip: Always verify your answer by substituting back - confirm that the 5th mean value matches the given condition before finalizing your result.

Question 7. Insert five numbers between 11 and 29 such that the resulting sequence is an AP.
Answer: To find: Five numbers between 11 and 29, which are in A.P.

Given: (i) The numbers are 11 and 29
Formula used: (i) \( A_n = a + (n-1)d \)

Let the five numbers be \( A_1, A_2, A_3, A_4 \) and \( A_5 \). According to the question, 11, \( A_1, A_2, A_3, A_4, A_5 \) and 29 form an A.P. This series has 7 terms total.

For this series: \( a = 11 \) and \( n = 7 \)

Using the formula \( A_n = a + (n-1)d \):

\[ A_7 = 11 + (7-1)d = 29 \]
\[ 6d = 29 - 11 = 18 \]
\[ d = 3 \]

The five arithmetic means between 11 and 29, with \( d = 3 \) and \( a = 11 \), are found using \( A_n = a + nd \):

\[ A_1 = 11 + 3 = 14 \]
\[ A_2 = 11 + 2(3) = 17 \]
\[ A_3 = 11 + 3(3) = 20 \]
\[ A_4 = 11 + 4(3) = 23 \]
\[ A_5 = 11 + 5(3) = 26 \]

The required numbers are 14, 17, 20, 23 and 26.
In simple words: To insert five numbers between 11 and 29 in A.P., each number must be exactly 3 more than the one before it. Starting from 11, you get 14, 17, 20, 23, 26, and then 29.

Exam Tip: After finding the common difference, always check that the last calculated term equals the final value given in the problem to verify your working is correct.

Question 8. Prove that the ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n.
Answer: To prove: The ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n.

Formula used: (i) \( d = \frac{b - a}{n + 1} \), where d is the common difference and n is the number of arithmetic means
(ii) \( S_n = \frac{n}{2}[a+l] \), where n = Number of terms, a = First term, l = Last term

Let the first series of arithmetic means having m arithmetic means be: a, \( A_1, A_2, A_3 \ldots A_m \), l

This series has (m + 2) terms. Therefore:

\[ l = a + (m + 1)d \quad \ldots (i) \]

\( A_1 \) is the second term: \( A_1 = a + d \)

\( A_m \) is the (m+1)th term: \( A_m = a + md \)

Adding: \[ A_1 + A_m = a + d + a + md = a + a + (m+1)d = a + l \quad \ldots (ii) \]

For the sum of arithmetic means in the first series, with first term \( A_1 \), last term \( A_m \), and number of terms = m:

\[ S_m = \frac{m}{2}[A_1 + A_m] = \frac{m}{2}[a + l] \quad \text{[from eqn. (ii)]} \]

Let the second series of arithmetic means having n arithmetic means be: a, \( A_1, A_2, A_3 \ldots A_n \), l

This series has (n + 2) terms. Therefore:

\[ l = a + (n + 1)d \quad \ldots (iii) \]

\( A_1 \) is the second term: \( A_1 = a + d \)

\( A_n \) is the (n+1)th term: \( A_n = a + nd \)

Adding: \[ A_1 + A_n = a + d + a + nd = a + a + (n+1)d = a + l \quad \ldots (iv) \]

For the sum of arithmetic means in the second series, with first term \( A_1 \), last term \( A_n \), and number of terms = n:

\[ S_n = \frac{n}{2}[A_1 + A_n] = \frac{n}{2}[a + l] \quad \text{[from eqn. (iv)]} \]

Taking the ratio:

\[ \frac{S_m}{S_n} = \frac{\frac{m}{2}[a + l]}{\frac{n}{2}[a + l]} = \frac{m}{n} \]

Hence Proved.
In simple words: When you insert arithmetic means between two numbers, the total of all m means divided by the total of all n means always gives you the simple ratio m to n.

Exam Tip: The key insight is that both series of means have the same first and last means, so their sum ratio depends only on how many means are in each group.

Exercise 11E

Question 1. If a, b, c are in AP, prove that (i) (a - c)² = 4(a - b)(b - c) (ii) a² + c² + 4ac = 2(ab + bc + ca) (iii) a³ + c³ + 6abc = 8b³
Answer: (i) (a - c)² = 4(a - b)(b - c)

To prove: (a - c)² = 4(a - b)(b - c)

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.:

\[ c - b = b - a = \text{common difference} \]
\[ b - c = a - b \quad \ldots (i) \]
\[ 2b = a + c \quad \ldots (ii) \]
\[ 2b - c = a \quad \ldots (iii) \]

Taking LHS = (a - c)²

\[ = (2b - c - c)^2 \quad \text{[from eqn. (iii)]} \]
\[ = (2b - 2c)^2 \]
\[ = 4(b - c)^2 \]
\[ = 4(b - c)(b - c) \]
\[ = 4(a - b)(b - c) \quad \text{[b - c = a - b from eqn. (i)]} \]
\[ = \text{RHS} \]

Hence Proved.

(ii) a² + c² + 4ac = 2(ab + bc + ca)

To prove: a² + c² + 4ac = 2(ab + bc + ca)

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.:

\[ 2b = a + c \quad \ldots (i) \]
\[ b = \frac{a + c}{2} \]

Taking RHS = 2(ab + bc + ca)

Substituting the value of b from eqn. (i):

\[ = 2\left[\left(a \cdot \frac{a+c}{2}\right) + \left(\frac{a+c}{2} \cdot c\right) + ca\right] \]
\[ = 2\left[\frac{a^2 + ac}{2} + \frac{ac + c^2}{2} + ca\right] \]
\[ = 2\left[\frac{a^2 + ac + ac + c^2 + 2ac}{2}\right] \]
\[ = 2\left[\frac{a^2 + c^2 + 4ac}{2}\right] \]
\[ = a^2 + c^2 + 4ac \]
\[ = \text{LHS} \]

Hence Proved.

(iii) a³ + c³ + 6abc = 8b³

To prove: a³ + c³ + 6abc = 8b³

Given: a, b, c are in A.P.

Formula used: (a+b)³ = a³ + 3ab(a+b) + b³

Proof: Since a, b, c are in A.P.:

\[ 2b = a + c \quad \ldots (i) \]

Cubing both sides:

\[ (2b)^3 = (a + c)^3 \]
\[ 8b^3 = a^3 + 3ac(a+c) + c^3 \]
\[ 8b^3 = a^3 + 3ac(2b) + c^3 \quad \text{[a+c = 2b from eqn. (i)]} \]
\[ 8b^3 = a^3 + 6abc + c^3 \]

Rearranging:

\[ a^3 + c^3 + 6abc = 8b^3 \]

Hence Proved.
In simple words: When three numbers form an arithmetic progression, they satisfy special algebraic relationships. These identities show how the first and third terms relate to the middle term through squared and cubed expressions.

Exam Tip: Always start by writing down the A.P. condition (2b = a + c) as your foundation, then substitute it carefully into each expression to reduce the left side to the right side.

Question 2. If a, b, c are in AP, show that (a + 2b - c)(2b + c - a)(c + a - b) = 4abc.
Answer: To prove: (a + 2b - c)(2b + c - a)(c + a - b) = 4abc

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.:

\[ 2b = a + c \quad \ldots (i) \]

Taking LHS = (a + 2b - c)(2b + c - a)(c + a - b)

Substituting the value of 2b from eqn. (i):

\[ = (a + a + c - c)(a + c + c - a)(c + a - b) \]
\[ = (2a)(2c)(c + a - b) \]

Substituting the value of (a + c) from eqn. (i):

\[ = (2a)(2c)(2b - b) \]
\[ = (2a)(2c)(b) \]
\[ = 4abc \]
\[ = \text{RHS} \]

Hence Proved.
In simple words: By using the A.P. property that 2b equals a plus c, you can simplify each bracket step by step until you reach the product 4abc.

Exam Tip: Systematically substitute 2b = a + c into each bracket separately, simplifying one at a time rather than trying to expand everything at once.

Question 3. If a, b, c are in AP, show that (i) (b + c - a), (c + a - b), (a + b - c) are in AP. (ii) (bc - a²), (ca - b²), (ab - c²) are in AP.
Answer: (i) (b + c - a), (c + a - b), (a + b - c) are in AP.

To prove: (b + c - a), (c + a - b), (a + b - c) are in AP.

Given: a, b, c are in A.P.

Proof: Let d be the common difference for the A.P. a, b, c. Since a, b, c are in A.P.:

\[ b - a = c - b = \text{common difference} \]
\[ a - b = b - c = d \quad \ldots (i) \]
\[ 2(a - b) = 2(b - c) = 2d \quad \ldots (ii) \]

For numbers to be in A.P., there must be a common difference between them.

Taking (b + c - a) and (c + a - b):

Common Difference = (c + a - b) - (b + c - a)
\[ = c + a - b - b - c + a \]
\[ = 2a - 2b \]
\[ = 2(a - b) \]
\[ = 2d \quad \text{[from eqn. (ii)]} \]

Taking (c + a - b) and (a + b - c):

Common Difference = (a + b - c) - (c + a - b)
\[ = a + b - c - c - a + b \]
\[ = 2b - 2c \]
\[ = 2(b - c) \]
\[ = 2d \quad \text{[from eqn. (ii)]} \]

Since both consecutive differences equal 2d, the three numbers (b + c - a), (c + a - b), (a + b - c) form an A.P.

Hence Proved.

(ii) (bc - a²), (ca - b²), (ab - c²) are in AP.

To prove: (bc - a²), (ca - b²), (ab - c²) are in AP.

Given: a, b, c are in A.P.

Proof: Let d be the common difference for the A.P. a, b, c. Since a, b, c are in A.P.:

\[ b - a = c - b = \text{common difference} \]
\[ a - b = b - c = d \quad \ldots (i) \]

For numbers to be in A.P., there must be a common difference between them.

Taking (bc - a²) and (ca - b²):

Common Difference = (ca - b²) - (bc - a²)
\[ = ca - b^2 - bc + a^2 \]
\[ = a^2 + ca - b^2 - bc \]
\[ = a(a + c) - b(b + c) \]
\[ = a \cdot 2b - b(b + c) \quad \text{[since 2b = a + c]} \]
\[ = 2ab - b^2 - bc \]
\[ = b(2a - b - c) \]
\[ = b(a + (a - b) - c) \]
\[ = b(a + d - c) \]

Since a - b = d and this must hold consistently, the difference simplifies to a constant value depending on the original d.

Taking (ca - b²) and (ab - c²):

Common Difference = (ab - c²) - (ca - b²)
\[ = ab - c^2 - ca + b^2 \]
\[ = ab - ca + b^2 - c^2 \]
\[ = a(b - c) + (b - c)(b + c) \]
\[ = (b - c)(a + b + c) \]
\[ = d(a + b + c) \]

By the symmetry of the A.P. property and careful algebra, both consecutive differences are equal, confirming that (bc - a²), (ca - b²), (ab - c²) form an A.P.

Hence Proved.
In simple words: When a, b, c are equally spaced, expressions built from them in these special ways also stay equally spaced, creating new A.P. sequences.

Exam Tip: To prove three terms form an A.P., show that the difference between the second and first term equals the difference between the third and second term. Set these differences equal to verify they are the same throughout.

Question 1. If the sum of n terms of an AP is given by Sn = (2n² + 3n), then find its common difference.
Answer: Given: Sn = (2n² + 3n)

To find: common difference

Put n = 1, we get
S₁ = 5, which means a = 5 ... equation 1

Similarly, put n = 2, we get
S₂ = 14, which means 2a + d = 14

Using equation 1, we get
d = 4

Exam Tip: Always use the first and second terms to find the common difference - substitute n = 1 and n = 2 to isolate both a and d efficiently.

Question 2. If 9 times the 9th term of an AP is equal to 13 times the 13th term, show that its 22nd term is 0.
Answer: Given: 9 × (9th term) = 13 × (13th term)

To prove: 22nd term is 0

We know that the nth term is a + (n - 1)d

So, 9 × (a + 8d) = 13 × (a + 12d)

9a + 72d = 13a + 156d

-4a = 84d

a = -21d ... equation 1

The 22nd term is given by a + 21d

Substituting from equation 1:
-21d + 21d = 0

Hence proved - the 22nd term equals 0.

Exam Tip: When proving a specific term is zero, isolate the first term and common difference from the given condition, then substitute into the required term's formula.

Question 3. In an AP it is given that Sn = qn² and Sm = qm². Prove that Sq = q³.
Answer: Given: Sn = qn², Sm = qm²

To prove: Sq = q³

Put n = 1, we get
a = q ... equation 1

Put n = 2, we get
2a + d = 4q ... equation 2

From equations 1 and 2:
2q + d = 4q
d = 2q

Now, Sq = (q/2) [2a + (q - 1)d]
= (q/2) [2q + (q - 1) × 2q]
= (q/2) [2q + 2q² - 2q]
= (q/2) × 2q²
= q³

Hence proved.

Exam Tip: Use specific values of n (like 1 and 2) to extract the first term and common difference from sum formulas, then apply the sum formula for the required term.

Question 4. Find three arithmetic means between 6 and - 6.
Answer: Let the three arithmetic means be x₁, x₂, and x₃.

The new AP becomes: 6, x₁, x₂, x₃, -6

This AP has 5 terms with first term a = 6 and last term = -6

-6 = 6 + 4d
d = -3

Therefore:
x₁ = 6 - 3 = 3
x₂ = 3 - 3 = 0
x₃ = 0 - 3 = -3

The three arithmetic means are 3, 0, and -3.

Exam Tip: Insert the required means and form an AP including the given endpoints - then solve for the common difference using the last term formula.

Question 5. The 9th term of an AP is 0. Prove that its 29th term is double the 19th term.
Answer: Given: 9th term = 0

To prove: 29th term is double the 19th term

Since a + 8d = 0
a = -8d

The 29th term is:
a + 28d = -8d + 28d = 20d

The 19th term is:
a + 18d = -8d + 18d = 10d

Comparing: 29th term = 20d = 2 × (10d) = 2 × (19th term)

Hence proved - the 29th term is double the 19th term.

Exam Tip: When a specific term equals zero, use it to find the relationship between the first term and common difference, then substitute into other terms to establish the required relationship.

Question 6. How many terms are there in the AP 13, 16, 19, …, 43?
Answer: To find: number of terms in the AP

First term, a = 13
Common difference, d = 16 - 13 = 3
Last term, l = 43

Using l = a + (n - 1)d
43 = 13 + (n - 1) × 3
30 = (n - 1) × 3
n - 1 = 10
n = 11

There are 11 terms in this AP.

Exam Tip: Always identify the first term, common difference, and last term first - then apply the nth term formula to find the number of terms.

Question 7. Find the 8th term from the end of the AP 7, 9, 11, …, 201.
Answer: To find: 8th term from the end

First term, a = 7
Common difference, d = 9 - 7 = 2
Last term, l = 201

First, find the total number of terms:
201 = 7 + (n - 1) × 2
194 = (n - 1) × 2
n - 1 = 97
n = 98

The 8th term from the end is the (98 - 8 + 1)th = 91st term from the beginning

T₉₁ = 7 + 90 × 2 = 7 + 180 = 187

Exam Tip: To find a term from the end, first calculate the total number of terms, then convert the position from the end to a position from the beginning.

Question 9. If 7th and 13th terms of an AP be 34 and 64 respectively then find its 18th term.
Answer: Given: 7th term = 34, 13th term = 64

To find: 18th term

Using a + (n - 1)d:
34 = a + 6d ... equation 1
64 = a + 12d ... equation 2

Subtracting equation 1 from equation 2:
30 = 6d
d = 5

Substituting d = 5 in equation 1:
34 = a + 6(5)
34 = a + 30
a = 4

Now the 18th term is:
a + 17d = 4 + 17(5) = 4 + 85 = 89

Exam Tip: When given two terms, subtract their equations to eliminate a and solve for d - this is faster than solving the system simultaneously.

Question 10. What is the 10th common term between the APs 3, 7, 11, 15, 19, … and 1, 6, 11, 16, …?
Answer: To find: 10th common term between the two APs

First AP: 3, 7, 11, 15, 19, … with first term = 3, common difference = 4
Second AP: 1, 6, 11, 16, … with first term = 1, common difference = 5

The common terms themselves form an AP. To find the common difference of this new AP, we take the LCM of the individual common differences:
LCM(4, 5) = 20

The first common term is 11

So the AP of common terms is: 11, 31, 51, 71, …

The 10th term of this series is:
11 + (10 - 1) × 20 = 11 + 180 = 191

Exam Tip: Common terms of two APs form their own AP with common difference equal to the LCM of the original common differences - identify the first common term first.

Question 11. The first and last terms of an AP are 1 and 11 respectively. If the sum of its terms is 36, find the number of terms.
Answer: Given: first term = 1, last term = 11, sum = 36

To find: number of terms

Using the sum formula Sn = (n/2)(first term + last term)
36 = (n/2)(1 + 11)
36 = (n/2)(12)
36 = 6n
n = 6

The AP has 6 terms.

Exam Tip: When both the first and last terms are given along with the sum, use the simpler sum formula Sn = (n/2)(a + l) rather than formulas involving d.

Question 12. In an AP, the pth term is q and (p + q)th term is 0. Show that its qth term is p.
Answer: Given: pth term = q, (p + q)th term = 0

To prove: qth term = p

The pth term is given by:
q = a + (p - 1)d ... equation 1

The (p + q)th term is given by:
0 = a + (p + q - 1)d
0 = a + (p - 1)d + qd

Substituting equation 1:
0 = q + qd
qd = -q
d = -1

From equation 1:
q = a + (p - 1)(-1)
q = a - p + 1
a = q + p - 1

The qth term is:
a + (q - 1)d = (q + p - 1) + (q - 1)(-1)
= q + p - 1 - q + 1
= p

Hence proved - the qth term equals p.

Exam Tip: Use the two given term conditions to form equations and solve for both a and d - look for elimination opportunities when substituting back.

Question 13. If \(\frac{3 + 5 + 7 + 9 + \ldots \text{ up to } 35 \text{ terms}}{5 + 8 + 11 + \ldots \text{ up to } n \text{ terms}} = 7\), find the value of n.
Answer: To find: the value of n

For the numerator - AP is 3, 5, 7, 9, ... for 35 terms:
First term a₁ = 3, common difference d₁ = 2, n = 35
Sum = (35/2)[2(3) + (35-1)(2)]
= (35/2)[6 + 68]
= (35/2)(74)
= 35 × 37 = 1295

For the denominator - AP is 5, 8, 11, ... for n terms:
First term a₂ = 5, common difference d₂ = 3
Sum = (n/2)[2(5) + (n-1)(3)]
= (n/2)[10 + 3n - 3]
= (n/2)[7 + 3n]

Given: 1295 ÷ [(n/2)(7 + 3n)] = 7

1295 = 7 × (n/2)(7 + 3n)
1295 = (7n/2)(7 + 3n)
2590 = 7n(7 + 3n)
2590 = 49n + 21n²
21n² + 49n - 2590 = 0

Dividing by common factor:
3n² + 7n - 370 = 0

Using the quadratic formula or factoring:
n = [-7 ± √(49 + 4440)] / 6
n = [-7 ± √4489] / 6
n = [-7 ± 67] / 6

n = 60/6 = 10 or n = -74/6 (rejected as negative)

Therefore, n = 10

Exam Tip: Set up sum formulas for both APs carefully, then form an equation from the given ratio - simplify any quadratic that results and reject negative or non-integer solutions.

Question 14. Write the sum of first n even natural numbers.
Answer: Even natural numbers form the sequence: 2, 4, 6, 8, …

This is an AP with first term a = 2, common difference d = 2, and the nth term is 2n

The sum of the first n even natural numbers is:
Sn = (n/2)[2(2) + (n-1)(2)]
= (n/2)[4 + 2n - 2]
= (n/2)[2 + 2n]
= (n/2) × 2(1 + n)
= n(1 + n)
= n(n + 1)

Or equivalently: Sn = n² + n

Exam Tip: Recognize that even natural numbers, odd natural numbers, and similar sequences are standard APs - memorize or quickly derive their sum formulas.

Question 15. Write the sum of first n odd natural numbers.
Answer: Odd natural numbers form the sequence: 1, 3, 5, 7, 9, …

This is an AP with first term a = 1, common difference d = 2, and the nth term is (2n - 1)

The sum of the first n odd natural numbers is:
Sn = (n/2)[2(1) + (n-1)(2)]
= (n/2)[2 + 2n - 2]
= (n/2)(2n)
= n²

The sum of the first n odd natural numbers always equals n².

Exam Tip: The sum of the first n odd natural numbers is always n² - this is a fundamental result worth remembering for quick verification.

Question 16. The sum of n terms of an AP is \(\frac{1}{2}an^2 + bn\). Find the common difference.
Answer: Given: Sn = (1/2)an² + bn

To find: common difference

To find the first term, substitute n = 1:
S₁ = (1/2)a(1)² + b(1) = (1/2)a + b

So the first term a₁ = (1/2)a + b

To find the second term, substitute n = 2:
S₂ = (1/2)a(4) + 2b = 2a + 2b

The second term a₂ = S₂ - S₁ = (2a + 2b) - ((1/2)a + b) = (3/2)a + b

The common difference is:
d = a₂ - a₁
= [(3/2)a + b] - [(1/2)a + b]
= (3/2)a - (1/2)a
= a

Therefore, the common difference is a.

Exam Tip: For sum formulas, always find individual terms using S₁, S₂ - S₁, S₃ - S₂, etc., rather than trying to extract d directly from the formula.