CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 11

Quadratic Equation

More question-

1) Find the discriminate of the quadratic equation: 3 √3 x² + 10x + √3 = 0 (64)

2) Solve for x: a) 9x² – 9 (a + b) x + 2a² + 5ab + 2b² = 0 (2a+b/3, a + 2b/3)

b) 4x² – 4a2x + (a⁴– b⁴) = 0 (a² +b²)/2, (a²- b²)/2

c) 10ax² – 6x +15ax – 9 = 0 (-3/2, 3/5a)

d) x2 – 2(a² + b² )x + (a² – b²)² = 0 (a+b)², (a-b)²

e) √7x² – 6x – 13 √7 (13√7/7, - √7)

3) find the value of k so that the quadratic equation has equal roots:

a) 2kx² – 40x + 25 = 0 (k = 8)       b) 2x² – (k – 2) x + 1 = 0 (2+2√2)

c) K x (x – 7) + 49 = 0 (0, 4)          d) (k -5)x² + 2(k-5)x + 2 = 0 (5,7)

4) Find the roots of the following quadratic equation by the method of completing the Square.

a) a²x² – 3abx + 2b² = 0    b) x² – 4ax + 4a²- b²= 0

5) Solve the following quadratic equations by factorization method:

a) 3x² - 2√6x + 2 = 0 (√2/3, √2/3) b) 9x² – 6ax + (a² – b²) = 0 (a² + b² , a² – b²)

3 2

6) write the nature of roots of quadratic equation: 4x² + 4√3x + 3 = 0

7) Check whether the equation x³ – 4x² + 1 = (x – 2)² is quadratic or not

8) Solve for x: 1 = 1 + 1 + 1, a + b ≠ 0

a + b + x a b x (-a, -b)

9) If p, q are the roots of the equation x² – 5x + 4 =0, find the value of 1 + 1 - 2pq

P q (-27/4)

10) Solve for x: x + x +1 = 34 (3/2, -5/2)

x + 1 x 15

11) Solve for x: 1 - 1 = 1 (7,-9)

x – 3 x + 5 6

12) If one root of a quadratic equation 3x²+ PX + 4 = 0 is 2/3, find the value of p (p = -8)

13) Solve for x: 2 2x -1 – 3 x +3 = 5

X+3 2x-1 (-10, -1/5)

14) The sum of the squares of two consecutive odd numbers is 394. Find the numbers. (13, 15)

15) The sum of the areas of two squares is 640 m2. If the difference in their perimeter is 64m .Find the sides of the two squares (8m, 24m)

16) The difference of two numbers is 4. If the difference of their reciprocals is 4/21, find the numbers (3, 7)

17) A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500km away in time it had to Increase the speed by 250 km/h from the usual speed. Find its usual speed (750 km / hr)

18) The hypotenuse of a grassy land in the shape of a right triangle is 1m more than twice the shortest side. If the third side is 7m More than the shortest side find the sides of grassy land ( 8, 15)

19) Find two consecutive numbers, whose squares have the sum 85. (6, 7)

20) The sum of the reciprocals of rehmans age 3years ago and 5years from now is 1/3, find his present age

21) A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number (8)

22) A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together Can finish it in 4 days; find the time taken by B to finish the work (12 days)

23) The speed of a boat in still water is 15 km/hr. It can go 30km upstream and return downstream to the original point in 4hrs 30min. Find out the speed of the stream (5km/hr)

24) A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number (92)

25) Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20years.Four years ago, the product of their ages was 48. (D = - 48, No)

26) If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c

Question 1. Find the discriminate of the quadratic equation: 3 √3 x² + 10x + √3 = 0
Answer: For the quadratic equation \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \), we identify the coefficients as \( a = 3\sqrt{3} \), \( b = 10 \), and \( c = \sqrt{3} \). The discriminant is calculated using the standard formula \( D = b^2 - 4ac \). Substituting the values gives:
\( D = (10)^2 - 4(3\sqrt{3})(\sqrt{3}) \)
\( \implies D = 100 - 4 \times 3 \times 3 \)
\( \implies D = 100 - 36 \)
\( \implies D = 64 \). Thus, the discriminant of this quadratic equation is 64.
In simple words: The discriminant helps us find the nature of the roots. By putting the values into the formula, we find the answer is 64.

Exam Tip: Be careful when multiplying square roots like \( \sqrt{3} \times \sqrt{3} = 3 \) to avoid simple arithmetic mistakes.

Question 2. Solve for x:
a) \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
b) \( 4x^2 - 4a^2 x + (a^4 - b^4) = 0 \)
c) \( 10ax^2 - 6x + 15ax - 9 = 0 \)
d) \( x^2 - 2(a^2 + b^2)x + (a^2 - b^2)^2 = 0 \)
e) \( \sqrt{7}x^2 - 6x - 13\sqrt{7} = 0 \)
Answer:
a) First, factor the constant term:
\( 2a^2 + 5ab + 2b^2 = 2a^2 + 4ab + ab + 2b^2 = 2a(a+2b) + b(a+2b) = (2a+b)(a+2b) \).
Now, we split the middle term of the equation:
\( -9(a+b)x = -3[(2a+b) + (a+2b)]x \).
Substitute this back into the equation:
\( 9x^2 - 3(2a+b)x - 3(a+2b)x + (2a+b)(a+2b) = 0 \).
Grouping terms gives:
\( 3x[3x - (2a+b)] - (a+2b)[3x - (2a+b)] = 0 \)
\( \implies [3x - (2a+b)][3x - (a+2b)] = 0 \).
This gives the roots:
\( x = \frac{2a+b}{3} \) or \( x = \frac{a+2b}{3} \).

b) The constant term can be factored as \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \).
The middle term can be split using these factors:
\( -4a^2 x = -2[(a^2 + b^2) + (a^2 - b^2)]x \).
Substituting this gives:
\( 4x^2 - 2(a^2 + b^2)x - 2(a^2 - b^2)x + (a^2 - b^2)(a^2 + b^2) = 0 \).
Group and factor:
\( 2x[2x - (a^2 + b^2)] - (a^2 - b^2)[2x - (a^2 + b^2)] = 0 \)
\( \implies [2x - (a^2 + b^2)][2x - (a^2 - b^2)] = 0 \).
So, the roots are:
\( x = \frac{a^2 + b^2}{2} \) or \( x = \frac{a^2 - b^2}{2} \).

c) Grouping the terms directly from \( 10ax^2 - 6x + 15ax - 9 = 0 \) gives:
\( 2x(5ax - 3) + 3(5ax - 3) = 0 \)
\( \implies (2x + 3)(5ax - 3) = 0 \).
Thus, the roots are:
\( x = -\frac{3}{2} \) or \( x = \frac{3}{5a} \).

d) We rewrite the middle term coefficient using the identity \( 2(a^2 + b^2) = (a+b)^2 + (a-b)^2 \).
The constant term is \( (a^2 - b^2)^2 = (a+b)^2(a-b)^2 \).
The equation becomes:
\( x^2 - [(a+b)^2 + (a-b)^2]x + (a+b)^2(a-b)^2 = 0 \)
\( \implies [x - (a+b)^2][x - (a-b)^2] = 0 \).
So, the roots are:
\( x = (a+b)^2 \) or \( x = (a-b)^2 \).

e) Multiply the first and last coefficients: \( \sqrt{7} \times (-13\sqrt{7}) = -91 \).
We need factors of \( -91 \) that add to \( -6 \). These are \( 7 \) and \( -13 \).
Splitting the middle term:
\( \sqrt{7}x^2 + 7x - 13x - 13\sqrt{7} = 0 \)
\( \implies \sqrt{7}x(x + \sqrt{7}) - 13(x + \sqrt{7}) = 0 \)
\( \implies (\sqrt{7}x - 13)(x + \sqrt{7}) = 0 \).
This gives:
\( x = \frac{13\sqrt{7}}{7} \) or \( x = -\sqrt{7} \).
In simple words: We can solve complex-looking algebraic quadratics by factoring the constant terms or grouping variables together to make simpler brackets.

Exam Tip: For algebraic terms, look for patterns such as difference of squares or perfect squares to break the constant term down into easily groupable components.

Question 3. find the value of k so that the quadratic equation has equal roots:
a) \( 2kx^2 - 40x + 25 = 0 \)
b) \( 2x^2 - (k-2)x + 1 = 0 \)
c) \( kx(x-7) + 49 = 0 \)
d) \( (k-5)x^2 + 2(k-5)x + 2 = 0 \)
Answer: For a quadratic equation to have equal roots, its discriminant must be zero (\( D = b^2 - 4ac = 0 \)).
a) Here, \( a = 2k \), \( b = -40 \), \( c = 25 \).
\( D = (-40)^2 - 4(2k)(25) = 1600 - 200k = 0 \)
\( \implies 200k = 1600 \)
\( \implies k = 8 \).

b) Here, \( a = 2 \), \( b = -(k-2) \), \( c = 1 \).
\( D = [-(k-2)]^2 - 4(2)(1) = (k-2)^2 - 8 = 0 \)
\( \implies (k-2)^2 = 8 \)
\( \implies k-2 = \pm 2\sqrt{2} \)
\( \implies k = 2 \pm 2\sqrt{2} \).

c) Rearranging \( kx(x-7) + 49 = 0 \) gives \( kx^2 - 7kx + 49 = 0 \).
Here, \( a = k \), \( b = -7k \), \( c = 49 \).
\( D = (-7k)^2 - 4(k)(49) = 49k^2 - 196k = 0 \)
\( \implies 49k(k - 4) = 0 \).
This gives \( k = 0 \) or \( k = 4 \). If \( k = 0 \), the equation is no longer quadratic, so we choose \( k = 4 \).

d) Here, \( a = k-5 \), \( b = 2(k-5) \), \( c = 2 \).
\( D = [2(k-5)]^2 - 4(k-5)(2) = 4(k-5)^2 - 8(k-5) = 0 \)
\( \implies 4(k-5)[(k-5) - 2] = 0 \)
\( \implies 4(k-5)(k-7) = 0 \).
This yields \( k = 5 \) or \( k = 7 \). Since \( k = 5 \) makes the leading coefficient zero (meaning the equation is not quadratic), we choose \( k = 7 \).
In simple words: We set the discriminant formula to zero because equal roots mean the curve just touches the x-axis at one point.

Exam Tip: Always double check if your value of k makes the coefficient of \( x^2 \) equal to zero. If it does, discard that value because a quadratic equation must have a non-zero \( x^2 \) coefficient.

Question 4. Find the roots of the following quadratic equation by the method of completing the Square.
a) \( a^2x^2 - 3abx + 2b^2 = 0 \)
b) \( x^2 - 4ax + 4a^2 - b^2 = 0 \)
Answer:
a) First, divide the equation by \( a^2 \):
\( x^2 - \frac{3b}{a}x + \frac{2b^2}{a^2} = 0 \).
Isolate the x-terms:
\( x^2 - \frac{3b}{a}x = -\frac{2b^2}{a^2} \).
Add the square of half the coefficient of \( x \), which is \( \left(\frac{3b}{2a}\right)^2 = \frac{9b^2}{4a^2} \), to both sides:
\( x^2 - \frac{3b}{a}x + \frac{9b^2}{4a^2} = -\frac{2b^2}{a^2} + \frac{9b^2}{4a^2} \)
\( \implies \left(x - \frac{3b}{2a}\right)^2 = \frac{-8b^2 + 9b^2}{4a^2} \)
\( \implies \left(x - \frac{3b}{2a}\right)^2 = \frac{b^2}{4a^2} \).
Taking the square root of both sides:
\( x - \frac{3b}{2a} = \pm \frac{b}{2a} \).
So, \( x = \frac{3b}{2a} + \frac{b}{2a} = \frac{2b}{a} \) or \( x = \frac{3b}{2a} - \frac{b}{2a} = \frac{b}{a} \).

b) Rewrite the equation as:
\( (x^2 - 4ax + 4a^2) = b^2 \).
Recognize the left side as a perfect square:
\( (x - 2a)^2 = b^2 \).
Taking square roots on both sides:
\( x - 2a = \pm b \)
\( \implies x = 2a \pm b \).
So, the roots are \( x = 2a + b \) and \( x = 2a - b \).
In simple words: Completing the square means rearranging the equation into a perfect square bracket on one side so we can easily take the square root.

Exam Tip: When taking square roots on both sides during this process, never forget to write the \( \pm \) sign, as quadratic equations must have two solutions.

Question 5. Solve the following quadratic equations by factorization method:
a) \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
b) \( 9x^2 - 6ax + (a^2 - b^2) = 0 \)
Answer:
a) Let us rewrite the equation \( 3x^2 - 2\sqrt{6}x + 2 = 0 \) as a perfect square:
\( (\sqrt{3}x)^2 - 2(\sqrt{3}x)(\sqrt{2}) + (\sqrt{2})^2 = 0 \)
\( \implies (\sqrt{3}x - \sqrt{2})^2 = 0 \).
So, the factors are identical:
\( x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \).
The identical roots are \( \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}} \).

b) Split the middle term \( -6a \) into \( -3(a+b) - 3(a-b) \):
\( 9x^2 - 3(a+b)x - 3(a-b)x + (a+b)(a-b) = 0 \).
Factor by grouping:
\( 3x[3x - (a+b)] - (a-b)[3x - (a+b)] = 0 \)
\( \implies [3x - (a+b)][3x - (a-b)] = 0 \).
This gives the roots:
\( x = \frac{a+b}{3} \) and \( x = \frac{a-b}{3} \).
In simple words: Factorization involves breaking the middle term into two parts so we can group and pull out common factors.

Exam Tip: In factorization, the terms inside the parentheses must match exactly after grouping. If they do not, re-evaluate how you split the terms.

Question 6. write the nature of roots of quadratic equation: 4x2 + 4√3x + 3 = 0
Answer: Let us find the discriminant (\( D = b^2 - 4ac \)) for the given equation where \( a = 4 \), \( b = 4\sqrt{3} \), and \( c = 3 \):
\( D = (4\sqrt{3})^2 - 4(4)(3) \)
\( \implies D = 48 - 48 \)
\( \implies D = 0 \).
Since the discriminant is equal to zero, the quadratic equation has real and equal roots.
In simple words: When the discriminant calculation yields exactly zero, it means the equation has two roots that are the same real number.

Exam Tip: Clearly state the condition \( D = 0 \) before concluding that the roots are "real and equal" to secure full step-marks.

Question 7. Check whether the equation x3 – 4x2 + 1 = (x – 2)2 is quadratic or not
Answer: Let us expand the right-hand side of the equation:
\( (x-2)^2 = x^2 - 4x + 4 \).
Now write the original equation:
\( x^3 - 4x^2 + 1 = x^2 - 4x + 4 \).
Move all terms to one side:
\( x^3 - 4x^2 - x^2 + 4x + 1 - 4 = 0 \)
\( \implies x^3 - 5x^2 + 4x - 3 = 0 \).
Since the highest power of \( x \) in this equation is 3 (cubic), it is not a quadratic equation.
In simple words: A quadratic equation must have its highest power of x as exactly 2. Here, the power of 3 does not get cancelled out, so it is not quadratic.

Exam Tip: Do not just look at the raw equation and guess. Always simplify it completely by expanding and bringing all terms to one side before making a decision.

Question 8. Solve for x: 1 / (a + b + x) = 1/a + 1/b + 1/x, a + b ≠ 0
Answer: Let us rearrange the equation by moving \( \frac{1}{x} \) to the left side:
\( \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \).
Taking the common denominator on both sides:
\( \frac{x - (a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \)
\( \implies \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \).
Since \( a+b \neq 0 \), we can divide both sides by \( (a+b) \):
\( \frac{-1}{x(a+b+x)} = \frac{1}{ab} \)
\( \implies -ab = x(a+b+x) \)
\( \implies x^2 + (a+b)x + ab = 0 \)
\( \implies x(x+a) + b(x+a) = 0 \)
\( \implies (x+a)(x+b) = 0 \).
Therefore, \( x = -a \) or \( x = -b \).
In simple words: Grouping the x-terms on one side allows us to simplify the fractions and solve for the values of x easily.

Exam Tip: Grouping terms strategically (like bringing \( 1/x \) to the left) makes algebraic equations significantly easier to factorize.

Question 9. If p, q are the roots of the equation x2 – 5x + 4 =0, find the value of 1/p + 1/q - 2pq
Answer: For the quadratic equation \( x^2 - 5x + 4 = 0 \), comparing with \( ax^2 + bx + c = 0 \), we get \( a=1, b=-5, c=4 \).
The sum of roots is:
\( p + q = -\frac{b}{a} = 5 \).
The product of roots is:
\( pq = \frac{c}{a} = 4 \).
We need to evaluate:
\( \frac{1}{p} + \frac{1}{q} - 2pq = \frac{p+q}{pq} - 2pq \).
Substitute the values into the expression:
\( \frac{5}{4} - 2(4) = \frac{5}{4} - 8 = \frac{5 - 32}{4} = -\frac{27}{4} \).
The final value is \( -\frac{27}{4} \).
In simple words: We find the sum and product of the roots using simple formulas and then plug them into our expression.

Exam Tip: Avoid calculating individual values for p and q. Using sum and product formulas directly is much faster and less prone to errors.

Question 10. Solve for x: x / (x + 1) + (x + 1) / x = 34 / 15
Answer: Let us substitute \( \frac{x}{x+1} = y \). This implies \( \frac{x+1}{x} = \frac{1}{y} \).
The equation becomes:
\( y + \frac{1}{y} = \frac{34}{15} \)
\( \implies \frac{y^2 + 1}{y} = \frac{34}{15} \)
\( \implies 15(y^2 + 1) = 34y \)
\( \implies 15y^2 - 34y + 15 = 0 \).
Splitting the middle term of this quadratic equation:
\( 15y^2 - 25y - 9y + 15 = 0 \)
\( \implies 5y(3y - 5) - 3(3y - 5) = 0 \)
\( \implies (5y - 3)(3y - 5) = 0 \).
This gives \( y = \frac{3}{5} \) or \( y = \frac{5}{3} \).

Case 1: If \( y = \frac{3}{5} \), then:
\( \frac{x}{x+1} = \frac{3}{5} \)
\( \implies 5x = 3(x+1) \)
\( \implies 2x = 3 \)
\( \implies x = \frac{3}{2} \).

Case 2: If \( y = \frac{5}{3} \), then:
\( \frac{x}{x+1} = \frac{5}{3} \)
\( \implies 3x = 5(x+1) \)
\( \implies -2x = 5 \)
\( \implies x = -\frac{5}{2} \).
So, the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{5}{2} \).
In simple words: We make a substitution to turn a complicated fraction equation into a simpler quadratic equation first.

Exam Tip: Substituting complex repeating terms with a single letter like 'y' makes the problem much cleaner and easier to manage.

Question 11. Solve for x: 1 / (x – 3) - 1 / (x + 5) = 1 / 6
Answer: Let us take the common denominator on the left side:
\( \frac{(x+5) - (x-3)}{(x-3)(x+5)} = \frac{1}{6} \)
\( \implies \frac{8}{x^2 + 2x - 15} = \frac{1}{6} \).
Cross-multiplying gives:
\( 48 = x^2 + 2x - 15 \)
\( \implies x^2 + 2x - 63 = 0 \).
Factoring the quadratic equation:
\( (x+9)(x-7) = 0 \).
This yields the roots \( x = 7 \) or \( x = -9 \).
In simple words: By simplifying the fractions on the left, we get a standard quadratic equation that we can easily solve.

Exam Tip: Be careful with signs when subtracting in the numerator: \( (x+5) - (x-3) \) simplifies to \( 8 \), not \( 2 \).

Question 12. If one root of a quadratic equation 3x2+ PX + 4 = 0 is 2/3, find the value of p
Answer: Since \( x = \frac{2}{3} \) is a root of the given equation, it must satisfy it. Substituting this value gives:
\( 3\left(\frac{2}{3}\right)^2 + P\left(\frac{2}{3}\right) + 4 = 0 \)
\( \implies 3\left(\frac{4}{9}\right) + \frac{2P}{3} + 4 = 0 \)
\( \implies \frac{4}{3} + \frac{2P}{3} + 4 = 0 \).
Multiplying the entire equation by 3 to clear denominators:
\( 4 + 2P + 12 = 0 \)
\( \implies 2P + 16 = 0 \)
\( \implies P = -8 \).
The value of P is \( -8 \).
In simple words: Since we know one of the answers for x, we can plug it back into the equation to figure out the missing number P.

Exam Tip: Plugging the given root directly into the equation is the most direct way to solve for an unknown constant.

Question 13. Solve for x: 2 (2x-1)/(x+3) – 3 (x+3)/(2x-1) = 5
Answer: Let us substitute \( \frac{2x-1}{x+3} = y \). This gives its reciprocal as \( \frac{x+3}{2x-1} = \frac{1}{y} \).
Substituting this into the equation:
\( 2y - \frac{3}{y} = 5 \)
\( \implies 2y^2 - 3 = 5y \)
\( \implies 2y^2 - 5y - 3 = 0 \).
Factoring the quadratic expression:
\( 2y^2 - 6y + y - 3 = 0 \)
\( \implies 2y(y - 3) + 1(y - 3) = 0 \)
\( \implies (2y+1)(y-3) = 0 \).
So, \( y = 3 \) or \( y = -\frac{1}{2} \).

Case 1: If \( y = 3 \), then:
\( \frac{2x-1}{x+3} = 3 \)
\( \implies 2x - 1 = 3x + 9 \)
\( \implies x = -10 \).

Case 2: If \( y = -\frac{1}{2} \), then:
\( \frac{2x-1}{x+3} = -\frac{1}{2} \)
\( \implies 2(2x-1) = -(x+3) \)
\( \implies 4x - 2 = -x - 3 \)
\( \implies 5x = -1 \)
\( \implies x = -\frac{1}{5} \).
Thus, \( x = -10 \) or \( x = -\frac{1}{5} \).
In simple words: Substitution transforms a complicated fractional equation into an easy-to-solve quadratic equation.

Exam Tip: When using the substitution method, remember to perform the back-substitution step to solve for the original variable 'x'.

Question 14. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: Let the two consecutive odd numbers be \( x \) and \( x+2 \).
According to the problem:
\( x^2 + (x+2)^2 = 394 \)
\( \implies x^2 + x^2 + 4x + 4 = 394 \)
\( \implies 2x^2 + 4x - 390 = 0 \).
Dividing by 2:
\( x^2 + 2x - 195 = 0 \).
Factoring the quadratic expression:
\( (x+15)(x-13) = 0 \).
This gives \( x = 13 \) (ignoring negative integer \( -15 \) for standard positive odd numbers).
The two consecutive odd numbers are 13 and 15.
In simple words: We write consecutive odd numbers as x and x+2, set up our equation based on their squares, and solve for x.

Exam Tip: If the question does not specify "positive" integers, mention that \( -15 \) and \( -13 \) are also a valid mathematical pair to earn full credit.

Question 15. The sum of the areas of two squares is 640 m2. If the difference in their perimeter is 64m .Find the sides of the two squares
Answer: Let the sides of the two squares be \( x \) meters and \( y \) meters, where \( x > y \).
The sum of their areas is given by:
\( x^2 + y^2 = 640 \).
The difference in their perimeters is:
\( 4x - 4y = 64 \)
\( \implies x - y = 16 \)
\( \implies x = y + 16 \).
Substitute this expression for \( x \) into the area equation:
\( (y+16)^2 + y^2 = 640 \)
\( \implies y^2 + 32y + 256 + y^2 = 640 \)
\( \implies 2y^2 + 32y - 384 = 0 \).
Dividing by 2:
\( y^2 + 16y - 192 = 0 \).
Factoring gives:
\( (y+24)(y-8) = 0 \).
Since a side length cannot be negative, we reject \( y = -24 \). Therefore, \( y = 8 \) meters.
Thus, the side of the larger square is \( x = 8 + 16 = 24 \) meters.
The sides of the squares are 8m and 24m.
In simple words: We find a relationship between the two sides using the perimeter difference and use it to solve the area equation.

Exam Tip: State clearly that a side length cannot be negative when discarding the negative solution during calculation.

Question 16. The difference of two numbers is 4. If the difference of their reciprocals is 4/21, find the numbers
Answer: Let the two numbers be \( x \) and \( y \), with \( x > y \).
The difference of the numbers is:
\( x - y = 4 \)
\( \implies x = y + 4 \).
The difference of their reciprocals is:
\( \frac{1}{y} - \frac{1}{x} = \frac{4}{21} \).
Substitute \( x = y+4 \):
\( \frac{1}{y} - \frac{1}{y+4} = \frac{4}{21} \)
\( \implies \frac{(y+4) - y}{y(y+4)} = \frac{4}{21} \)
\( \implies \frac{4}{y^2 + 4y} = \frac{4}{21} \).
Dividing by 4 on both sides:
\( y^2 + 4y = 21 \)
\( \implies y^2 + 4y - 21 = 0 \)
\( \implies (y+7)(y-3) = 0 \).
This gives \( y = 3 \) or \( y = -7 \).
If \( y = 3 \), then \( x = 7 \). (Numbers are 3 and 7).
If \( y = -7 \), then \( x = -3 \). (Numbers are -3 and -7).
The numbers are 3 and 7.
In simple words: The smaller number has the larger reciprocal, so we subtract in that order to set up the equation correctly.

Exam Tip: Remember that \( 1/y \) is larger than \( 1/x \) when \( x > y \) (for positive numbers). Writing the subtraction in the wrong order is a very common mistake.

Question 17. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500km away in time it had to Increase the speed by 250 km/h from the usual speed. Find its usual speed
Answer: Let the usual speed of the plane be \( v \) km/h.
The distance to the destination is 1500 km.
The usual time taken is \( \frac{1500}{v} \) hours.
The new speed is \( v + 250 \) km/h, and the new time is \( \frac{1500}{v+250} \) hours.
The difference in time is 30 minutes, which is \( \frac{1}{2} \) hour:
\( \frac{1500}{v} - \frac{1500}{v+250} = \frac{1}{2} \)
\( \implies 1500\left(\frac{(v+250) - v}{v(v+250)}\right) = \frac{1}{2} \)
\( \implies \frac{1500 \times 250}{v^2 + 250v} = \frac{1}{2} \).
Cross-multiplying gives:
\( v^2 + 250v = 750000 \)
\( \implies v^2 + 250v - 750000 = 0 \)
\( \implies (v + 1000)(v - 750) = 0 \).
Since speed cannot be negative, we discard \( v = -1000 \).
The usual speed of the plane is 750 km/h.
In simple words: To make up for the 30-minute delay, the plane flew faster. We use the time difference to set up our equation.

Exam Tip: Convert time from minutes to hours first before using it in equations where speed is in km/h.

Question 18. The hypotenuse of a grassy land in the shape of a right triangle is 1m more than twice the shortest side. If the third side is 7m More than the shortest side find the sides of grassy land
Answer: Let the shortest side of the right-angled triangular land be \( x \) meters.
Then, the hypotenuse is \( 2x + 1 \) meters, and the third side is \( x + 7 \) meters.
Using Pythagoras' theorem:
\( (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \)
\( \implies (2x+1)^2 = x^2 + (x+7)^2 \)
\( \implies 4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49 \)
\( \implies 2x^2 - 10x - 48 = 0 \).
Dividing by 2:
\( x^2 - 5x - 24 = 0 \)
\( \implies (x-8)(x+3) = 0 \).
Since a length cannot be negative, we reject \( x = -3 \). Thus, \( x = 8 \) meters.
The sides are:
Shortest side = 8m
Third side = \( 8 + 7 = 15 \)m
Hypotenuse = \( 2(8) + 1 = 17 \)m.
The sides of the grassy land are 8m and 15m.
In simple words: We write all sides in terms of the shortest side and then use the Pythagoras formula to solve for x.

Exam Tip: Be accurate when expanding terms like \( (2x+1)^2 \) and do not forget the middle term \( 4x \).

Question 19. Find two consecutive numbers, whose squares have the sum 85.
Answer: Let the two consecutive numbers be \( n \) and \( n+1 \).
According to the given condition:
\( n^2 + (n+1)^2 = 85 \)
\( \implies n^2 + n^2 + 2n + 1 = 85 \)
\( \implies 2n^2 + 2n - 84 = 0 \).
Dividing by 2:
\( n^2 + n - 42 = 0 \)
\( \implies (n+7)(n-6) = 0 \).
This gives \( n = 6 \) or \( n = -7 \).
If we consider positive integers, the numbers are 6 and 7.
In simple words: We write the consecutive numbers as n and n+1, square them, add them together to equal 85, and solve.

Exam Tip: State both pairs \( (6, 7) \) and \( (-7, -6) \) in your final answer unless the question specifies positive integers.

Question 20. The sum of the reciprocals of rehmans age 3years ago and 5years from now is 1/3, find his present age
Answer: Let Rehman's present age be \( x \) years.
His age 3 years ago was \( x - 3 \) years, and his age 5 years from now will be \( x + 5 \) years.
According to the question:
\( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \)
\( \implies \frac{(x+5) + (x-3)}{(x-3)(x+5)} = \frac{1}{3} \)
\( \implies \frac{2x+2}{x^2 + 2x - 15} = \frac{1}{3} \).
Cross-multiplying yields:
\( 3(2x + 2) = x^2 + 2x - 15 \)
\( \implies 6x + 6 = x^2 + 2x - 15 \)
\( \implies x^2 - 4x - 21 = 0 \)
\( \implies (x-7)(x+3) = 0 \).
Since age cannot be negative, we discard \( x = -3 \).
Rehman's present age is 7 years.
In simple words: We write down Rehman's age in the past and future as fractions, add them to equal 1/3, and solve for his current age.

Exam Tip: Always state that age cannot be a negative value when rejecting the negative root in age-related problems.

Question 21. A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number
Answer: Let the natural number be \( n \).
According to the given condition:
\( n + 12 = 160 \times \frac{1}{n} \)
\( \implies n(n + 12) = 160 \)
\( \implies n^2 + 12n - 160 = 0 \)
\( \implies (n+20)(n-8) = 0 \).
Since \( n \) must be a natural number, \( n = 8 \) (rejecting the negative root \( n = -20 \)).
The required number is 8.
In simple words: We set up a simple equation where adding 12 to a number is the same as dividing 160 by that same number, then solve it.

Exam Tip: Remind yourself that "natural numbers" must be positive integers, which helps you easily rule out any negative values.

Question 22. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together Can finish it in 4 days; find the time taken by B to finish the work
Answer: Let B take \( x \) days to finish the work alone.
Then A takes \( x - 6 \) days to finish the work alone.
The work done by B in one day is \( \frac{1}{x} \), and by A in one day is \( \frac{1}{x-6} \).
Together, they complete the work in 4 days, so they complete \( \frac{1}{4} \) of the work in one day:
\( \frac{1}{x} + \frac{1}{x-6} = \frac{1}{4} \)
\( \implies \frac{(x-6) + x}{x(x-6)} = \frac{1}{4} \)
\( \implies \frac{2x - 6}{x^2 - 6x} = \frac{1}{4} \).
Cross-multiplying gives:
\( 4(2x - 6) = x^2 - 6x \)
\( \implies 8x - 24 = x^2 - 6x \)
\( \implies x^2 - 14x + 24 = 0 \)
\( \implies (x - 12)(x - 2) = 0 \).
If \( x = 2 \), then A would take \( 2 - 6 = -4 \) days, which is impossible. Thus, we select \( x = 12 \).
B takes 12 days to complete the work alone.
In simple words: We calculate how much work each person does in a single day and add them together to solve the problem.

Exam Tip: Be sure to verify both roots. Here, \( x = 2 \) makes A's work days negative, so it must be discarded with a brief explanation.

Question 23. The speed of a boat in still water is 15 km/hr. It can go 30km upstream and return downstream to the original point in 4hrs 30min. Find out the speed of the stream
Answer: Let the speed of the stream be \( s \) km/hr.
The speed of the boat upstream is \( 15 - s \) km/hr, and its speed downstream is \( 15 + s \) km/hr.
The total time taken for the trip is 4 hours and 30 minutes, which is \( 4.5 \) hours or \( \frac{9}{2} \) hours:
\( \frac{30}{15-s} + \frac{30}{15+s} = \frac{9}{2} \)
\( \implies 30\left(\frac{(15+s) + (15-s)}{(15-s)(15+s)}\right) = \frac{9}{2} \)
\( \implies 30\left(\frac{30}{225 - s^2}\right) = \frac{9}{2} \)
\( \implies \frac{900}{225 - s^2} = \frac{9}{2} \).
Cross-multiplying gives:
\( 1800 = 9(225 - s^2) \)
\( \implies 200 = 225 - s^2 \)
\( \implies s^2 = 25 \)
\( \implies s = 5 \) km/hr (since speed cannot be negative).
The speed of the stream is 5 km/hr.
In simple words: The stream slows the boat down going up but speeds it up coming back. We use this to set up a time equation.

Exam Tip: Using the identity \( (a-b)(a+b) = a^2 - b^2 \) in the denominator simplifies the calculation significantly.

Question 24. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number
Answer: Let the tens digit be \( x \) and the units digit be \( y \).
The product of the digits is:
\( xy = 18 \)
\( \implies y = \frac{18}{x} \).
The original number is \( 10x + y \).
When we reverse the digits, the new number is \( 10y + x \).
According to the given condition:
\( (10x + y) - 63 = 10y + x \)
\( \implies 9x - 9y = 63 \)
\( \implies x - y = 7 \).
Substitute \( y = \frac{18}{x} \) into this equation:
\( x - \frac{18}{x} = 7 \)
\( \implies x^2 - 7x - 18 = 0 \)
\( \implies (x - 9)(x + 2) = 0 \).
Since \( x \) is a single-digit positive number, we choose \( x = 9 \).
This gives \( y = \frac{18}{9} = 2 \).
The original number is 92.
In simple words: We write the tens and units digits as x and y. Solving the equations shows us the digits are 9 and 2, making the number 92.

Exam Tip: A two-digit number must always be represented algebraically as \( 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit.

Question 25. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20years. Four years ago, the product of their ages was 48.
Answer: Let the present age of one friend be \( x \) years.
Then, the present age of the other friend is \( 20 - x \) years.
Four years ago, their ages were \( x - 4 \) years and \( (20 - x) - 4 = 16 - x \) years.
The product of their ages four years ago was 48:
\( (x - 4)(16 - x) = 48 \)
\( \implies 16x - x^2 - 64 + 4x = 48 \)
\( \implies -x^2 + 20x - 64 - 48 = 0 \)
\( \implies x^2 - 20x + 112 = 0 \).
Let us check the discriminant \( D = b^2 - 4ac \):
\( D = (-20)^2 - 4(1)(112) \)
\( \implies D = 400 - 448 \)
\( \implies D = -48 \).
Since the discriminant is negative (\( D < 0 \)), there are no real roots. Therefore, the given situation is not possible.
In simple words: We check if the math works by calculating the discriminant. Since it is negative, this scenario cannot happen.

Exam Tip: Whenever asked "Is this situation possible?", calculate the discriminant. If \( D < 0 \), the situation is impossible.

Question 26. If the roots of the equation (b – c) x2 + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c
Answer: Let us substitute \( x = 1 \) into the quadratic expression \( (b-c)x^2 + (c-a)x + (a-b) \):
\( (b-c)(1)^2 + (c-a)(1) + (a-b) = b - c + c - a + a - b = 0 \).
Since \( x = 1 \) satisfies the equation, it is one of its roots.
Because the problem states that the roots are equal, both roots must be equal to 1.
The product of the roots for a quadratic equation \( Ax^2 + Bx + C = 0 \) is \( \frac{C}{A} \):
\( \text{Product of roots} = \frac{a-b}{b-c} \).
Since both roots are equal to 1, the product of the roots is \( 1 \times 1 = 1 \):
\( \frac{a-b}{b-c} = 1 \)
\( \implies a - b = b - c \)
\( \implies a + c = 2b \).
Hence proved.
In simple words: Since putting 1 in place of x makes the equation zero, 1 is a root. Since both roots are equal, their product is also 1, leading directly to the proof.

Exam Tip: Recognizing that the sum of coefficients \( (b-c) + (c-a) + (a-b) = 0 \) instantly reveals that \( x=1 \) is a root is a very elegant shortcut for proofs.

Question 27. A peacock is sitting on the top of a pillar, which is 9m high. From a point 27m away from the bottom of the pillar, a snake is Coming to its hole at the base of the pillar .Seeing the snake the peacock pounces on it. If their speeds are equal, at what Distance from the hole is the snake caught?
Answer: Let \( AB \) be the pillar of height 9m with the peacock at the top \( A \). Let the hole be at the base \( B \).
The snake is initially at point \( C \), which is 27m from the base \( B \), so \( BC = 27 \)m.
Let the peacock catch the snake at point \( D \) at a distance \( x \) meters from the hole \( B \), so \( BD = x \).
The distance covered by the snake is \( CD = 27 - x \).
Since their speeds and times are equal, they must travel the exact same distance. Thus, the peacock's distance \( AD = CD = 27 - x \).
In the right-angled triangle \( ABD \), using Pythagoras' theorem:
\( AD^2 = AB^2 + BD^2 \)
\( \implies (27 - x)^2 = 9^2 + x^2 \)
\( \implies 729 - 54x + x^2 = 81 + x^2 \)
\( \implies 54x = 729 - 81 \)
\( \implies 54x = 648 \)
\( \implies x = 12 \) meters.
The snake is caught at a distance of 12 meters from the hole.
In simple words: Because they run at the same speed, the distance the peacock flies is the same as the distance the snake slithers. We use Pythagoras to find where they meet.

Exam Tip: Draw a neat right-angled triangle diagram to help visualize and set up the Pythagoras equation correctly.

Question 28. Two train leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5km/hr faster than the second train. If after two hours, they are 50km apart, find the average speed of each train
Answer: Let the speed of the second train (traveling north) be \( v \) km/hr.
Then, the speed of the first train (traveling west) is \( v + 5 \) km/hr.
In 2 hours:
Distance traveled by the second train (north) = \( 2v \) km.
Distance traveled by the first train (west) = \( 2(v + 5) = 2v + 10 \) km.
Since north and west directions are perpendicular, the distance between them forms a right-angled triangle with hypotenuse 50 km.
Using Pythagoras' theorem:
\( (2v)^2 + (2v+10)^2 = 50^2 \)
\( \implies 4v^2 + 4v^2 + 40v + 100 = 2500 \)
\( \implies 8v^2 + 40v - 2400 = 0 \).
Dividing by 8:
\( v^2 + 5v - 300 = 0 \)
\( \implies (v + 20)(v - 15) = 0 \).
Since speed cannot be negative, we reject \( v = -20 \). Thus, \( v = 15 \) km/hr.
Speed of the second train = 15 km/hr.
Speed of the first train = \( 15 + 5 = 20 \) km/hr.
In simple words: One train goes north and one goes west, forming a right triangle. We use Pythagoras to find their speeds.

Exam Tip: Don't forget to multiply the speeds by the time (2 hours) to get the distance sides before using Pythagoras' theorem.

Question 29. A journey of 192km from station A to station B takes 2hours less by a superfast train that by an ordinary train If the average Speed of the slower train is 16km/hr than that of the faster train, determine their average speed
Answer: Let the speed of the slower train be \( v \) km/hr. This means the speed of the faster train is \( v + 16 \) km/hr.
The total distance is 192 km.
The time taken by the slower train is \( \frac{192}{v} \) hours, and by the faster train is \( \frac{192}{v+16} \) hours.
The difference in travel times is 2 hours:
\( \frac{192}{v} - \frac{192}{v+16} = 2 \)
\( \implies 192 \left(\frac{(v+16) - v}{v(v+16)}\right) = 2 \)
\( \implies \frac{192 \times 16}{v^2 + 16v} = 2 \).
Dividing by 2 on both sides:
\( v^2 + 16v = 1536 \)
\( \implies v^2 + 16v - 1536 = 0 \)
\( \implies (v + 48)(v - 32) = 0 \).
Since speed cannot be negative, we choose \( v = 32 \) km/hr.
The average speed of the slower train is 32 km/hr, and the speed of the faster train is \( 32 + 16 = 48 \) km/hr.
In simple words: Slower trains take longer to cover the same distance. We use this time difference to solve for their speeds.

Exam Tip: Be consistent with your variables: if you define \( v \) as the slower speed, the faster is \( v + 16 \). If you define \( v \) as the faster speed, the slower is \( v - 16 \).

Question 30. Two water taps together can fill a tank in 6 hrs. The tap of larger diameter takes 9 hrs less than the smaller one to fill the Tank separately. Find the time in which each tap can separately fill the tank
Answer: Let the smaller tap take \( x \) hours to fill the tank alone.
Then, the larger tap takes \( x - 9 \) hours to fill the tank alone.
In 1 hour, the smaller tap fills \( \frac{1}{x} \) of the tank, and the larger tap fills \( \frac{1}{x-9} \) of the tank.
Together, they fill the tank in 6 hours, so in 1 hour they fill \( \frac{1}{6} \) of the tank:
\( \frac{1}{x} + \frac{1}{x-9} = \frac{1}{6} \)
\( \implies \frac{(x-9) + x}{x(x-9)} = \frac{1}{6} \)
\( \implies \frac{2x - 9}{x^2 - 9x} = \frac{1}{6} \).
Cross-multiplying gives:
\( 6(2x - 9) = x^2 - 9x \)
\( \implies 12x - 54 = x^2 - 9x \)
\( \implies x^2 - 21x + 54 = 0 \)
\( \implies (x - 18)(x - 3) = 0 \).
If \( x = 3 \), then the larger tap would take \( 3 - 9 = -6 \) hours, which is impossible. Thus, \( x = 18 \).
The smaller tap takes 18 hours to fill the tank separately, and the larger tap takes \( 18 - 9 = 9 \) hours.
In simple words: We calculate how much of the tank each tap fills in one hour, and add those fractions to find the answer.

Exam Tip: Always verify that both calculated times are positive and realistic. A time of 3 hours for the small tap gives a negative time for the larger tap, so it must be discarded.

Question 31. The product of Bilals age five years ago and eight years later is 198. Find his present age
Answer: Let Bilal's present age be \( x \) years.
His age 5 years ago was \( x - 5 \) years, and his age 8 years later will be \( x + 8 \) years.
According to the problem:
\( (x - 5)(x + 8) = 198 \)
\( \implies x^2 + 3x - 40 = 198 \)
\( \implies x^2 + 3x - 238 = 0 \)
\( \implies (x + 17)(x - 14) = 0 \).
Since age cannot be negative, we reject \( x = -17 \).
Bilal's present age is 14 years.
In simple words: We write formulas for Bilal's past and future age, multiply them to equal 198, and solve the equation.

Exam Tip: Practice finding factors of larger numbers like 238 quickly using prime factorization (e.g., \( 238 = 2 \times 7 \times 17 = 14 \times 17 \)).

Question 32. The age of father is equal to the square of the age of his son. The sum of the age of father and five times the age of the son Is 6 years. Find their ages
Answer: Let the present age of the son be \( y \) years.
Then, the present age of the father is \( y^2 \) years.
According to standard mathematical context (where a realistic sum of their ages with the son being 6 and father being 36 is 66, correcting a common typo of 6 for 66 in the textbook text):
\( y^2 + 5y = 66 \)
\( \implies y^2 + 5y - 66 = 0 \)
\( \implies (y + 11)(y - 6) = 0 \).
Since age cannot be negative, we reject \( y = -11 \). Thus, the son's age is \( y = 6 \) years.
The father's age is \( y^2 = 6^2 = 36 \) years.
The son's age is 6 years, and the father's age is 36 years.
In simple words: By correcting a minor printing typo from 6 to 66, we find that the son is 6 years old and the father is 36.

Exam Tip: When you run into a printing typo in an exam, write down your assumption clearly (e.g. "assuming sum is 66 to match realistic ages") and carry on with your solution.

Question 33. A train travels 180km at a uniform speed. If the speed had been 9 km/ hr more, it would have taken 1 hour less for the same Journey. Find the speed of the train.
Answer: Let the uniform speed of the train be \( v \) km/hr.
The total journey distance is 180 km.
The time taken at uniform speed is \( \frac{180}{v} \) hours, and at the increased speed is \( \frac{180}{v+9} \) hours.
The difference in times is 1 hour:
\( \frac{180}{v} - \frac{180}{v+9} = 1 \)
\( \implies 180\left(\frac{(v+9) - v}{v(v+9)}\right) = 1 \)
\( \implies \frac{180 \times 9}{v^2 + 9v} = 1 \).
Cross-multiplying gives:
\( v^2 + 9v = 1620 \)
\( \implies v^2 + 9v - 1620 = 0 \)
\( \implies (v + 45)(v - 36) = 0 \).
Since speed cannot be negative, we choose \( v = 36 \) km/hr.
The speed of the train is 36 km/hr.
In simple words: Increasing the speed of the train cuts down travel time. We use this relation to solve for the speed.

Exam Tip: Be sure to keep units consistent. Speed is in km/hr, distance in km, and time in hours.

Question 34. Rs 1200 were distributed equally among certain number of students. Had there been 8 more students, each would have Received Rs 5 less. Find the number of students.
Answer: Let the initial number of students be \( x \).
The share of each student initially is \( \frac{1200}{x} \) Rupees.
If there are 8 more students, the share of each student becomes \( \frac{1200}{x+8} \) Rupees.
According to the problem:
\( \frac{1200}{x} - \frac{1200}{x+8} = 5 \)
\( \implies 1200\left(\frac{(x+8) - x}{x(x+8)}\right) = 5 \)
\( \implies \frac{1200 \times 8}{x^2 + 8x} = 5 \).
Dividing by 5 on both sides:
\( \frac{240 \times 8}{x^2 + 8x} = 1 \)
\( \implies x^2 + 8x = 1920 \)
\( \implies x^2 + 8x - 1920 = 0 \)
\( \implies (x + 48)(x - 40) = 0 \).
Since the number of students cannot be negative, we reject \( x = -48 \).
The number of students is 40.
In simple words: When more students share the money, each gets a smaller share. We use this difference to find the count of students.

Exam Tip: Dividing both sides by a common factor (like dividing by 5 here) makes the numbers smaller and much easier to calculate.

Topic: Arithmetic Progressions

Question 1. For what value of p, are 2p-1, 7 and 3p three consecutive terms of an A.P?
Answer: Since \( 2p-1 \), \( 7 \), and \( 3p \) are in Arithmetic Progression, their consecutive differences must be equal:
\( 7 - (2p - 1) = 3p - 7 \)
\( \implies 7 - 2p + 1 = 3p - 7 \)
\( \implies 8 - 2p = 3p - 7 \)
\( \implies 5p = 15 \)
\( \implies p = 3 \).
The value of p is 3.
In simple words: For three numbers to form an AP, the middle number minus the first must equal the third number minus the middle.

Exam Tip: Use the property \( 2b = a + c \) for three consecutive terms \( a, b, c \) in an AP to write the equation directly.

Question 2. Find the value of k, so that 3k + 7, 2k +5, 2k + 7 are in A.P
Answer: Since the terms \( 3k + 7 \), \( 2k + 5 \), and \( 2k + 7 \) are in AP, the common difference is constant:
\( (2k + 5) - (3k + 7) = (2k + 7) - (2k + 5) \)
\( \implies 2k + 5 - 3k - 7 = 2k + 7 - 2k - 5 \)
\( \implies -k - 2 = 2 \)
\( \implies -k = 4 \)
\( \implies k = -4 \).
The value of k is \( -4 \).
In simple words: We find k by ensuring the step size from the first to the second term is equal to the step size from the second to the third.

Exam Tip: Be careful with signs when subtracting binomials: \( (2k+5) - (3k+7) \) simplifies to \( -k-2 \).

Question 3. Find the 15th term from the end of the A.P: 3, 5, 7,………, 201
Answer: For the given AP: \( 3, 5, 7, \dots, 201 \), we have:
First term \( a = 3 \), common difference \( d = 2 \), and last term \( l = 201 \).
The formula for the \( n \)-th term from the end is:
\( a_n = l - (n - 1)d \).
For the 15th term from the end:
\( a_{15} = 201 - (15 - 1)2 \)
\( \implies a_{15} = 201 - 14(2) \)
\( \implies a_{15} = 201 - 28 \)
\( \implies a_{15} = 173 \).
The 15th term from the end is 173.
In simple words: To find a term from the end, we start at the last term and jump backwards using the common difference.

Exam Tip: Alternatively, you can reverse the AP so that it starts at 201 and goes down by 2 each time, then find the 15th term normally.

Question 4. Find the 11th term from the end of the A.P: 10, 7, 4,……, - 62
Answer: For the given AP: \( 10, 7, 4, \dots, -62 \), we have:
First term \( a = 10 \), common difference \( d = -3 \), and last term \( l = -62 \).
The formula for the \( n \)-th term from the end is:
\( a_n = l - (n - 1)d \).
For the 11th term from the end:
\( a_{11} = -62 - (11 - 1)(-3) \)
\( \implies a_{11} = -62 - 10(-3) \)
\( \implies a_{11} = -62 + 30 \)
\( \implies a_{11} = -32 \).
The 11th term from the end is \( -32 \).
In simple words: We calculate backwards starting from \( -62 \), adding 3 for each of the 10 steps to find the 11th term.

Exam Tip: Pay close attention to double negatives like \( - (10)(-3) \) which simplifies to \( +30 \).

Question 5. If Sn, the sum of first n terms of an A.P is given by Sn = 3n2 – 4n, then find its nth term
Answer: The sum of the first \( n \) terms is \( S_n = 3n^2 - 4n \).
The \( n \)-th term \( a_n \) can be found using the formula \( a_n = S_n - S_{n-1} \).
First, find \( S_{n-1} \):
\( S_{n-1} = 3(n-1)^2 - 4(n-1) \)
\( \implies S_{n-1} = 3(n^2 - 2n + 1) - 4n + 4 \)
\( \implies S_{n-1} = 3n^2 - 6n + 3 - 4n + 4 \)
\( \implies S_{n-1} = 3n^2 - 10n + 7 \).
Now compute \( a_n \):
\( a_n = (3n^2 - 4n) - (3n^2 - 10n + 7) \)
\( \implies a_n = 3n^2 - 4n - 3n^2 + 10n - 7 \)
\( \implies a_n = 6n - 7 \).
The \( n \)-th term is \( 6n - 7 \).
In simple words: Subtracting the sum of \( n-1 \) terms from the sum of \( n \) terms leaves us with just the last (\( n \)-th) term.

Exam Tip: A quicker check is to find \( S_1 = -1 \) and \( S_2 = 4 \), which gives \( a_1 = -1 \), \( a_2 = 5 \), so \( d = 6 \). This verifies \( a_n = a_1 + (n-1)d = -1 + 6(n-1) = 6n - 7 \).

Question 6. The sum of n terms of an A.P. is 3n2 + 5n. Find the A.P. Hence, find its 16th term
Answer: Let the sum of \( n \) terms be \( S_n = 3n^2 + 5n \).
The first term of the AP is:
\( a_1 = S_1 = 3(1)^2 + 5(1) = 8 \).
The sum of the first two terms is:
\( S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 \).
The second term is:
\( a_2 = S_2 - S_1 = 22 - 8 = 14 \).
The common difference is:
\( d = a_2 - a_1 = 14 - 8 = 6 \).
The arithmetic progression is \( 8, 14, 20, \dots \)
The \( n \)-th term of this AP is:
\( a_n = a + (n - 1)d = 8 + (n - 1)6 = 6n + 2 \).
To find the 16th term:
\( a_{16} = 6(16) + 2 = 96 + 2 = 98 \).
The AP is \( 8, 14, 20, \dots \) and its 16th term is 98.
In simple words: We find the first term and second term by using the sum formula, which gives us the starting number and the gap size.

Exam Tip: Remember that \( S_1 \) is always equal to the first term \( a \). This is a crucial starting point for sum-based AP questions.

Question 7. In the following A.P. find the missing term: *, 38, *, *, *, -22
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
We are given:
Second term \( a_2 = a + d = 38 \)
Sixth term \( a_6 = a + 5d = -22 \).
Subtracting the first equation from the second:
\( (a + 5d) - (a + d) = -22 - 38 \)
\( \implies 4d = -60 \)
\( \implies d = -15 \).
Substitute \( d = -15 \) back into the first equation:
\( a + (-15) = 38 \)
\( \implies a = 53 \).
Now we find the remaining terms:
First term = 53
Third term = \( 38 + (-15) = 23 \)
Fourth term = \( 23 + (-15) = 8 \)
Fifth term = \( 8 + (-15) = -7 \).
The complete AP is \( 53, 38, 23, 8, -7, -22 \).
In simple words: By using the two given terms, we set up simultaneous equations to find the start value and step size.

Exam Tip: Since the terms are decreasing, make sure your calculated common difference \( d \) is a negative number.

Question 8. Find the sum of all natural numbers less than 100 which are divisible by 6
Answer: The natural numbers less than 100 that are divisible by 6 are:
\( 6, 12, 18, \dots, 96 \).
This forms an AP with:
First term \( a = 6 \), common difference \( d = 6 \), and last term \( a_n = 96 \).
Let us find the number of terms \( n \):
\( a_n = a + (n - 1)d \)
\( \implies 96 = 6 + (n - 1)6 \)
\( \implies 90 = (n - 1)6 \)
\( \implies n - 1 = 15 \)
\( \implies n = 16 \).
Now, find the sum using the formula \( S_n = \frac{n}{2}(a + a_n) \):
\( S_{16} = \frac{16}{2}(6 + 96) \)
\( \implies S_{16} = 8(102) \)
\( \implies S_{16} = 816 \).
The sum of these numbers is 816.
In simple words: We list out the multiples of 6 up to 100, count how many there are, and use the sum formula to add them up quickly.

Exam Tip: Use the formula \( S_n = \frac{n}{2}(a + l) \) instead of \( S_n = \frac{n}{2}[2a + (n-1)d] \) when the last term is already known to save calculation time.

Question 9. Find the sum of 3 digit numbers which are not divisible by 7
Answer: The three-digit numbers range from 100 to 999. The total number of terms is \( 900 \).
The sum of all three-digit numbers is:
\( S_{\text{all}} = \frac{900}{2}(100 + 999) = 450 \times 1099 = 494,550 \).
Now, we find the sum of three-digit numbers that are divisible by 7:
The first three-digit multiple of 7 is 105, and the last is 994.
Using \( a_n = a + (n-1)d \) to find the number of terms \( m \):
\( 994 = 105 + (m - 1)7 \)
\( \implies 889 = (m - 1)7 \)
\( \implies m - 1 = 127 \)
\( \implies m = 128 \).
The sum of these multiples of 7 is:
\( S_7 = \frac{128}{2}(105 + 994) = 64 \times 1099 = 70,336 \).
The sum of three-digit numbers not divisible by 7 is:
\( S_{\text{required}} = S_{\text{all}} - S_7 = 494,550 - 70,336 = 424,214 \).
In simple words: We calculate the sum of all 3-digit numbers and subtract the sum of those that can be divided by 7.

Exam Tip: Breaking a problem into "total minus unwanted parts" is a powerful strategy for counting or summation questions.

Question 10. Find the sum of all three digit numbers which leave the remainder 3 when divided by 5
Answer: Three-digit numbers range from 100 to 999.
The first three-digit number leaving a remainder of 3 when divided by 5 is 103.
The last such number is 998.
These numbers form an AP: \( 103, 108, 113, \dots, 998 \) with first term \( a = 103 \) and common difference \( d = 5 \).
Let us find the number of terms \( n \):
\( 998 = 103 + (n - 1)5 \)
\( \implies 895 = (n - 1)5 \)
\( \implies n - 1 = 179 \)
\( \implies n = 180 \).
Now, find the sum:
\( S_{180} = \frac{180}{2}(103 + 998) \)
\( \implies S_{180} = 90(1101) \)
\( \implies S_{180} = 99,090 \).
The required sum is 99,090.
In simple words: We find the first and last three-digit numbers that fit the rule, count them, and sum them up using the AP formula.

Exam Tip: To quickly find the first term, take the smallest 3-digit number (100), divide by 5 (remainder 0), and add 3 to get 103.

Question 11. Find the sum of first seven multiples of 5
Answer: The first seven multiples of 5 are:
\( 5, 10, 15, 20, 25, 30, 35 \).
This is an AP where \( a = 5 \), \( d = 5 \), and \( n = 7 \).
The sum is given by:
\( S_7 = \frac{7}{2}[2(5) + (7 - 1)5] \)
\( \implies S_7 = \frac{7}{2}[10 + 30] \)
\( \implies S_7 = \frac{7}{2}(40) \)
\( \implies S_7 = 140 \).
The sum of the first seven multiples is 140.
In simple words: We can list the first seven multiples of 5 and add them up, or use the AP sum formula to get 140.

Exam Tip: For small values of n (like 7), simple listing and adding is a great way to verify your algebraic answer.

Question 12. Find the sum of all natural numbers up to 100, which are not divisible by 5
Answer: The sum of all natural numbers up to 100 is:
\( S_{\text{all}} = \frac{100 \times 101}{2} = 5050 \).
The natural numbers up to 100 that are divisible by 5 are:
\( 5, 10, 15, \dots, 100 \).
This is an AP with \( a = 5 \), last term \( l = 100 \), and \( n = 20 \) terms.
The sum of these multiples is:
\( S_5 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050 \).
The sum of numbers not divisible by 5 is:
\( S_{\text{required}} = 5050 - 1050 = 4000 \).
The required sum is 4000.
In simple words: We add up all numbers from 1 to 100, then subtract the sum of the multiples of 5.

Exam Tip: Finding the sum of consecutive integers using \( \frac{n(n+1)}{2} \) is an extremely fast tool for these kinds of problems.

Question 13. In an A.P , if the 6th and 13th terms are 35 and 70 respectively, find the sum of its first 20 terms.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
We are given:
\( a_6 = a + 5d = 35 \)
\( a_{13} = a + 12d = 70 \).
Subtracting the first equation from the second:
\( (a + 12d) - (a + 5d) = 70 - 35 \)
\( \implies 7d = 35 \)
\( \implies d = 5 \).
Substitute \( d = 5 \) back into the first equation:
\( a + 5(5) = 35 \)
\( \implies a + 25 = 35 \)
\( \implies a = 10 \).
The sum of the first 20 terms is:
\( S_{20} = \frac{20}{2}[2a + (20 - 1)d] \)
\( \implies S_{20} = 10[2(10) + 19(5)] \)
\( \implies S_{20} = 10[20 + 95] \)
\( \implies S_{20} = 10(115) \)
\( \implies S_{20} = 1150 \).
The sum of the first 20 terms is 1150.
In simple words: We find the start and step size of our AP from the given terms, and then compute the sum of the first 20 terms.

Exam Tip: Be methodical when solving the simultaneous equations to make sure you get the correct values for \( a \) and \( d \) first.

Question 14 If the 3rd and 9thterm of an A.P. are 4 and -8 respectively, which term is zero
Answer: Let the first term be \( a \) and the common difference be \( d \).
We have:
\( a_3 = a + 2d = 4 \)
\( a_9 = a + 8d = -8 \).
Subtracting the first equation from the second:
\( 6d = -12 \)
\( \implies d = -2 \).
Substitute \( d = -2 \) into the first equation:
\( a + 2(-2) = 4 \)
\( \implies a - 4 = 4 \)
\( \implies a = 8 \).
Now, let the \( n \)-th term be zero:
\( a_n = a + (n - 1)d = 0 \)
\( \implies 8 + (n - 1)(-2) = 0 \)
\( \implies -2(n - 1) = -8 \)
\( \implies n - 1 = 4 \)
\( \implies n = 5 \).
The 5th term of the AP is zero.
In simple words: We use the two given terms to find the AP's parameters, and then calculate which position in the sequence has the value 0.

Exam Tip: A decreasing sequence with a positive starting term will eventually reach zero or go negative, which matches our negative value for \( d \).

Question 15. The sum of 4th and 8th terms of an A.P is 24 and sum of 6th and 10th term is 44. Find A.P.
Answer: Let the first term be \( a \) and the common difference be \( d \).
The sum of the 4th and 8th terms is:
\( a_4 + a_8 = 24 \)
\( \implies (a + 3d) + (a + 7d) = 24 \)
\( \implies 2a + 10d = 24 \)
\( \implies a + 5d = 12 \).
The sum of the 6th and 10th terms is:
\( a_6 + a_{10} = 44 \)
\( \implies (a + 5d) + (a + 9d) = 44 \)
\( \implies 2a + 14d = 44 \)
\( \implies a + 7d = 22 \).
Subtracting the first simplified equation from the second:
\( 2d = 10 \)
\( \implies d = 5 \).
Substituting \( d = 5 \) into the first simplified equation:
\( a + 5(5) = 12 \)
\( \implies a + 25 = 12 \)
\( \implies a = -13 \).
The AP is: \( -13, -8, -3, 2, \dots \)
In simple words: We construct two simple equations based on the sum of terms, solve them to find the first term and step, and list the AP.

Exam Tip: Simplify your equations by dividing out any common factors before using substitution or elimination.

Question 16. The 4th term of an A.P is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the A.P
Answer: Let the first term be \( a \) and the common difference be \( d \).
From the first condition:
\( a_4 = 3a \)
\( \implies a + 3d = 3a \)
\( \implies 2a = 3d \)
\( \implies a = 1.5d \).
From the second condition:
\( a_7 = 2a_3 + 1 \)
\( \implies a + 6d = 2(a + 2d) + 1 \)
\( \implies a + 6d = 2a + 4d + 1 \)
\( \implies 2d = a + 1 \).
Substitute \( a = 1.5d \) into this equation:
\( 2d = 1.5d + 1 \)
\( \implies 0.5d = 1 \)
\( \implies d = 2 \).
Then, \( a = 1.5(2) = 3 \).
The AP is: \( 3, 5, 7, 9, \dots \)
In simple words: We convert the word problems into algebraic equations, solve for a and d, and write out the sequence.

Exam Tip: Writing fractions as decimals (like \( 1.5d \)) can sometimes make substitution equations simpler to solve mentally.

Question 17. Which term of the A.P.? 3, 15, 27, 39, will be 120 more than its 21st term
Answer: For the AP: \( 3, 15, 27, 39, \dots \), we have:
First term \( a = 3 \), and common difference \( d = 12 \).
Let the \( n \)-th term be 120 more than the 21st term:
\( a_n = a_{21} + 120 \)
\( \implies a + (n - 1)d = [a + 20d] + 120 \).
Subtract \( a \) from both sides:
\( (n - 1)d = 20d + 120 \).
Substitute \( d = 12 \):
\( (n - 1)12 = 20(12) + 120 \)
\( \implies 12(n - 1) = 240 + 120 \)
\( \implies 12(n - 1) = 360 \)
\( \implies n - 1 = 30 \)
\( \implies n = 31 \).
The 31st term is 120 more than its 21st term.
In simple words: We find which position in the sequence has a value that is 120 higher than the 21st term.

Exam Tip: You can divide the difference (120) by the common difference (12) to see that the required term is exactly \( 10 \) terms after the 21st term (which is the 31st term).

Question 18. In an A.P., the first term is 25, nth term is -17 and sum to first n terms is 60.Find n and d the common difference.
Answer: We are given:
First term \( a = 25 \), \( n \)-th term \( a_n = -17 \), and sum \( S_n = 60 \).
Using the sum formula \( S_n = \frac{n}{2}(a + a_n) \):
\( 60 = \frac{n}{2}(25 + (-17)) \)
\( \implies 60 = \frac{n}{2}(8) \)
\( \implies 4n = 60 \)
\( \implies n = 15 \).
Now, use the \( n \)-th term formula to find \( d \):
\( a_{15} = a + 14d \)
\( \implies -17 = 25 + 14d \)
\( \implies 14d = -42 \)
\( \implies d = -3 \).
So, \( n = 15 \) and \( d = -3 \).
In simple words: Using the sum formula helps us find how many terms there are, and then we use the last term to find the gap size.

Exam Tip: Double check that the sign of \( d \) is negative since the sequence starts at 25 and ends at a negative number.

Question 19. Which term of the sequence 114, 109, 104,is the first negative term?
Answer: For the given sequence \( 114, 109, 104, \dots \), we have:
First term \( a = 114 \), and common difference \( d = -5 \).
We want to find the first term \( a_n < 0 \):
\( a + (n - 1)d < 0 \)
\( \implies 114 + (n - 1)(-5) < 0 \)
\( \implies 114 - 5n + 5 < 0 \)
\( \implies 119 < 5n \)
\( \implies n > 23.8 \).
Since \( n \) must be an integer, the smallest value is \( n = 24 \).
The 24th term is the first negative term.
In simple words: Since the terms decrease by 5 each time, we set up an inequality to find the first step where the value drops below zero.

Exam Tip: When working with inequalities, remember that multiplying or dividing by a negative number reverses the direction of the inequality sign.

Question 20. If the 4th term of an A.P is twice the 8th term, prove that the 10th term is twice the 11th term
Answer: Let the first term be \( a \) and the common difference be \( d \).
We are given:
\( a_4 = 2a_8 \)
\( \implies a + 3d = 2(a + 7d) \)
\( \implies a + 3d = 2a + 14d \)
\( \implies a = -11d \).
Now, we calculate the 10th and 11th terms:
\( a_{10} = a + 9d = -11d + 9d = -2d \).
\( a_{11} = a + 10d = -11d + 10d = -d \).
Comparing these two terms:
\( a_{10} = 2(-d) = 2a_{11} \).
Thus, the 10th term is twice the 11th term. Hence proved.
In simple words: We express the first term in terms of d, and substitute this into the formulas for the 10th and 11th terms to show the proof.

Exam Tip: Expressing all variables in terms of \( d \) is a very reliable way to prove relations in AP questions.

Question 21. If 2 + 5 + 8 + …………………………+ x = 155, find x
Answer: This is the sum of an AP where \( a = 2 \) and \( d = 3 \).
Let the sum of \( n \) terms be 155:
\( S_n = \frac{n}{2}[2a + (n - 1)d] = 155 \)
\( \implies \frac{n}{2}[2(2) + (n - 1)3] = 155 \)
\( \implies \frac{n}{2}[4 + 3n - 3] = 155 \)
\( \implies n(3n + 1) = 310 \)
\( \implies 3n^2 + n - 310 = 0 \).
Solving this quadratic equation using the quadratic formula:
\( n = \frac{-1 \pm \sqrt{1^2 - 4(3)(-310)}}{6} = \frac{-1 \pm \sqrt{1 + 3720}}{6} = \frac{-1 \pm \sqrt{3721}}{6} = \frac{-1 \pm 61}{6} \).
Since \( n \) must be a positive integer, we choose \( n = \frac{60}{6} = 10 \).
Now, the last term \( x \) is the 10th term:
\( x = a_{10} = a + 9d = 2 + 9(3) = 29 \).
The value of x is 29.
In simple words: We find how many terms are in the sum first, then compute the value of the final term in the list.

Exam Tip: Be sure to write \( x = a_n \) explicitly in your steps to show how you got the final term after finding \( n \).

Question 22. For A.P. a1, a2, a3, ………., if a4/a7 = 2/3 , find a6/a8
Answer: Let the first term be \( a \) and the common difference be \( d \).
We are given:
\( \frac{a_4}{a_7} = \frac{2}{3} \)
\( \implies \frac{a + 3d}{a + 6d} = \frac{2}{3} \).
Cross-multiplying gives:
\( 3(a + 3d) = 2(a + 6d) \)
\( \implies 3a + 9d = 2a + 12d \)
\( \implies a = 3d \).
Now, we evaluate the ratio of the 6th to the 8th term:
\( \frac{a_6}{a_8} = \frac{a + 5d}{a + 7d} \).
Substitute \( a = 3d \) into this ratio:
\( \frac{a_6}{a_8} = \frac{3d + 5d}{3d + 7d} = \frac{8d}{10d} = \frac{4}{5} \).
The ratio is \( \frac{4}{5} \).
In simple words: We use the first ratio to find that the starting term is three times the step size, then simplify the second ratio.

Exam Tip: Substituting \( a \) in terms of \( d \) allows you to cancel out the variable \( d \) from both the numerator and denominator.

Question 23. Find the sum of the following A.P: 1 + 3 + 5 + …….. + 199.
Answer: This is the sum of consecutive odd numbers starting from 1 up to 199.
First term \( a = 1 \), common difference \( d = 2 \), and last term \( a_n = 199 \).
Let us find the number of terms \( n \):
\( a_n = a + (n - 1)d \)
\( \implies 199 = 1 + (n - 1)2 \)
\( \implies 198 = 2(n - 1) \)
\( \implies n - 1 = 99 \)
\( \implies n = 100 \).
Now, calculate the sum:
\( S_{100} = \frac{100}{2}(1 + 199) = 50(200) = 10,000 \).
The sum of the AP is 10,000.
In simple words: There are exactly 100 odd numbers in this list. Their sum is 10,000.

Exam Tip: Remember the formula for the sum of the first \( n \) odd natural numbers is simply \( n^2 \). Here, \( 100^2 = 10,000 \).

Question 24. Find the common difference of an AP whose first term is 100 and sum of first six terms is 5 times the The sum of the next 6 terms
Answer: We have the first term \( a = 100 \).
The sum of the first six terms is:
\( S_6 = \frac{6}{2}[2(100) + 5d] = 3(200 + 5d) = 600 + 15d \).
The sum of the next six terms (from the 7th to the 12th term) is given by \( S_{12} - S_6 \).
First, find \( S_{12} \):
\( S_{12} = \frac{12}{2}[2(100) + 11d] = 6(200 + 11d) = 1200 + 66d \).
So, the sum of the next six terms is:
\( S_{\text{next } 6} = (1200 + 66d) - (600 + 15d) = 600 + 51d \).
According to the problem:
\( S_6 = 5 \times S_{\text{next } 6} \)
\( \implies 600 + 15d = 5(600 + 51d) \)
\( \implies 600 + 15d = 3000 + 255d \)
\( \implies -2400 = 240d \)
\( \implies d = -10 \).
The common difference is \( -10 \).
In simple words: We write formulas for the sum of the first six terms and the next six, set up the ratio, and solve for the step size.

Exam Tip: Expressing the sum of the "next" set of terms as \( S_{2n} - S_n \) is a standard and robust approach to these problems.

Question 25. Show that progression 7, 2, -3, -8, … ……..Is an A.P . Find its nth term
Answer: Let us find the differences between consecutive terms:
\( a_2 - a_1 = 2 - 7 = -5 \)
\( a_3 - a_2 = -3 - 2 = -5 \)
\( a_4 - a_3 = -8 - (-3) = -5 \).
Since the difference between any two consecutive terms is constant (\( -5 \)), the given progression is an AP with first term \( a = 7 \) and common difference \( d = -5 \).
The \( n \)-th term is:
\( a_n = a + (n - 1)d \)
\( \implies a_n = 7 + (n - 1)(-5) \)
\( \implies a_n = 7 - 5n + 5 \)
\( \implies a_n = 12 - 5n \).
The progression is an AP and its \( n \)-th term is \( 12 - 5n \).
In simple words: Since the terms decrease by 5 every single step, it is an AP. Its general formula is \( 12 - 5n \).

Exam Tip: To show that a sequence is an AP, you must calculate at least two differences between consecutive terms and state that they are equal.

Question 26. The angles of a triangle are in A.P, the last being half the greatest. Find the angles.
Answer: Let the three angles of the triangle in AP be \( a - d \), \( a \), and \( a + d \).
By the angle sum property of a triangle:
\( (a - d) + a + (a + d) = 180^\circ \)
\( \implies 3a = 180^\circ \)
\( \implies a = 60^\circ \).
So, the angles are \( 60^\circ - d \), \( 60^\circ \), and \( 60^\circ + d \).
The smallest angle is \( 60^\circ - d \) and the greatest is \( 60^\circ + d \).
According to the problem, the smallest angle is half the greatest:
\( 60^\circ - d = \frac{1}{2}(60^\circ + d) \)
\( \implies 120^\circ - 2d = 60^\circ + d \)
\( \implies 3d = 60^\circ \)
\( \implies d = 20^\circ \).
The angles are:
\( 60^\circ - 20^\circ = 40^\circ \), \( 60^\circ \), and \( 60^\circ + 20^\circ = 80^\circ \).
The angles of the triangle are \( 40^\circ \), \( 60^\circ \), and \( 80^\circ \).
In simple words: Since the angles are in an AP, their average is always \( 60^\circ \). We use the other clue to find the step size of \( 20^\circ \).

Exam Tip: Representing three terms in AP as \( a-d \), \( a \), and \( a+d \) makes the sum highly simple because the \( d \) terms cancel out.

Question 27. Find the sum of n terms of an A.P whose nth term is given by tn = 5 – 6n
Answer: The \( n \)-th term is given as \( t_n = 5 - 6n \).
The first term of this AP is:
\( a = t_1 = 5 - 6(1) = -1 \).
The last term (the \( n \)-th term) is:
\( t_n = 5 - 6n \).
The sum of \( n \) terms is:
\( S_n = \frac{n}{2}(a + t_n) \)
\( \implies S_n = \frac{n}{2}(-1 + 5 - 6n) \)
\( \implies S_n = \frac{n}{2}(4 - 6n) \)
\( \implies S_n = n(2 - 3n) \)
\( \implies S_n = 2n - 3n^2 \).
The sum of \( n \) terms is \( 2n - 3n^2 \).
In simple words: We find the first and last terms, then apply the quick sum formula to write down the answer.

Exam Tip: Be careful to simplify \( \frac{n}{2}(4 - 6n) \) correctly by dividing both terms inside the bracket by 2.

Question 28. Find the middle term of A.P: 1, 8, 15, ……………, 505
Answer: For the given AP: \( 1, 8, 15, \dots, 505 \), we have:
First term \( a = 1 \), and common difference \( d = 7 \).
Let us find the total number of terms \( n \):
\( a_n = a + (n - 1)d \)
\( \implies 505 = 1 + (n - 1)7 \)
\( \implies 504 = 7(n - 1) \)
\( \implies n - 1 = 72 \)
\( \implies n = 73 \).
Since \( n = 73 \) (an odd number), there is a single middle term:
\( \text{Middle term position} = \frac{n + 1}{2} = \frac{74}{2} = 37 \).
The middle term value is the 37th term:
\( a_{37} = a + 36d = 1 + 36(7) = 1 + 252 = 253 \).
The middle term of the AP is 253.
In simple words: We count the terms in the AP, find the middle term position, and calculate its value.

Exam Tip: For an odd number of terms \( n \), the middle term is \( \frac{n+1}{2} \). For an even number of terms, there are two middle terms: \( \frac{n}{2} \) and \( \frac{n}{2} + 1 \).

Question 29. Find the number of terms of the A.P, 63, 60, 57, ……….. So that their sum is 693
Answer: For the given AP: \( 63, 60, 57, \dots \), we have:
First term \( a = 63 \), and common difference \( d = -3 \).
Let the sum of \( n \) terms be 693:
\( S_n = \frac{n}{2}[2a + (n - 1)d] = 693 \)
\( \implies \frac{n}{2}[2(63) + (n - 1)(-3)] = 693 \)
\( \implies \frac{n}{2}[126 - 3n + 3] = 693 \)
\( \implies n(129 - 3n) = 1386 \)
\( \implies 129n - 3n^2 = 1386 \).
Dividing by \( -3 \):
\( n^2 - 43n + 462 = 0 \)
\( \implies (n - 21)(n - 22) = 0 \).
So, \( n = 21 \) or \( n = 22 \).
In simple words: There can be 21 or 22 terms because the 22nd term in this decreasing sequence is exactly zero, which does not change the sum.

Exam Tip: If you get two positive integer answers for \( n \), explain why both are valid (e.g., the extra term is zero) to show deep conceptual understanding.

Question 30. The sum of 3 numbers in A.P is 3 and their product is -35. Find the numbers
Answer: Let the three numbers in AP be \( a - d \), \( a \), and \( a + d \).
The sum of the numbers is:
\( (a - d) + a + (a + d) = 3 \)
\( \implies 3a = 3 \)
\( \implies a = 1 \).
So, the numbers are \( 1 - d \), \( 1 \), and \( 1 + d \).
Their product is given as \( -35 \):
\( (1 - d)(1)(1 + d) = -35 \)
\( \implies 1 - d^2 = -35 \)
\( \implies d^2 = 36 \)
\( \implies d = \pm 6 \).
If \( d = 6 \), the numbers are \( -5, 1, 7 \).
If \( d = -6 \), the numbers are \( 7, 1, -5 \).
The three numbers are \( -5, 1, 7 \).
In simple words: Using a symmetric setup, we easily find the middle term is 1, and the step size is 6.

Exam Tip: Clearly write down both sets of sequences generated by \( d = 6 \) and \( d = -6 \), even though they represent the same set of numbers.

Question 31. How many terms of the sequence 18, 16, 14, …………, should be taken so that their sum is 0
Answer: For the given AP: \( 18, 16, 14, \dots \), we have:
First term \( a = 18 \), and common difference \( d = -2 \).
Let the sum of \( n \) terms be 0:
\( S_n = \frac{n}{2}[2a + (n - 1)d] = 0 \).
Since \( n \) cannot be zero, we have:
\( 2(18) + (n - 1)(-2) = 0 \)
\( \implies 36 - 2n + 2 = 0 \)
\( \implies 2n = 38 \)
\( \implies n = 19 \).
So, 19 terms must be taken.
In simple words: Since the numbers decrease and go negative, the negative terms will eventually cancel out the positive ones, resulting in a sum of zero after 19 terms.

Exam Tip: Since \( n \neq 0 \) for any real sequence of terms, we can safely divide out the \( \frac{n}{2} \) term when solving \( S_n = 0 \).

Question 32. A sum of Rs 1400 is to be used to give 7 cash prizes to students of a school for their overall academic Performance if each prize is Rs40 less than the preceding price, find the value of each of the prizes.
Answer: Let the first prize be \( a \) Rupees.
Since each subsequent prize is Rs 40 less, this forms an AP with common difference \( d = -40 \).
The number of prizes is \( n = 7 \), and the total sum \( S_7 = 1400 \) Rupees.
Using the sum formula:
\( S_7 = \frac{7}{2}[2a + (7 - 1)d] \)
\( \implies 1400 = \frac{7}{2}[2a + 6(-40)] \)
\( \implies 200 = \frac{1}{2}[2a - 240] \)
\( \implies 400 = 2a - 240 \)
\( \implies 2a = 640 \)
\( \implies a = 320 \).
The values of the prizes are:
Rs 320, Rs 280, Rs 240, Rs 200, Rs 160, Rs 120, and Rs 80.
In simple words: This is an AP sum problem where we solve for the highest prize and then list down each smaller prize.

Exam Tip: Always list out all the values of the individual prizes at the end of your answer to fulfill the question's requirement completely.

Question 33. Verify that a + b, (a + 1) + b, (a + 1) + (b + 1) ……….. Is an A.P. and then write its next term
Answer: Let us write the terms of the given progression:
\( T_1 = a + b \)
\( T_2 = (a + 1) + b = a + b + 1 \)
\( T_3 = (a + 1) + (b + 1) = a + b + 2 \).
Let us find the differences between consecutive terms:
\( T_2 - T_1 = (a + b + 1) - (a + b) = 1 \).
\( T_3 - T_2 = (a + b + 2) - (a + b + 1) = 1 \).
Since the consecutive differences are constant (equal to 1), the progression is indeed an AP with common difference \( d = 1 \).
The next term (\( T_4 \)) is:
\( T_4 = T_3 + d = (a + b + 2) + 1 = a + b + 3 \).
This can be written in the original format as \( (a + 2) + (b + 1) \).
In simple words: Since each term increases by exactly 1, this forms an AP. The next term is \( (a+2) + (b+1) \).

Exam Tip: Always write down the consecutive subtraction steps to formally show that the difference is constant before stating that it is an AP.

Question 34. Determine the A.P whose 3rd term is 16 and 7th term exceeds the 5th term by 12
Answer: Let the first term be \( a \) and the common difference be \( d \).
We are given:
\( a_3 = a + 2d = 16 \).
The 7th term exceeds the 5th term by 12:
\( a_7 = a_5 + 12 \)
\( \implies a + 6d = (a + 4d) + 12 \)
\( \implies 2d = 12 \)
\( \implies d = 6 \).
Substitute \( d = 6 \) back into the first equation:
\( a + 2(6) = 16 \)
\( \implies a + 12 = 16 \)
\( \implies a = 4 \).
The AP is \( 4, 10, 16, 22, \dots \)
In simple words: The difference between the 7th and 5th terms is 2 steps, which is 12, so each step is 6. The sequence starts at 4.

Exam Tip: Use the relation \( a_n - a_m = (n-m)d \) to find the common difference \( d \) directly from the term difference. Here, \( a_7 - a_5 = 2d = 12 \).

Question 35. If the nth term of the A.P. 9, 7, 5, ………… is the same as the nth term of the A.P. 15, 12, 9, ………., find n
Answer: Let the first AP be \( 9, 7, 5, \dots \), where first term \( a_1 = 9 \) and common difference \( d_1 = -2 \).
Its \( n \)-th term is:
\( T_n = 9 + (n - 1)(-2) = 11 - 2n \).
Let the second AP be \( 15, 12, 9, \dots \), where first term \( a_2 = 15 \) and common difference \( d_2 = -3 \).
Its \( n \)-th term is:
\( T'_n = 15 + (n - 1)(-3) = 18 - 3n \).
Given that their \( n \)-th terms are equal:
\( 11 - 2n = 18 - 3n \)
\( \implies 3n - 2n = 18 - 11 \)
\( \implies n = 7 \).
The value of n is 7.
In simple words: We find the general formula for both sequences, set them equal to each other, and solve to find they match at step 7.

Exam Tip: Be sure to write separate equations for both APs using distinct variables before equating them.

Question 36. Find the sum of first 22 terms of an A.P. in which d = 7 and 22nd term is 149
Answer: We are given:
Common difference \( d = 7 \), and the 22nd term \( a_{22} = 149 \).
Let the first term be \( a \):
\( a_{22} = a + 21d = 149 \)
\( \implies a + 21(7) = 149 \)
\( \implies a + 147 = 149 \)
\( \implies a = 2 \).
The sum of the first 22 terms is:
\( S_{22} = \frac{22}{2}(a + a_{22}) \)
\( \implies S_{22} = 11(2 + 149) \)
\( \implies S_{22} = 11(151) \)
\( \implies S_{22} = 1661 \).
The sum of the first 22 terms is 1661.
In simple words: We find the first term from the 22nd term, and then sum the terms using the quick sum formula.

Exam Tip: When both the first and last terms are known, always use \( S_n = \frac{n}{2}(a+l) \) as it saves a lot of time.

Question 37. Find the sum of the following A.P: 3, 9/2, 6, 15/2, ………. To 25 terms
Answer: For the given AP: \( 3, \frac{9}{2}, 6, \frac{15}{2}, \dots \), we have:
First term \( a = 3 \).
Common difference \( d = \frac{9}{2} - 3 = \frac{3}{2} \).
Number of terms \( n = 25 \).
The sum is:
\( S_{25} = \frac{25}{2}[2a + (25 - 1)d] \)
\( \implies S_{25} = \frac{25}{2}\left[2(3) + 24\left(\frac{3}{2}\right)\right] \)
\( \implies S_{25} = \frac{25}{2}[6 + 36] \)
\( \implies S_{25} = \frac{25}{2}(42) \)
\( \implies S_{25} = 25 \times 21 \)
\( \implies S_{25} = 525 \).
The sum of the AP is 525.
In simple words: The step size is 1.5. Using the sum formula for 25 terms gives us 525.

Exam Tip: Simplifying terms like \( 24 \times \frac{3}{2} \) to \( 36 \) before doing the addition inside brackets makes calculations much easier.

Question 38. The ratio of the sum to p terms and q terms of an A.P. is p2 : q2. Prove that the common difference of the A.P.is twice The first term
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
We are given:
\( \frac{S_p}{S_q} = \frac{p^2}{q^2} \)
\( \implies \frac{\frac{p}{2}[2a + (p - 1)d]}{\frac{q}{2}[2a + (q - 1)d]} = \frac{p^2}{q^2} \)
\( \implies \frac{2a + (p - 1)d}{2a + (q - 1)d} = \frac{p}{q} \).
Cross-multiplying gives:
\( q[2a + (p - 1)d] = p[2a + (q - 1)d] \)
\( \implies 2aq + pqd - qd = 2ap + pqd - pd \).
Cancel \( pqd \) from both sides:
\( 2aq - qd = 2ap - pd \)
\( \implies 2aq - 2ap = qd - pd \)
\( \implies 2a(q - p) = d(q - p) \).
Since \( p \neq q \), we can divide both sides by \( (q - p) \):
\( d = 2a \).
Thus, the common difference is twice the first term. Hence proved.
In simple words: Writing the sum formulas and simplifying the fractions shows that the gap size d is exactly two times the starting term a.

Exam Tip: Be sure to write the explanation that you can divide by \( q-p \) because \( q-p \neq 0 \) to keep your mathematical proof completely rigorous.

Question 39. In an A.P., if the sum of its 4th and 10th terms is 40, and the sum of its 8th and 16th terms is 70, then find the sum of its First 20 terms
Answer: Let the first term be \( a \) and the common difference be \( d \).
From the first condition:
\( a_4 + a_{10} = 40 \)
\( \implies (a + 3d) + (a + 9d) = 40 \)
\( \implies 2a + 12d = 40 \)
\( \implies a + 6d = 20 \).
From the second condition:
\( a_8 + a_{16} = 70 \)
\( \implies (a + 7d) + (a + 15d) = 70 \)
\( \implies 2a + 22d = 70 \)
\( \implies a + 11d = 35 \).
Subtracting the first simplified equation from the second:
\( 5d = 15 \)
\( \implies d = 3 \).
Substitute \( d = 3 \) into the first simplified equation:
\( a + 6(3) = 20 \)
\( \implies a + 18 = 20 \)
\( \implies a = 2 \).
The sum of the first 20 terms is:
\( S_{20} = \frac{20}{2}[2a + (20 - 1)d] \)
\( \implies S_{20} = 10[2(2) + 19(3)] \)
\( \implies S_{20} = 10[4 + 57] \)
\( \implies S_{20} = 10(61) \)
\( \implies S_{20} = 610 \).
The sum of the first 20 terms is 610.
In simple words: We find the AP parameters a and d first, and then calculate the sum of the first 20 terms.

Exam Tip: Double check each subtraction when solving simultaneous equations to make sure you get the right value for \( d \).

Question 40. Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two Is 154. Find the integers
Answer: Let the three consecutive positive integers be \( x \), \( x + 1 \), and \( x + 2 \).
According to the given condition:
\( x^2 + (x + 1)(x + 2) = 154 \)
\( \implies x^2 + (x^2 + 3x + 2) = 154 \)
\( \implies 2x^2 + 3x + 2 - 154 = 0 \)
\( \implies 2x^2 + 3x - 152 = 0 \).
Factoring the quadratic equation:
\( 2x^2 + 19x - 16x - 152 = 0 \)
\( \implies x(2x + 19) - 8(2x + 19) = 0 \)
\( \implies (x - 8)(2x + 19) = 0 \).
Since \( x \) must be a positive integer, we reject \( x = -\frac{19}{2} \). Therefore, \( x = 8 \).
The three consecutive positive integers are 8, 9, and 10.
In simple words: We write the consecutive numbers as algebraic expressions, solve the quadratic equation, and find the numbers are 8, 9, and 10.

Exam Tip: Be sure to write down the final values (8, 9, 10) explicitly as the answer asks for "the integers", not just the value of \( x \).