CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 10

Question 1. Solve for x:
a) 9x² - 9 (a + b) x + 2a² + 5ab + 2b² = 0
b) 4x² - 4a²x + (a⁴ - b⁴) = 0
c) 10ax² - 6x + 15ax - 9 = 0
d) x² - 2(a² + b²)x + (a² - b²)² = 0
e) √7x² - 6x - 13 √7 = 0
f) 10x² + 3bx + a² - 7ax - b² = 0
Answer:
a) Let us first factorize the constant term of the equation:
\( 2a^2 + 5ab + 2b^2 = 2a^2 + 4ab + ab + 2b^2 \)
\( \implies 2a(a + 2b) + b(a + 2b) = (2a + b)(a + 2b) \).
Now, we split the middle term \( -9(a+b)x \) as \( -3[(2a+b) + (a+2b)]x \).
The quadratic equation becomes:
\( 9x^2 - 3(2a+b)x - 3(a+2b)x + (2a+b)(a+2b) = 0 \)
\( \implies 3x[3x - (2a+b)] - (a+2b)[3x - (2a+b)] = 0 \)
\( \implies [3x - (2a+b)][3x - (a+2b)] = 0 \).
This gives the roots as:
\( x = \frac{2a+b}{3} \) or \( x = \frac{a+2b}{3} \).

b) Factorize the constant term as \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \).
We can split the middle term coefficient \( -4a^2 \) as \( -2[(a^2 + b^2) + (a^2 - b^2)] \).
The equation becomes:
\( 4x^2 - 2(a^2 + b^2)x - 2(a^2 - b^2)x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \implies 2x[2x - (a^2 + b^2)] - (a^2 - b^2)[2x - (a^2 + b^2)] = 0 \)
\( \implies [2x - (a^2 + b^2)][2x - (a^2 - b^2)] = 0 \).
This gives the roots as:
\( x = \frac{a^2 + b^2}{2} \) or \( x = \frac{a^2 - b^2}{2} \).

c) Group the terms in \( 10ax^2 - 6x + 15ax - 9 = 0 \):
\( 2x(5ax - 3) + 3(5ax - 3) = 0 \)
\( \implies (2x + 3)(5ax - 3) = 0 \).
This gives the roots as:
\( x = -\frac{3}{2} \) or \( x = \frac{3}{5a} \).

d) Let us split 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 is:
\( x^2 - [(a+b)^2 + (a-b)^2]x + (a+b)^2(a-b)^2 = 0 \)
\( \implies [x - (a+b)^2][x - (a-b)^2] = 0 \).
This gives the roots as:
\( x = (a+b)^2 \) or \( x = (a-b)^2 \).

e) Multiply the coefficients of the first and last terms: \( \sqrt{7} \times (-13\sqrt{7}) = -91 \).
We need two numbers that multiply to \( -91 \) and 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 the roots as:
\( x = \frac{13\sqrt{7}}{7} \) or \( x = -\sqrt{7} \).

f) Rearrange the terms of \( 10x^2 + 3bx + a^2 - 7ax - b^2 = 0 \) as:
\( 10x^2 - 7ax + a^2 + 3bx - b^2 = 0 \).
Let us rewrite this to group terms systematically:
\( 10x^2 - (7a - 3b)x + (a^2 - b^2) = 0 \).
We can split the middle term using the factors of \( 10(a^2 - b^2) \), which are \( 5(a-b) \) and \( 2(a+b) \):
\( -(7a - 3b) = -[5(a-b) + 2(a+b)] \).
Substituting this in:
\( 10x^2 - 5(a-b)x - 2(a+b)x + (a-b)(a+b) = 0 \)
\( \implies 5x[2x - (a-b)] - (a+b)[2x - (a-b)] = 0 \)
\( \implies [5x - (a+b)][2x - (a-b)] = 0 \).
This gives the roots as:
\( x = \frac{a+b}{5} \) or \( x = \frac{a-b}{2} \).
In simple words: To solve quadratic equations with alphabetic variables, we simplify and factorize the constant terms first, then split the middle term.

Exam Tip: For complex algebraic coefficients, look for common algebraic identities such as \( a^2 - b^2 = (a-b)(a+b) \) to help split the middle terms.

Question 2. find the value of k so that the quadratic equation has equal roots:
a) 2kx² – 40x + 25 = 0
b) 2x² – (k – 2) x + 1 = 0
c) ( k + 3 ) x² + 2 ( k + 3 )x + 4 = 0
Answer: A quadratic equation has equal roots when its discriminant is zero (\( D = b^2 - 4ac = 0 \)).
a) Here, \( a = 2k, b = -40, c = 25 \).
\( D = (-40)^2 - 4(2k)(25) = 0 \)
\( \implies 1600 - 200k = 0 \)
\( \implies 200k = 1600 \)
\( \implies k = 8 \).

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

c) Here, \( a = k+3, b = 2(k+3), c = 4 \).
\( D = [2(k+3)]^2 - 4(k+3)(4) = 0 \)
\( \implies 4(k+3)^2 - 16(k+3) = 0 \)
\( \implies 4(k+3)[(k+3) - 4] = 0 \)
\( \implies 4(k+3)(k-1) = 0 \).
This gives \( k = -3 \) or \( k = 1 \). If \( k = -3 \), the coefficient of \( x^2 \) becomes zero, meaning the equation is not quadratic. Thus, we select \( k = 1 \).
In simple words: Setting the discriminant formula to zero lets us calculate the exact value of k that gives equal roots.

Exam Tip: Always make sure your final value of k does not make the coefficient of \( x^2 \) equal to zero.

Question 3. For what value of p the equation (1 + p) x2 + 2(1 + 2p) x + (1 + p) = 0 has coincident roots
Answer: Coincident roots mean that the discriminant of the equation must be zero (\( D = 0 \)).
In this equation, \( a = 1+p \), \( b = 2(1+2p) \), and \( c = 1+p \).
\( D = [2(1+2p)]^2 - 4(1+p)(1+p) = 0 \)
\( \implies 4(1+2p)^2 - 4(1+p)^2 = 0 \).
Dividing both sides by 4:
\( (1+2p)^2 - (1+p)^2 = 0 \)
\( \implies [(1+2p) - (1+p)][(1+2p) + (1+p)] = 0 \)
\( \implies [p][2 + 3p] = 0 \).
This yields \( p = 0 \) or \( p = -\frac{2}{3} \).
In simple words: Coincident roots mean the roots are equal, so we set the discriminant to zero to find the values of p.

Exam Tip: Applying the identity \( X^2 - Y^2 = (X-Y)(X+Y) \) is a rapid way to solve difference-of-squares equations without fully expanding them.

Question 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
c) 6x² – 7x + 2 = 0
d) 4x² + 4√3x + 3 = 0
Answer:
a) Divide the equation by \( a^2 \):
\( x^2 - \frac{3b}{a}x + \frac{2b^2}{a^2} = 0 \)
\( \implies x^2 - \frac{3b}{a}x = -\frac{2b^2}{a^2} \).
Add \( \left(\frac{3b}{2a}\right)^2 = \frac{9b^2}{4a^2} \) to both sides:
\( x^2 - \frac{3b}{a}x + \frac{9b^2}{4a^2} = -\frac{2b^2}{a^2} + \frac{9b^2}{4a^2} \)
\( \implies \left(x - \frac{3b}{2a}\right)^2 = \frac{b^2}{4a^2} \).
Taking the square root:
\( x - \frac{3b}{2a} = \pm \frac{b}{2a} \)
\( \implies x = \frac{3b}{2a} \pm \frac{b}{2a} \).
So, \( x = \frac{2b}{a} \) or \( x = \frac{b}{a} \).

b) Rewrite the equation by grouping terms:
\( (x^2 - 4ax + 4a^2) = b^2 \)
\( \implies (x - 2a)^2 = b^2 \).
Taking the square root:
\( x - 2a = \pm b \)
\( \implies x = 2a \pm b \).

c) Divide the equation by 6:
\( x^2 - \frac{7}{6}x + \frac{2}{6} = 0 \)
\( \implies x^2 - \frac{7}{6}x = -\frac{1}{3} \).
Add \( \left(\frac{7}{12}\right)^2 = \frac{49}{144} \) to both sides:
\( x^2 - \frac{7}{6}x + \frac{49}{144} = -\frac{1}{3} + \frac{49}{144} \)
\( \implies \left(x - \frac{7}{12}\right)^2 = \frac{-48 + 49}{144} \)
\( \implies \left(x - \frac{7}{12}\right)^2 = \frac{1}{144} \).
Taking the square root:
\( x - \frac{7}{12} = \pm \frac{1}{12} \)
\( \implies x = \frac{7}{12} \pm \frac{1}{12} \).
So, \( x = \frac{8}{12} = \frac{2}{3} \) or \( x = \frac{6}{12} = \frac{1}{2} \).

d) Divide by 4:
\( x^2 + \sqrt{3}x + \frac{3}{4} = 0 \)
\( \implies x^2 + \sqrt{3}x = -\frac{3}{4} \).
Add \( \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \) to both sides:
\( x^2 + \sqrt{3}x + \frac{3}{4} = -\frac{3}{4} + \frac{3}{4} \)
\( \implies \left(x + \frac{\sqrt{3}}{2}\right)^2 = 0 \).
Taking the square root:
\( x + \frac{\sqrt{3}}{2} = 0 \)
\( \implies x = -\frac{\sqrt{3}}{2} \).
The identical roots are \( \pm \frac{\sqrt{3}}{2} \).
In simple words: Completing the square involves adding and subtracting a term to create a perfect square trinomial, which makes taking the square root straightforward.

Exam Tip: Always make the coefficient of \( x^2 \) equal to 1 by dividing the entire equation before starting the completing the square process.

Question 5. Solve the following quadratic equations by factorization method:
a) 3x² - 2√6x + 2 = 0
b) x² - 5√5x + 30 = 0
c) ax² + a = a²x + x
Answer:
a) We can write \( 3x^2 - 2\sqrt{6}x + 2 = 0 \) as a perfect square expression:
\( (\sqrt{3}x)^2 - 2(\sqrt{3}x)(\sqrt{2}) + (\sqrt{2})^2 = 0 \)
\( \implies (\sqrt{3}x - \sqrt{2})^2 = 0 \).
This gives identical roots:
\( x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \).
So, the roots are \( \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}} \).

b) We need to split the middle term \( -5\sqrt{5}x \) into two parts whose product is \( 30 \). These are \( -3\sqrt{5}x \) and \( -2\sqrt{5}x \):
\( x^2 - 3\sqrt{5}x - 2\sqrt{5}x + 30 = 0 \)
\( \implies x(x - 3\sqrt{5}) - 2\sqrt{5}(x - 3\sqrt{5}) = 0 \)
\( \implies (x - 2\sqrt{5})(x - 3\sqrt{5}) = 0 \).
This gives the roots as:
\( x = 2\sqrt{5} \) or \( x = 3\sqrt{5} \).

c) Rearranging \( ax^2 + a = a^2x + x \):
\( ax^2 - a^2x - x + a = 0 \)
\( \implies ax(x - a) - 1(x - a) = 0 \)
\( \implies (ax - 1)(x - a) = 0 \).
This gives the roots as:
\( x = a \) or \( x = \frac{1}{a} \).
In simple words: Factorization involves grouping the quadratic terms into brackets to solve for x.

Exam Tip: When factoring terms with square roots like \( 30 \), write them as \( 6 \times 5 = 6 \times (\sqrt{5})^2 \) to help visualize the factors easily.

Question 6. write the nature of roots of quadratic equation: a) 4x2 + 4√3x + 3 = 0 b) x2 - b2 - a (2x – a) = 0
Answer:
a) For \( 4x^2 + 4\sqrt{3}x + 3 = 0 \), here \( a = 4, b = 4\sqrt{3}, c = 3 \).
The discriminant is:
\( D = b^2 - 4ac = (4\sqrt{3})^2 - 4(4)(3) \)
\( \implies D = 48 - 48 = 0 \).
Since \( D = 0 \), the roots are real and equal.

b) For \( x^2 - b^2 - a(2x - a) = 0 \), let us simplify the equation:
\( x^2 - 2ax + (a^2 - b^2) = 0 \).
Here, the coefficients are \( A = 1, B = -2a, C = a^2 - b^2 \).
The discriminant is:
\( D = B^2 - 4AC = (-2a)^2 - 4(1)(a^2 - b^2) \)
\( \implies D = 4a^2 - 4a^2 + 4b^2 \)
\( \implies D = 4b^2 \).
Since \( 4b^2 \geq 0 \), the roots are real (and they are distinct if \( b \neq 0 \)).
In simple words: The sign of the discriminant tells us whether the roots are real, equal, or complex.

Exam Tip: Since a squared term like \( 4b^2 \) is always non-negative, you can confidently state that the roots are real.

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 down the full equation:
\( x^3 - 4x^2 + 1 = x^2 - 4x + 4 \).
Move all terms to the left side:
\( x^3 - 4x^2 - x^2 + 4x + 1 - 4 = 0 \)
\( \implies x^3 - 5x^2 + 4x - 3 = 0 \).
Since the highest exponent of the variable \( x \) is 3, this is a cubic equation, not a quadratic equation.
In simple words: A quadratic equation must have an \( x^2 \) term as its highest power. Here, the \( x^3 \) term does not cancel out, so it is not quadratic.

Exam Tip: Always expand and simplify both sides of an equation completely before determining its degree.

Question 8. Solve for x: 1 / (a + b + x) = 1/a + 1/b + 1/x, a + b ≠ 0
Answer: Let us rearrange the terms by moving \( \frac{1}{x} \) to the left side:
\( \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \).
Find a common denominator on both sides:
\( \frac{x - (a+b+x)}{x(a+b+x)} = \frac{b+a}{ab} \)
\( \implies \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \).
Since \( a+b \neq 0 \), we divide both sides by \( (a+b) \):
\( \frac{-1}{x(a+b+x)} = \frac{1}{ab} \)
\( \implies -ab = x(a+b+x) \)
\( \implies x^2 + (a+b)x + ab = 0 \)
\( \implies x(x+a) + b(x+a) = 0 \)
\( \implies (x+a)(x+b) = 0 \).
This gives: \( x = -a \) or \( x = -b \).
In simple words: Moving the x term to the left side lets us simplify the fractions and solve for the roots easily.

Exam Tip: Grouping terms of similar variables (like x-terms) is a critical step in solving fractional algebraic equations.

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 the roots is:
\( p + q = -\frac{b}{a} = 5 \).
The product of the roots is:
\( pq = \frac{c}{a} = 4 \).
Now, we write the expression we need to evaluate:
\( \frac{1}{p} + \frac{1}{q} - 2pq = \frac{p+q}{pq} - 2pq \).
Substitute the sum and product values in:
\( \frac{5}{4} - 2(4) = \frac{5}{4} - 8 = \frac{5 - 32}{4} = -\frac{27}{4} \).
The final value is \( -\frac{27}{4} \).
In simple words: We find the sum and product of the roots using quick formulas, and then put them into the expression.

Exam Tip: Avoid finding the actual values of the roots p and q individually. Using the sum and product relationships directly is faster and less prone to errors.

Question 10. Solve for x: x / (x + 1) + (x + 1) / x = 34 / 15
Answer: Let us substitute \( y = \frac{x}{x+1} \). This means its reciprocal is \( \frac{1}{y} = \frac{x+1}{x} \).
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:
\( 15y^2 - 25y - 9y + 15 = 0 \)
\( \implies 5y(3y - 5) - 3(3y - 5) = 0 \)
\( \implies (5y - 3)(3y - 5) = 0 \).
So, \( y = \frac{3}{5} \) or \( y = \frac{5}{3} \).

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

Case 2: If \( y = \frac{5}{3} \), then:
\( \frac{x}{x+1} = \frac{5}{3} \)
\( \implies 3x = 5x + 5 \)
\( \implies -2x = 5 \)
\( \implies x = -\frac{5}{2} \).
So, the roots are \( x = \frac{3}{2} \) and \( x = -\frac{5}{2} \).
In simple words: Substituting a complex term with a simple letter like y helps us find the answers much more easily.

Exam Tip: Substituting repeating reciprocal terms with \( y \) and \( 1/y \) is a highly effective way to simplify fractional equations.

Question 11. Solve for x: 1 / (x – 3) - 1 / (x + 5) = 1 / 6
Answer: Take a common denominator on the left side of the equation:
\( \frac{(x+5) - (x-3)}{(x-3)(x+5)} = \frac{1}{6} \)
\( \implies \frac{8}{x^2 + 2x - 15} = \frac{1}{6} \).
Cross-multiplying gives:
\( 48 = x^2 + 2x - 15 \)
\( \implies x^2 + 2x - 63 = 0 \)
\( \implies (x+9)(x-7) = 0 \).
This gives the roots as \( x = 7 \) or \( x = -9 \).
In simple words: Simplifying the fractions leads us to a simple quadratic equation that we can easily factor.

Exam Tip: Be careful with signs 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 equation, it must satisfy the equation. Substitute this value into the equation:
\( 3\left(\frac{2}{3}\right)^2 + P\left(\frac{2}{3}\right) + 4 = 0 \)
\( \implies 3\left(\frac{4}{9}\right) + \frac{2P}{3} + 4 = 0 \)
\( \implies \frac{4}{3} + \frac{2P}{3} + 4 = 0 \).
Multiply 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 root, we plug it in for x and solve the equation to find p.

Exam Tip: Plugging a given root back into the equation is the quickest way to find an unknown constant.

Question 13. If x = √2 is a solution of quadratic equation x2 + k x – 4 = 0, then find the value of k
Answer: Since \( x = \sqrt{2} \) is a root of the equation, it must satisfy it. Substitute this value into the equation:
\( (\sqrt{2})^2 + k(\sqrt{2}) - 4 = 0 \)
\( \implies 2 + k\sqrt{2} - 4 = 0 \)
\( \implies k\sqrt{2} - 2 = 0 \)
\( \implies k\sqrt{2} = 2 \)
\( \implies k = \frac{2}{\sqrt{2}} = \sqrt{2} \).
The value of k is \( \sqrt{2} \).
In simple words: We plug the root value in for x, which lets us solve the equation for the unknown constant k.

Exam Tip: Simplify the fraction \( \frac{2}{\sqrt{2}} \) to \( \sqrt{2} \) to present the final answer in its simplest form.

Question 14. Solve for x: 2 (2x-1)/(x+3) – 3 (x+3)/(2x-1) = 5
Answer: Let us substitute \( y = \frac{2x-1}{x+3} \). Its reciprocal is \( \frac{1}{y} = \frac{x+3}{2x-1} \).
The equation becomes:
\( 2y - \frac{3}{y} = 5 \)
\( \implies 2y^2 - 3 = 5y \)
\( \implies 2y^2 - 5y - 3 = 0 \).
Factoring this quadratic equation:
\( 2y^2 - 6y + y - 3 = 0 \)
\( \implies 2y(y - 3) + 1(y - 3) = 0 \)
\( \implies (2y + 1)(y - 3) = 0 \).
This gives \( y = 3 \) 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:
\( \frac{2x-1}{x+3} = -\frac{1}{2} \)
\( \implies 2(2x - 1) = -(x + 3) \)
\( \implies 4x - 2 = -x - 3 \)
\( \implies 5x = -1 \)
\( \implies x = -\frac{1}{5} \).
The roots are \( x = -10 \) and \( x = -\frac{1}{5} \).
In simple words: Substitution transforms a complicated fractional equation into a standard quadratic equation.

Exam Tip: Remember to solve for the original variable x at the end of the substitution process.

Question 15. Solve the equation: 2(x – 3)2 + 3(x – 2) (2x – 3) = 8(x + 4) (x – 4) - 1
Answer: Let us expand each term of the equation:
\( 2(x-3)^2 = 2(x^2 - 6x + 9) = 2x^2 - 12x + 18 \).
\( 3(x-2)(2x-3) = 3(2x^2 - 7x + 6) = 6x^2 - 21x + 18 \).
\( 8(x+4)(x-4) - 1 = 8(x^2 - 16) - 1 = 8x^2 - 128 - 1 = 8x^2 - 129 \).
Now substitute these expansions back into the equation:
\( (2x^2 - 12x + 18) + (6x^2 - 21x + 18) = 8x^2 - 129 \)
\( \implies 8x^2 - 33x + 36 = 8x^2 - 129 \).
Subtract \( 8x^2 \) from both sides:
\( -33x + 36 = -129 \)
\( \implies -33x = -165 \)
\( \implies x = 5 \).
The solution is \( x = 5 \).
In simple words: Expanding and combining the terms simplifies the equation into a simple linear equation.

Exam Tip: Watch for terms that cancel out, like \( 8x^2 \) here, which simplifies the quadratic equation into a simple linear one.

Question 16. 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:
\( (b-c)(1)^2 + (c-a)(1) + (a-b) = b - c + c - a + a - b = 0 \).
Since \( x = 1 \) satisfies the equation, it is a root.
Since the roots are equal, both roots must be equal to 1.
The product of the roots is:
\( \text{Product of roots} = \frac{a-b}{b-c} \).
Since both roots are 1, their product 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 the sum of the coefficients is zero, 1 is a root. Because the roots are equal, both roots are 1, which helps us prove the relation.

Exam Tip: Recognizing that the sum of the coefficients is zero is a very elegant way to find a root instantly.

Question 17. 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 the entire equation by 2:
\( x^2 + 2x - 195 = 0 \)
\( \implies (x+15)(x-13) = 0 \).
This gives \( x = 13 \) (rejecting negative integer \( -15 \) for standard positive odd numbers).
The two consecutive odd numbers are 13 and 15.
In simple words: We write the consecutive odd numbers as x and x+2, set up our equation based on their squares, and solve.

Exam Tip: If the problem does not specify "positive" integers, mention that the pair \( (-15, -13) \) is also a mathematically valid solution.

Question 18. Find two consecutive numbers, whose squares have the sum 85.
Answer: Let the two consecutive numbers be \( n \) and \( n + 1 \).
According to the problem:
\( n^2 + (n+1)^2 = 85 \)
\( \implies n^2 + n^2 + 2n + 1 = 85 \)
\( \implies 2n^2 + 2n - 84 = 0 \).
Dividing the entire equation by 2:
\( n^2 + n - 42 = 0 \)
\( \implies (n+7)(n-6) = 0 \).
Since they are consecutive positive integers, we choose \( n = 6 \).
The two consecutive numbers are 6 and 7.
In simple words: We write consecutive numbers as n and n+1, square them, and set up an equation to find the values.

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

Question 19. The product of 3 consecutive even numbers is equal to 20 times their sum. Find the numbers
Answer: Let the three consecutive even numbers be \( y - 2 \), \( y \), and \( y + 2 \), where \( y \) is an even integer.
Their sum is:
\( (y-2) + y + (y+2) = 3y \).
Their product is:
\( (y-2)(y)(y+2) = y(y^2 - 4) \).
According to the problem:
\( y(y^2 - 4) = 20(3y) \)
\( \implies y(y^2 - 4) = 60y \).
Since the numbers are positive even integers, \( y \neq 0 \). Dividing by \( y \) on both sides:
\( y^2 - 4 = 60 \)
\( \implies y^2 = 64 \)
\( \implies y = 8 \) (since \( y \) must be positive).
The three consecutive even numbers are \( 8 - 2 = 6 \), \( 8 \), and \( 8 + 2 = 10 \).
The numbers are 6, 8, and 10.
In simple words: Setting up the terms symmetrically as y-2, y, and y+2 makes the math much simpler to solve.

Exam Tip: Symmetrical term choices like \( y-2, y, y+2 \) simplify both the sum and the product calculations immensely.

Question 20. 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:
\( x^2 + y^2 = 640 \).
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:
\( y^2 + 16y - 192 = 0 \)
\( \implies (y+24)(y-8) = 0 \).
Since a side length cannot be negative, we reject \( y = -24 \). Therefore, \( y = 8 \) meters.
This gives \( x = 8 + 16 = 24 \) meters.
The sides of the squares are 8m and 24m.
In simple words: We find a relationship between the sides from the perimeter difference and use it to solve the area equation.

Exam Tip: Always state clearly that lengths cannot be negative when discarding negative roots in geometry problems.

Question 21. 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 \( y \), where \( x > y \).
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 both sides by 4:
\( y^2 + 4y = 21 \)
\( \implies y^2 + 4y - 21 = 0 \)
\( \implies (y+7)(y-3) = 0 \).
This gives \( y = 3 \) or \( y = -7 \).
If \( y = 3 \), then \( x = 7 \). (The numbers are 3 and 7).
If \( y = -7 \), then \( x = -3 \). (The numbers are -3 and -7).
The numbers are 3 and 7.
In simple words: Since the smaller number has the larger reciprocal, 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). Reversing this order is a very common mistake.

Question 22. The sum of two numbers is 15 and sum of their reciprocals is 3/10. Find the numbers
Answer: Let the two numbers be \( x \) and \( y \).
Their sum is:
\( x + y = 15 \)
\( \implies y = 15 - x \).
The sum of their reciprocals is:
\( \frac{1}{x} + \frac{1}{y} = \frac{3}{10} \)
\( \implies \frac{x+y}{xy} = \frac{3}{10} \).
Substitute \( x+y = 15 \) into the equation:
\( \frac{15}{xy} = \frac{3}{10} \)
\( \implies 3xy = 150 \)
\( \implies xy = 50 \).
Substitute \( y = 15-x \):
\( x(15 - x) = 50 \)
\( \implies 15x - x^2 = 50 \)
\( \implies x^2 - 15x + 50 = 0 \)
\( \implies (x-5)(x-10) = 0 \).
This gives \( x = 5 \) or \( x = 10 \).
The numbers are 5 and 10.
In simple words: We find the sum and product of the two numbers from the given clues, and solve the resulting quadratic equation.

Exam Tip: Substituting the sum directly into the numerator of the reciprocal sum fraction \( \frac{x+y}{xy} \) makes the calculation much easier.

Topic: Quadratic Equations (Continued)

Question 23. 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 grassy land be \( x \) meters.
According to the problem, the hypotenuse is \( 2x + 1 \) meters, and the third side is \( x + 7 \) meters.
Using Pythagoras' theorem:
\( (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \)
\( \implies (2x + 1)^2 = x^2 + (x + 7)^2 \)
\( \implies 4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49 \)
\( \implies 4x^2 + 4x + 1 = 2x^2 + 14x + 49 \)
\( \implies 2x^2 - 10x - 48 = 0 \).
Dividing by 2:
\( x^2 - 5x - 24 = 0 \)
\( \implies (x - 8)(x + 3) = 0 \).
Since a side length cannot be negative, we reject the root \( x = -3 \). 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 express all the sides of the triangle in terms of the shortest side, and then apply Pythagoras' theorem to calculate the side lengths.

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

Question 24. The perimeter of a right angled triangle is 70units and its hypotenuse is 29 units. Find the lengths of the other sides
Answer: Let the other two sides of the right-angled triangle be \( a \) and \( b \) units, and the hypotenuse be \( c = 29 \) units.
The perimeter of the triangle is 70 units:
\( a + b + c = 70 \)
\( \implies a + b + 29 = 70 \)
\( \implies a + b = 41 \)
\( \implies b = 41 - a \).
According to Pythagoras' theorem:
\( a^2 + b^2 = c^2 \)
\( \implies a^2 + (41 - a)^2 = 29^2 \)
\( \implies a^2 + 1681 - 82a + a^2 = 841 \)
\( \implies 2a^2 - 82a + 840 = 0 \).
Dividing by 2:
\( a^2 - 41a + 420 = 0 \)
\( \implies (a - 20)(a - 21) = 0 \).
This gives \( a = 20 \) or \( a = 21 \).
If \( a = 20 \), then \( b = 41 - 20 = 21 \) units.
If \( a = 21 \), then \( b = 41 - 21 = 20 \) units.
The lengths of the other sides are 20 units and 21 units.
In simple words: Using the perimeter, we write one side in terms of the other and then use Pythagoras' theorem to find both sides.

Exam Tip: Factoring quadratic equations like \( a^2 - 41a + 420 = 0 \) is much easier if you look for factors of 420 that add up to 41, which are 20 and 21.

Question 25. The length of the sides forming a right angled ∆ is 5x cm and (3x – 1) cm. Area of the triangle is 60 cm2. Find the hypotenuse
Answer: The sides forming the right angle are \( 5x \) cm and \( (3x - 1) \) cm. These represent the base and height of the right-angled triangle.
The area of a triangle is given by:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( \implies 60 = \frac{1}{2}(5x)(3x - 1) \)
\( \implies 120 = 5x(3x - 1) \)
\( \implies 120 = 15x^2 - 5x \)
\( \implies 15x^2 - 5x - 120 = 0 \).
Dividing by 5:
\( 3x^2 - x - 24 = 0 \)
\( \implies 3x^2 - 9x + 8x - 24 = 0 \)
\( \implies 3x(x - 3) + 8(x - 3) = 0 \)
\( \implies (3x + 8)(x - 3) = 0 \).
Since length must be positive, we reject \( x = -\frac{8}{3} \). Thus, \( x = 3 \).
The sides are:
\( 5(3) = 15 \) cm,
\( 3(3) - 1 = 8 \) cm.
Using Pythagoras' theorem to find the hypotenuse:
\( \text{Hypotenuse} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \) cm.
The hypotenuse of the triangle is 17 cm.
In simple words: We find x from the area formula, use it to get the two sides, and then calculate the hypotenuse using Pythagoras' theorem.

Exam Tip: Be sure to calculate the final hypotenuse value at the end instead of just solving for \( x \), as that is what the question asks for.

Question 26. The length of the hypotenuse of a right angled triangle exceeds the base by 1cm and also exceeds twice the length of the altitude by 3cm. Find the length of each side of ∆
Answer: Let the hypotenuse of the right-angled triangle be \( h \) cm.
According to the problem:
The hypotenuse exceeds the base by 1 cm:
\( h = \text{base} + 1 \)
\( \implies \text{base} = h - 1 \).
The hypotenuse also exceeds twice the altitude by 3 cm:
\( h = 2(\text{altitude}) + 3 \)
\( \implies \text{altitude} = \frac{h - 3}{2} \).
Using Pythagoras' theorem:
\( (\text{altitude})^2 + (\text{base})^2 = (\text{hypotenuse})^2 \)
\( \implies \left(\frac{h - 3}{2}\right)^2 + (h - 1)^2 = h^2 \)
\( \implies \frac{h^2 - 6h + 9}{4} + h^2 - 2h + 1 = h^2 \).
Multiplying the entire equation by 4 to clear the denominator:
\( (h^2 - 6h + 9) + 4(h^2 - 2h + 1) = 4h^2 \)
\( \implies h^2 - 6h + 9 + 4h^2 - 8h + 4 = 4h^2 \)
\( \implies 5h^2 - 14h + 13 = 4h^2 \)
\( \implies h^2 - 14h + 13 = 0 \)
\( \implies (h - 13)(h - 1) = 0 \).
If \( h = 1 \) cm, the altitude becomes negative: \( \frac{1-3}{2} = -1 \) cm, which is impossible. Thus, \( h = 13 \) cm.
The sides of the triangle are:
Hypotenuse = 13 cm
Base = \( 13 - 1 = 12 \) cm
Altitude = \( \frac{13 - 3}{2} = 5 \) cm.
In simple words: We express both the base and altitude in terms of the hypotenuse, and then use Pythagoras' theorem to find all the side lengths.

Exam Tip: Setting the hypotenuse as the main variable \( h \) makes the fractional calculations simpler than setting the altitude or base as \( x \).

Question 27. 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 \( n \) is a natural number (positive integer), we reject \( n = -20 \). Therefore, \( n = 8 \).
The required natural number is 8.
In simple words: We construct an equation where adding 12 to a number is the same as dividing 160 by it, and then solve the quadratic equation.

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

Question 28. 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.
In 1 day, B does \( \frac{1}{x} \) of the work, and A does \( \frac{1}{x-6} \) of the work.
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 29. 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 of the number 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 30. A two – digit number is such that the product of its digits is 14. When 45 is added to the number, the digits interchange their Places. Find the number
Answer: Let the tens digit of the number be \( x \) and the units digit be \( y \).
The product of the digits is:
\( xy = 14 \)
\( \implies y = \frac{14}{x} \).
The original number is \( 10x + y \).
When we reverse the digits, the new number is \( 10y + x \).
According to the given condition:
\( (10x + y) + 45 = 10y + x \)
\( \implies 9y - 9x = 45 \)
\( \implies y - x = 5 \).
Substitute \( y = \frac{14}{x} \) into this equation:
\( \frac{14}{x} - x = 5 \)
\( \implies 14 - x^2 = 5x \)
\( \implies x^2 + 5x - 14 = 0 \)
\( \implies (x + 7)(x - 2) = 0 \).
Since \( x \) is a single-digit positive number, we choose \( x = 2 \).
This gives \( y = \frac{14}{2} = 7 \).
The original number is 27.
In simple words: Representing the digits as x and y helps us create a system of equations, revealing the digits are 2 and 7, making the number 27.

Exam Tip: If \( y-x = 5 \), then the units digit is larger than the tens digit. This means adding 45 increases the value, confirming the solution makes sense.

Question 31. 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 32. 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 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. 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 average 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 35. 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 36. 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 37. 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 66 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):
\( y^2 + 5y = 66 \)
\( \implies y^2 + 5y - 66 = 0 \)
\( \implies (y + 11)(y - 6) = 0 \).
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: Solving the quadratic equation shows that the son is 6 years old and the father is 36.

Exam Tip: Always make sure your variable substitution is clear (e.g., son's age as \( y \) and father's age as \( y^2 \)) before setting up the equation.

Question 38. 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 39. 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 40. 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 41. Two pipes running together can fill a tank in 3 1/13 minutes. If one pipe takes 3 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately
Answer: Let the faster pipe take \( x \) minutes to fill the tank separately.
Then, the slower pipe takes \( x + 3 \) minutes to fill the tank separately.
Together, they fill the tank in \( 3 \frac{1}{13} = \frac{40}{13} \) minutes.
In 1 minute, the faster pipe fills \( \frac{1}{x} \), the slower fills \( \frac{1}{x+3} \), and together they fill \( \frac{13}{40} \) of the tank:
\( \frac{1}{x} + \frac{1}{x+3} = \frac{13}{40} \)
\( \implies \frac{x+3 + x}{x(x+3)} = \frac{13}{40} \)
\( \implies \frac{2x+3}{x^2+3x} = \frac{13}{40} \).
Cross-multiplying yields:
\( 40(2x + 3) = 13(x^2 + 3x) \)
\( \implies 80x + 120 = 13x^2 + 39x \)
\( \implies 13x^2 - 41x - 120 = 0 \).
Using factorization, let us split the middle term:
\( 13x^2 - 65x + 24x - 120 = 0 \)
\( \implies 13x(x - 5) + 24(x - 5) = 0 \)
\( \implies (13x + 24)(x - 5) = 0 \).
Since time cannot be negative, we reject \( x = -\frac{24}{13} \). Thus, \( x = 5 \) minutes.
The faster pipe takes 5 minutes, and the slower pipe takes \( 5 + 3 = 8 \) minutes to fill the tank separately.
In simple words: We calculate what fraction of the tank each pipe fills in one minute, add them together, and solve the quadratic equation to find the times.

Exam Tip: Be sure to write the mixed fraction \( 3 \frac{1}{13} \) as an improper fraction \( \frac{40}{13} \) first before taking its reciprocal as \( \frac{13}{40} \).

Question 42. 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.

Question 43. By increasing the list price of a book by Rs 10 a person can buy 10 less books for Rs 1200. Find the original list price of the book
Answer: Let the original list price of each book be Rs \( x \).
The number of books that can be purchased for Rs 1200 is \( \frac{1200}{x} \).
When the price increases by Rs 10, the new price is Rs \( x + 10 \), and the number of books becomes \( \frac{1200}{x + 10} \).
According to the problem:
\( \frac{1200}{x} - \frac{1200}{x + 10} = 10 \).
Dividing the entire equation by 10:
\( \frac{120}{x} - \frac{120}{x+10} = 1 \)
\( \implies 120 \left(\frac{(x+10) - x}{x(x+10)}\right) = 1 \)
\( \implies \frac{120 \times 10}{x^2 + 10x} = 1 \)
\( \implies x^2 + 10x = 1200 \)
\( \implies x^2 + 10x - 1200 = 0 \)
\( \implies (x + 40)(x - 30) = 0 \).
Since the price of a book cannot be negative, we reject \( x = -40 \). Therefore, \( x = 30 \).
The original list price of the book is Rs 30.
In simple words: If the book becomes more expensive, you can buy fewer books. We write this difference as a quadratic equation to find the original price.

Exam Tip: Always verify that the units of the final answer are clearly stated (e.g. Rs 30).

Question 44. One – fourth of a herd of camels was seen in the forest. Twice the square root of the herd gone to mountains and the Remaining 15 camels were seen on the bank of a river. Find the total number of camels
Answer: Let the total number of camels in the herd be \( x \).
According to the problem:
Camels seen in the forest = \( \frac{x}{4} \)
Camels gone to the mountains = \( 2\sqrt{x} \)
Camels on the river bank = \( 15 \).
The total sum of these groups must equal \( x \):
\( \frac{x}{4} + 2\sqrt{x} + 15 = x \).
Multiplying the entire equation by 4 to clear the fraction:
\( x + 8\sqrt{x} + 60 = 4x \)
\( \implies 3x - 8\sqrt{x} - 60 = 0 \).
Let us substitute \( y = \sqrt{x} \) (where \( y \geq 0 \)), so the equation becomes:
\( 3y^2 - 8y - 60 = 0 \)
\( \implies 3y^2 - 18y + 10y - 60 = 0 \)
\( \implies 3y(y - 6) + 10(y - 6) = 0 \)
\( \implies (3y + 10)(y - 6) = 0 \).
Since \( y \) must be positive, we reject \( y = -\frac{10}{3} \). Thus, \( y = 6 \).
This gives:
\( \sqrt{x} = 6 \)
\( \implies x = 36 \).
The total number of camels is 36.
In simple words: Substituting \( y = \sqrt{x} \) allows us to solve this tricky equation as a standard quadratic equation.

Exam Tip: Using substitution for radical equations (like setting \( y = \sqrt{x} \)) is a very reliable way to avoid complicated algebraic errors.

Question 45. 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 46. If ½ is the root of the equation x² + kx – 5/4 =0 then the value 0f k is
(a) 2
(b) – 2
(c) ¼
(d) ½
Answer: (a) 2
In simple words: Since 1/2 is a root, we plug it into the equation for x and solve to find that k is equal to 2.

Exam Tip: Substituting a given root directly into a quadratic equation is the fastest way to find any missing constant coefficient.

Question 47. Value of k for which the quadratic equation 2x² – kx + k = 0 has equal roots
(a) 0
(b) 4
(c) 8
(d) 0, 8
Answer: (d) 0, 8
In simple words: For equal roots, we set the discriminant \( b^2 - 4ac = 0 \). This gives a quadratic equation in k, which has solutions 0 and 8.

Exam Tip: Do not forget that \( k = 0 \) is a mathematically valid root for equal roots here, so the correct option is (d), not just (c).

Question 48. If the discriminant of the equation 6x² – b x+ 2 = 0 is 1, then the value of b is
(a) 7
(b) – 7
(c) ± 7
(d) ±√7
Answer: (c) ± 7
In simple words: The discriminant is given as \( b^2 - 4ac = 1 \). Plugging in the coefficients gives \( b^2 = 49 \), so b can be either positive or negative 7.

Exam Tip: When taking the square root of a positive number in discriminant problems, always remember to include both the positive and negative signs.

Question 49. One root of the quadratic equation 2x² – x + 1/8 = 0 , is ¼. The other root is
(a) 0
(b) ¼
(c) 1/8
(d) – ¼
Answer: (b) ¼
In simple words: The sum of the roots is equal to \( -b/a = 1/2 \). Since one root is 1/4, the other root must also be 1/4 to add up to 1/2.

Exam Tip: Using the sum of roots relation \( \alpha + \beta = -b/a \) is a incredibly fast shortcut to find a second root when the first is given.

Question 50. The roots of the equation x² - 3x – m(m+3) = 0 , where m is a constant , are
(a) m, m+3
(b) –m, m+3
(c) m, - (m+3)
(d) –m, -(m+3)
Answer: (b) –m, m+3
In simple words: We split the middle term \( -3x \) into \( -(m+3)x + mx \). Factorizing the equation reveals the roots are \( -m \) and \( m+3 \).

Exam Tip: For constant terms containing brackets like \( m(m+3) \), look to see how the factors \( m \) and \( m+3 \) relate to the middle coefficient \( 3 \).

Question 51. The quadratic equation ax² + bx + c = 0 has equal roots if
(a) b2 = 4ac
(b) b2 < 4ac
(c) b2 > 4ac
(d) b2 = ac
Answer: (a) b2 = 4ac
In simple words: Equal roots occur when the discriminant \( b^2 - 4ac \) is exactly equal to zero, which can be rewritten as \( b^2 = 4ac \).

Exam Tip: This is a standard identity question. Simply equate the discriminant to zero to choose the correct formula.

Question 1. For what value of p, are 2p-1, 7 and 3p three consecutive terms of an A.P?
Answer: Let the three consecutive terms of the A.P. be \( a_1 = 2p - 1 \), \( a_2 = 7 \), and \( a_3 = 3p \). Since these terms form an Arithmetic Progression, the difference between consecutive terms must be equal: \( a_2 - a_1 = a_3 - a_2 \)
\( \implies 7 - (2p - 1) = 3p - 7 \)
\( \implies 7 - 2p + 1 = 3p - 7 \)
\( \implies 8 - 2p = 3p - 7 \)
\( \implies 8 + 7 = 3p + 2p \)
\( \implies 15 = 5p \)
\( \implies p = 3 \) Thus, the value of p is 3.
In simple words: For three numbers to form an arithmetic sequence, the step from the first to the second must be the same as the step from the second to the third. Setting these steps equal helps us find that p must be 3.

Exam Tip: Remember that if three terms a, b, and c are in A.P., then \( 2b = a + c \). This relation is a very useful shortcut to solve for unknown variables quickly.

Question 2. Find the value of k, so that 3k + 7, 2k +5, 2k + 7 are in A.P
Answer: Let the given terms be \( a_1 = 3k + 7 \), \( a_2 = 2k + 5 \), and \( a_3 = 2k + 7 \). For these terms to be in A.P., the common difference must be constant: \( 2 \cdot a_2 = a_1 + a_3 \)
\( \implies 2(2k + 5) = (3k + 7) + (2k + 7) \)
\( \implies 4k + 10 = 5k + 14 \)
\( \implies 4k - 5k = 14 - 10 \)
\( \implies -k = 4 \)
\( \implies k = -4 \) Thus, the value of k is -4.
In simple words: The middle term is the average of the first and last terms. Solving this relation shows that k must be -4.

Exam Tip: Always double-check your answer by putting the value of the variable back into the original terms to verify they form a valid progression.

Question 3. If \( \frac{1}{x + 2} \), \( \frac{1}{x + 3} \) and \( \frac{1}{x + 5} \) are in A.P , find the value of x
Answer: Let the consecutive terms of the A.P. be \( a_1 = \frac{1}{x + 2} \), \( a_2 = \frac{1}{x + 3} \), and \( a_3 = \frac{1}{x + 5} \). For these terms to be in A.P., we have: \( 2 \cdot a_2 = a_1 + a_3 \)
\( \implies \frac{2}{x + 3} = \frac{1}{x + 2} + \frac{1}{x + 5} \)
\( \implies \frac{2}{x + 3} = \frac{(x + 5) + (x + 2)}{(x + 2)(x + 5)} \)
\( \implies \frac{2}{x + 3} = \frac{2x + 7}{x^2 + 7x + 10} \) Cross-multiplying: \( 2(x^2 + 7x + 10) = (2x + 7)(x + 3) \)
\( \implies 2x^2 + 14x + 20 = 2x^2 + 13x + 21 \)
\( \implies 14x - 13x = 21 - 20 \)
\( \implies x = 1 \) Note that if \( x = -3 \), the second denominator becomes zero, which is undefined. Therefore, the only valid solution is \( x = 1 \).
In simple words: Since twice the middle fraction equals the sum of the other two, we cross-multiply and solve the equation to find that x is 1.

Exam Tip: In questions involving rational terms, check that your solved variable does not make any of the original denominators equal to zero.

Question 4. How many two digit numbers are divisible by 7
Answer: The two-digit natural numbers start at 10 and end at 99. The first two-digit number divisible by 7 is 14, and the last is 98. These numbers form an A.P.: 14, 21, 28, ..., 98. Here, the first term is \( a = 14 \), the common difference is \( d = 7 \), and the last term is \( a_n = 98 \). Using the general term formula: \( a_n = a + (n - 1)d \)
\( \implies 98 = 14 + (n - 1)7 \)
\( \implies 84 = (n - 1)7 \)
\( \implies n - 1 = 12 \)
\( \implies n = 13 \) Thus, there are 13 two-digit numbers divisible by 7.
In simple words: The numbers start at 14 and go up by 7 until we reach 98. Counting them shows there are 13 such numbers.

Exam Tip: Correctly identifying the first term \( a \) and last term \( a_n \) is crucial. Always divide the boundary numbers by the divisor to find these values.

Question 5. Find the number of integers between 50 and 500 which are divisible by 7
Answer: The integers between 50 and 500 divisible by 7 form an A.P. starting with 56 (since \( 7 \times 8 = 56 \)) and ending with 497 (since \( 500 = 7 \times 71 + 3 \)). The sequence is: 56, 63, 70, ..., 497. Here, \( a = 56 \), \( d = 7 \), and \( a_n = 497 \). Using the general term formula: \( a_n = a + (n - 1)d \)
\( \implies 497 = 56 + (n - 1)7 \)
\( \implies 441 = (n - 1)7 \)
\( \implies n - 1 = 63 \)
\( \implies n = 64 \) Thus, there are 64 such integers.
In simple words: The first multiple of 7 after 50 is 56, and the last before 500 is 497. Setting up our formula, we find there are 64 numbers in total.

Exam Tip: Pay close attention to the word "between", which means the boundary numbers (50 and 500) are excluded from the count.

Question 6. Find the 15th term from the end of the A.P: 3, 5, 7,………... ,201
Answer: The given A.P. is 3, 5, 7, ..., 201. Here, the first term is \( a = 3 \), the common difference is \( d = 2 \), and the last term is \( L = 201 \). The formula for the nth term from the end of an A.P. is: \( a_n' = L - (n - 1)d \) To find the 15th term from the end: \( a_{15}' = 201 - (15 - 1)2 \)
\( \implies a_{15}' = 201 - 14(2) \)
\( \implies a_{15}' = 201 - 28 \)
\( \implies a_{15}' = 173 \) Thus, the 15th term from the end is 173.
In simple words: We start from the back at 201 and count backwards 14 steps of 2 each, giving us 173.

Exam Tip: You can also reverse the A.P. to become 201, 199, ..., 3. Here, the first term is 201 and the common difference is -2. Finding the 15th term of this reversed A.P. will yield the same result.

Question 7. Find the 11th term from the end of the A.P: 10, 7, 4,…….., - 62
Answer: The given A.P. is 10, 7, 4, ..., -62. Here, the common difference is \( d = 7 - 10 = -3 \), and the last term is \( L = -62 \). The formula for the nth term from the end is: \( a_n' = L - (n - 1)d \) To find the 11th term from the end: \( a_{11}' = -62 - (11 - 1)(-3) \)
\( \implies a_{11}' = -62 - 10(-3) \)
\( \implies a_{11}' = -62 + 30 \)
\( \implies a_{11}' = -32 \) Thus, the 11th term from the end is -32.
In simple words: Starting from the last number (-62) and moving 10 steps back in the opposite direction of the common difference yields -32.

Exam Tip: Be very careful with signs when subtracting a negative common difference: \( L - (n-1)d \) becomes a positive addition if \( d \) is negative.

Question 8. Find the middle term of A.P: 1, 8, 15, ……………, 505
Answer: The given A.P. is 1, 8, 15, ..., 505. Here, \( a = 1 \), \( d = 7 \), and the last term is \( a_n = 505 \). First, find the total number of terms, \( n \): \( a_n = a + (n - 1)d \)
\( \implies 505 = 1 + (n - 1)7 \)
\( \implies 504 = (n - 1)7 \)
\( \implies n - 1 = 72 \)
\( \implies n = 73 \) Since the total number of terms is odd, there is a single middle term: Middle term position = \( \frac{n + 1}{2} = \frac{73 + 1}{2} = 37\text{-th term} \) Now, find the value of the 37th term: \( a_{37} = a + 36d \)
\( \implies a_{37} = 1 + 36(7) \)
\( \implies a_{37} = 1 + 252 = 253 \) Thus, the middle term is 253.
In simple words: First, we find that the list has 73 numbers. The exact middle number is the 37th one, which is calculated to be 253.

Exam Tip: If \( n \) is odd, there is one middle term at position \( \frac{n+1}{2} \). If \( n \) is even, there are two middle terms at positions \( \frac{n}{2} \) and \( \frac{n}{2} + 1 \).

Question 9. In an A.P. the first term is 8 and the common difference is 7. If the last term of the A.P is 218, find its middle term
Answer: We are given: First term, \( a = 8 \) Common difference, \( d = 7 \) Last term, \( a_n = 218 \) Let's find the total number of terms, \( n \): \( a_n = a + (n - 1)d \)
\( \implies 218 = 8 + (n - 1)7 \)
\( \implies 210 = (n - 1)7 \)
\( \implies n - 1 = 30 \)
\( \implies n = 31 \) Since \( n = 31 \) is odd, the middle term is the \( \frac{31 + 1}{2} = 16\text{-th term} \). The value of the 16th term is: \( a_{16} = a + 15d \)
\( \implies a_{16} = 8 + 15(7) \)
\( \implies a_{16} = 8 + 105 = 113 \) Thus, the middle term is 113.
In simple words: The progression contains 31 numbers. The middle number is the 16th one, and its value is 113.

Exam Tip: Always show the step where you calculate the position of the middle term before computing its final numerical value.

Question 10. Which term of the sequence 114, 109, 104… is the first negative term?
Answer: The given sequence is 114, 109, 104, ... Here, first term \( a = 114 \), and common difference \( d = 109 - 114 = -5 \). Let the nth term be the first negative term. Thus, we require \( a_n < 0 \): \( a + (n - 1)d < 0 \)
\( \implies 114 + (n - 1)(-5) < 0 \)
\( \implies 114 - 5n + 5 < 0 \)
\( \implies 119 < 5n \)
\( \implies n > \frac{119}{5} \)
\( \implies n > 23.8 \) Since \( n \) must be a natural number, the smallest integer value for \( n \) is 24. Thus, the 24th term is the first negative term.
In simple words: Since the numbers decrease by 5 each time, we find when they will fall below zero. Solving the inequality shows this happens starting from the 24th number.

Exam Tip: Remember to reverse the inequality sign if you multiply or divide both sides by a negative number during calculations.

Question 11. Which term of the sequence 121, 117, 113… is the first negative term?
Answer: The given sequence is 121, 117, 113, ... Here, first term \( a = 121 \), and common difference \( d = 117 - 121 = -4 \). Let the nth term be the first negative term, so \( a_n < 0 \): \( a + (n - 1)d < 0 \)
\( \implies 121 + (n - 1)(-4) < 0 \)
\( \implies 121 - 4n + 4 < 0 \)
\( \implies 125 < 4n \)
\( \implies n > \frac{125}{4} \)
\( \implies n > 31.25 \) Since the term number \( n \) must be a natural number, the smallest integer value is 32. Thus, the 32nd term is the first negative term.
In simple words: The numbers decrease by 4 each step. Working out when they go below zero shows that the 32nd term is the first negative value.

Exam Tip: State the inequality clearly as \( a_n < 0 \) and solve it systematically to earn full marks for steps.

Question 12. 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: For the first A.P.: 9, 7, 5, ... First term \( a_1 = 9 \), common difference \( d_1 = 7 - 9 = -2 \). The nth term is: \( a_n = 9 + (n - 1)(-2) = 11 - 2n \) For the second A.P.: 15, 12, 9, ... First term \( a_2 = 15 \), common difference \( d_2 = 12 - 15 = -3 \). The nth term is: \( a_n' = 15 + (n - 1)(-3) = 18 - 3n \) Equating both nth terms: \( 11 - 2n = 18 - 3n \)
\( \implies -2n + 3n = 18 - 11 \)
\( \implies n = 7 \) Thus, the value of n is 7.
In simple words: We find the formula for the nth term of both lists, set them equal to each other, and solve to find that their 7th terms are the same.

Exam Tip: Formulating the expressions for both general terms separately before equating them prevents algebraic mix-ups.

Question 13. If the 3rd and 9thterm of an A.P. are 4 and -8 respectively, which term is zero
Answer: Let \( a \) be the first term and \( d \) be the common difference. We are given: \( a_3 = 4 \implies a + 2d = 4 \) (Equation 1) \( a_9 = -8 \implies a + 8d = -8 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( (a + 8d) - (a + 2d) = -8 - 4 \)
\( \implies 6d = -12 \)
\( \implies d = -2 \) Substituting \( d = -2 \) in Equation 1: \( a + 2(-2) = 4 \)
\( \implies a - 4 = 4 \)
\( \implies a = 8 \) Now, let the nth term be zero: \( a_n = 0 \)
\( \implies a + (n - 1)d = 0 \)
\( \implies 8 + (n - 1)(-2) = 0 \)
\( \implies -2(n - 1) = -8 \)
\( \implies n - 1 = 4 \)
\( \implies n = 5 \) Thus, the 5th term is zero.
In simple words: Using the 3rd and 9th terms, we calculate that our sequence starts at 8 and decreases by 2 each time. It takes 5 steps to reach zero.

Exam Tip: Solving simultaneous equations to find the values of \( a \) and \( d \) is a standard procedure that appears frequently in board exams.

Question 14. Determine the A.P whose 3rd term is 16 and 7th term exceeds the 5th term by 12
Answer: Let \( a \) be the first term and \( d \) be the common difference. We are given: \( a_3 = 16 \implies a + 2d = 16 \) (Equation 1) According to the second condition: \( a_7 - a_5 = 12 \)
\( \implies (a + 6d) - (a + 4d) = 12 \)
\( \implies 2d = 12 \)
\( \implies d = 6 \) Substitute \( d = 6 \) in Equation 1: \( a + 2(6) = 16 \)
\( \implies a + 12 = 16 \)
\( \implies a = 4 \) The first term is 4 and the common difference is 6. Thus, the A.P. is 4, 10, 16, 22, ...
In simple words: The 7th term being 12 larger than the 5th term tells us that two steps of the progression equal 12, so each step is 6. This helps us find the starting term of 4.

Exam Tip: Expressing the differences between terms directly as multiples of \( d \) (e.g., \( a_7 - a_5 = 2d \)) is much faster than writing out the full formulas.

Question 15. 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 \( a \) be the first term and \( d \) be the common difference. From the first condition: \( a_4 = 3a \)
\( \implies a + 3d = 3a \)
\( \implies 2a = 3d \)
\( \implies a = \frac{3}{2}d \) (Equation 1) From the second condition: \( a_7 = 2 \cdot a_3 + 1 \)
\( \implies a + 6d = 2(a + 2d) + 1 \)
\( \implies a + 6d = 2a + 4d + 1 \)
\( \implies 2d - 1 = a \) (Equation 2) Equating both expressions for \( a \): \( \frac{3}{2}d = 2d - 1 \)
\( \implies 3d = 4d - 2 \)
\( \implies d = 2 \) Using Equation 1 to find \( a \): \( a = \frac{3}{2}(2) = 3 \) The A.P. is 3, 5, 7, 9, ...
In simple words: We write equations using the clues provided. Solving them together shows the sequence starts at 3 and increases by 2 each time.

Exam Tip: Be careful when converting verbal statements like "exceeds twice the third term by 1" into the equation \( a_7 = 2a_3 + 1 \).

Question 16. 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 \( a \) be the first term and \( d \) be the common difference. From the first sum: \( a_4 + a_8 = 24 \)
\( \implies (a + 3d) + (a + 7d) = 24 \)
\( \implies 2a + 10d = 24 \)
\( \implies a + 5d = 12 \) (Equation 1) From the second sum: \( a_6 + a_{10} = 44 \)
\( \implies (a + 5d) + (a + 9d) = 44 \)
\( \implies 2a + 14d = 44 \)
\( \implies a + 7d = 22 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( (a + 7d) - (a + 5d) = 22 - 12 \)
\( \implies 2d = 10 \)
\( \implies d = 5 \) Substitute \( d = 5 \) in Equation 1: \( a + 5(5) = 12 \)
\( \implies a + 25 = 12 \)
\( \implies a = -13 \) The A.P. is -13, -8, -3, 2, ...
In simple words: Adding the given terms helps us create two simple equations. Solving them tells us the starting number is -13 and we add 5 at each step.

Exam Tip: Always divide equations by their greatest common divisor (like dividing \( 2a + 10d = 24 \) by 2) to simplify the calculations.

Question 17. Find the A.P whose nth term is 10 – 3n
Answer: The nth term of the A.P. is given by \( a_n = 10 - 3n \). To find the terms of the A.P., we substitute \( n = 1, 2, 3, \dots \): For \( n = 1 \): \( a_1 = 10 - 3(1) = 7 \) For \( n = 2 \): \( a_2 = 10 - 3(2) = 4 \) For \( n = 3 \): \( a_3 = 10 - 3(3) = 1 \) For \( n = 4 \): \( a_4 = 10 - 3(4) = -2 \) Thus, the A.P. is 7, 4, 1, -2, ...
In simple words: To find the sequence, we substitute 1, 2, 3, and so on, in place of n. This gives us the numbers 7, 4, 1, and so on.

Exam Tip: The coefficient of \( n \) in any linear expression for the general term of an A.P. represents its common difference. Here, the common difference is -3.

Question 18. Determine the 2nd term and nth term of an A.P whose 6th term is 12 and 8th term is 22
Answer: Let \( a \) be the first term and \( d \) be the common difference. We are given: \( a_6 = 12 \implies a + 5d = 12 \) (Equation 1) \( a_8 = 22 \implies a + 7d = 22 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 2d = 10 \implies d = 5 \) Substitute \( d = 5 \) in Equation 1: \( a + 5(5) = 12 \)
\( \implies a + 25 = 12 \implies a = -13 \) Now, find the 2nd term: \( a_2 = a + d = -13 + 5 = -8 \) Now, find the nth term: \( a_n = a + (n - 1)d \)
\( \implies a_n = -13 + (n - 1)5 \)
\( \implies a_n = -13 + 5n - 5 = 5n - 18 \) Thus, the 2nd term is -8 and the nth term is \( 5n - 18 \) (or \( -18 + 5n \)).
In simple words: We find that our sequence starts at -13 and goes up by 5 each time. This helps us find that the second term is -8 and write the general expression.

Exam Tip: Always write the final expression for the nth term in its simplest form \( An + B \).

Question 19. If 6 times the sixth term of an A.P is equal to 15 times the fifteenth term, find its 21st term
Answer: Let \( a \) be the first term and \( d \) be the common difference. According to the given condition: \( 6 \cdot a_6 = 15 \cdot a_{15} \)
\( \implies 6(a + 5d) = 15(a + 14d) \) Divide both sides by 3: \( 2(a + 5d) = 5(a + 14d) \)
\( \implies 2a + 10d = 5a + 70d \)
\( \implies 3a + 60d = 0 \) Divide both sides by 3: \( a + 20d = 0 \) Since the 21st term of an A.P. is \( a_{21} = a + 20d \), we have: \( a_{21} = 0 \) Thus, the 21st term is 0.
In simple words: Simplifying the relation given in the question shows that the formula for the 21st term is equal to 0.

Exam Tip: A useful shortcut is that if \( m \cdot a_m = n \cdot a_n \) for an A.P., then the \( (m+n) \)-th term is always 0.

Question 20. Which term of the A.P.? 3, 15, 27, 39, will be 120 more than its 21st term
Answer: The given A.P. is 3, 15, 27, 39, ... Here, first term \( a = 3 \), and common difference \( d = 15 - 3 = 12 \). The 21st term is: \( a_{21} = a + 20d = 3 + 20(12) = 3 + 240 = 243 \) We want to find the term \( a_n \) that is 120 more than \( a_{21} \): \( a_n = 243 + 120 = 363 \) Using the general term formula: \( a + (n - 1)d = 363 \)
\( \implies 3 + (n - 1)12 = 363 \)
\( \implies (n - 1)12 = 360 \)
\( \implies n - 1 = 30 \)
\( \implies n = 31 \) Thus, the 31st term is 120 more than the 21st term.
In simple words: The 21st term is 243. Adding 120 gives 363. Calculating which position in the sequence has the value of 363 gives us the 31st term.

Exam Tip: Alternatively, since each step increases the term by \( d = 12 \), adding 120 is equivalent to taking \( \frac{120}{12} = 10 \) more steps. Thus, the term is \( 21 + 10 = 31 \)-st term.

Question 21. Show that progression 7, 2, -3, -8,…..… Is an A.P. Find its nth term
Answer: The given progression is 7, 2, -3, -8, ... Let's 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 differences between consecutive terms are constant, the progression is an A.P. with first term \( a = 7 \) and common difference \( d = -5 \). The nth term is: \( a_n = a + (n - 1)d \)
\( \implies a_n = 7 + (n - 1)(-5) \)
\( \implies a_n = 7 - 5n + 5 = 12 - 5n \) Thus, the nth term is \( 12 - 5n \).
In simple words: The numbers always decrease by exactly 5, which proves it is an arithmetic progression. Its general formula is 12 - 5n.

Exam Tip: To prove a progression is an A.P., show that \( a_{k+1} - a_k \) remains constant for at least two consecutive pairs of terms.

Question 22. Verify that a + b, (a + 1) + b, (a + 1) + (b + 1) ……….. Is an A.P. and then write its next term
Answer: Let the terms be: \( t_1 = a + b \) \( t_2 = (a + 1) + b = a + b + 1 \) \( t_3 = (a + 1) + (b + 1) = a + b + 2 \) Now, find the differences: \( t_2 - t_1 = (a + b + 1) - (a + b) = 1 \) \( t_3 - t_2 = (a + b + 2) - (a + b + 1) = 1 \) Since the difference is constant (\( d = 1 \)), the given progression is an A.P. The next term (4th term) is: \( t_4 = t_3 + d = (a + b + 2) + 1 = a + b + 3 \) This can be written in the same pattern as: \( t_4 = (a + 2) + (b + 1) \)
In simple words: Each term increases by exactly 1 compared to the previous term. This means it is an AP, and the next term is (a + 2) + (b + 1).

Exam Tip: Group variables in brackets to match the visual pattern shown in the question for the final term.

Question 23. In the following A.P. find the missing term: *, 38, *, *, *, -22
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \). We are given: \( a_2 = 38 \implies a + d = 38 \) (Equation 1) \( a_6 = -22 \implies a + 5d = -22 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 4d = -60 \implies d = -15 \) Substitute \( d = -15 \) in Equation 1: \( a + (-15) = 38 \implies a = 53 \) Now, we find the missing terms: 1st term: \( a_1 = 53 \) 3rd term: \( a_3 = a_2 + d = 38 - 15 = 23 \) 4th term: \( a_4 = a_3 + d = 23 - 15 = 8 \) 5th term: \( a_5 = a_4 + d = 8 - 15 = -7 \) The complete sequence is: 53, 38, 23, 8, -7, -22.
In simple words: Using the 2nd and 6th numbers, we find that the sequence starts at 53 and goes down by 15 each time. This allows us to fill in the gaps.

Exam Tip: List out all the missing values explicitly at the end of your answer to make it easy for the examiner to award full marks.

Question 24. For A.P. a1, a2, a3… if a4/a7 = 2/3 , find a6/a8
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P. We are given: \( \frac{a_4}{a_7} = \frac{2}{3} \)
\( \implies \frac{a + 3d}{a + 6d} = \frac{2}{3} \) Cross-multiplying: \( 3(a + 3d) = 2(a + 6d) \)
\( \implies 3a + 9d = 2a + 12d \)
\( \implies a = 3d \) (Equation 1) Now, find the ratio of \( a_6 \) and \( a_8 \): \( \frac{a_6}{a_8} = \frac{a + 5d}{a + 7d} \) Substituting \( a = 3d \) from Equation 1: \( \frac{a_6}{a_8} = \frac{3d + 5d}{3d + 7d} = \frac{8d}{10d} = \frac{4}{5} \) Thus, the ratio is 4/5.
In simple words: The given ratio helps us find that the starting term is equal to 3 times the step size. Substituting this into the final ratio gives us 4/5.

Exam Tip: Expressing one variable in terms of the other is a robust algebraic method when dealing with rational ratios.

Question 25. The angles of a triangle are in A.P, the last being half the greatest. Find the angles.
Answer: Let the angles of the triangle in A.P. be \( a - d \), \( a \), and \( a + d \). The sum of the angles in a triangle is \( 180^\circ \): \( (a - d) + a + (a + d) = 180^\circ \)
\( \implies 3a = 180^\circ \)
\( \implies a = 60^\circ \) The angles are \( 60^\circ - d \), \( 60^\circ \), and \( 60^\circ + d \). The smallest (last) angle is half the greatest angle: \( 60^\circ - d = \frac{1}{2}(60^\circ + d) \)
\( \implies 120^\circ - 2d = 60^\circ + d \)
\( \implies 3d = 60^\circ \implies d = 20^\circ \) The angles are: \( 60^\circ - 20^\circ = 40^\circ \), \( 60^\circ \), and \( 60^\circ + 20^\circ = 80^\circ \). Thus, the angles are \( 40^\circ \), \( 60^\circ \), and \( 80^\circ \).
In simple words: Since the angles form an A.P. and sum to 180, the middle angle is 60. Using the other clue, we find the step is 20, giving us 40, 60, and 80 degrees.

Exam Tip: Using \( a - d \), \( a \), and \( a + d \) for three terms in A.P. simplifies equations involving sums, as \( d \) cancels out immediately.

Question 26. The sum of 3 numbers in A.P is 3 and their product is -35. Find the numbers
Answer: Let the three numbers in A.P. be \( a - d \), \( a \), and \( a + d \). According to the first condition: \( (a - d) + a + (a + d) = 3 \)
\( \implies 3a = 3 \implies a = 1 \) According to the second condition: \( (a - d) \cdot a \cdot (a + d) = -35 \) Substituting \( a = 1 \): \( (1 - d)(1)(1 + d) = -35 \)
\( \implies 1 - d^2 = -35 \)
\( \implies d^2 = 36 \implies d = \pm 6 \) If \( d = 6 \), the numbers are: \( 1 - 6 = -5 \), \( 1 \), and \( 1 + 6 = 7 \). If \( d = -6 \), the numbers are: 7, 1, and -5. Thus, the three numbers are -5, 1, and 7.
In simple words: We find the middle number is 1 from the sum. The product condition tells us that the step size is 6, which gives the numbers as -5, 1, and 7.

Exam Tip: Remember to write both positive and negative values when taking a square root (\( d = \pm 6 \)), even though they result in the same set of numbers.

Question 27. 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 \( a \) be the first term and \( d \) be the common difference. We are given: \( a_4 = 2 \cdot a_8 \)
\( \implies a + 3d = 2(a + 7d) \)
\( \implies a + 3d = 2a + 14d \)
\( \implies a = -11d \) (Equation 1) Now, find the 10th term: \( a_{10} = a + 9d \) Substituting \( a = -11d \): \( a_{10} = -11d + 9d = -2d \) (Equation 2) Next, find the 11th term: \( a_{11} = a + 10d \) Substituting \( a = -11d \): \( a_{11} = -11d + 10d = -d \) (Equation 3) Comparing Equation 2 and Equation 3: \( a_{10} = -2d = 2(-d) = 2 \cdot a_{11} \) Thus, the 10th term is twice the 11th term. Hence proved.
In simple words: The clue about the 4th and 8th terms tells us the starting term is -11 times the step size. Using this, we find the 10th term is double the 11th.

Exam Tip: Clearly write down "Hence Proved" at the end of verification questions to signal a complete solution.

Question 28. Find a30 – a20 for the A.P : -9, -14, -19, -24, ……………
Answer: The given A.P. is -9, -14, -19, -24, ... Here, first term \( a = -9 \) and common difference \( d = -14 - (-9) = -5 \). We need to find \( a_{30} - a_{20} \): \( a_{30} - a_{20} = (a + 29d) - (a + 19d) \)
\( \implies a_{30} - a_{20} = 10d \) Substituting the value of \( d = -5 \): \( a_{30} - a_{20} = 10(-5) = -50 \) Thus, the value of \( a_{30} - a_{20} \) is -50.
In simple words: The gap between the 30th and 20th terms is exactly 10 times the step size. Since we go down by 5 each time, the difference is -50.

Exam Tip: Avoid calculating individual values for \( a_{30} \) and \( a_{20} \). Subtracting their general formulas directly saves valuable time and prevents errors.

Question 29. If the nth term of an A.P is (5n – 2), find its first term and common difference
Answer: The general term is \( a_n = 5n - 2 \). To find the first term (\( a_1 \)), substitute \( n = 1 \): \( a_1 = 5(1) - 2 = 3 \) To find the second term (\( a_2 \)), substitute \( n = 2 \): \( a_2 = 5(2) - 2 = 8 \) The common difference (\( d \)) is: \( d = a_2 - a_1 = 8 - 3 = 5 \) Thus, the first term is 3 and the common difference is 5.
In simple words: Placing 1 and 2 in the formula gives the first two numbers as 3 and 8. The difference between them is 5.

Exam Tip: For any linear nth term expression \( An + B \), the common difference is always the coefficient \( A \), and the first term is \( A + B \).

Question 30. 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 \( a \) be the first term and \( d \) be the common difference. We are given: \( a_6 = 35 \implies a + 5d = 35 \) (Equation 1) \( a_{13} = 70 \implies a + 12d = 70 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 7d = 35 \implies d = 5 \) Substitute \( d = 5 \) in Equation 1: \( a + 5(5) = 35 \implies a = 10 \) Now, find the sum of the first 20 terms (\( S_{20} \)): \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \implies S_{20} = \frac{20}{2} [2(10) + (20 - 1)5] \)
\( \implies S_{20} = 10 [20 + 19(5)] \)
\( \implies S_{20} = 10 [20 + 95] \)
\( \implies S_{20} = 10(115) = 1150 \) Thus, the sum is 1150.
In simple words: We find our sequence starts at 10 and increases by 5 each time. Adding the first 20 terms using the formula gives 1,150.

Exam Tip: Always state both the basic formulas used (general term and sum of n terms) before substituting values to ensure step-wise marking.

Question 31. In an A.P., if the sum of its 4th and 10th terms is 40, and sum of its 8th and 16th terms is 70, then find the sum of its First20 terms
Answer: Let \( a \) be the first term and \( d \) be the common difference. From the first sum: \( a_4 + a_{10} = 40 \)
\( \implies (a + 3d) + (a + 9d) = 40 \)
\( \implies 2a + 12d = 40 \)
\( \implies a + 6d = 20 \) (Equation 1) From the second sum: \( a_8 + a_{16} = 70 \)
\( \implies (a + 7d) + (a + 15d) = 70 \)
\( \implies 2a + 22d = 70 \)
\( \implies a + 11d = 35 \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 5d = 15 \implies d = 3 \) Substitute \( d = 3 \) in Equation 1: \( a + 6(3) = 20 \implies a = 2 \) Now, find the sum of the first 20 terms (\( S_{20} \)): \( S_{20} = \frac{20}{2} [2a + 19d] \)
\( \implies S_{20} = 10 [2(2) + 19(3)] \)
\( \implies S_{20} = 10 [4 + 57] \)
\( \implies S_{20} = 10(61) = 610 \) Thus, the sum is 610.
In simple words: The given clues help us find that the sequence starts at 2 and increases by 3 each time. Adding up the first 20 numbers gives 610.

Exam Tip: Simplify your simultaneous equations by dividing by 2 to make subtraction steps easier.

Question 32. 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 \( a = 25 \), nth term \( a_n = -17 \), and sum \( S_n = 60 \). Using the sum formula: \( S_n = \frac{n}{2} (a + a_n) \)
\( \implies 60 = \frac{n}{2} (25 - 17) \)
\( \implies 60 = \frac{n}{2} (8) \)
\( \implies 4n = 60 \implies n = 15 \) Now, find \( d \) using the general term formula for the 15th term: \( a_{15} = a + 14d \)
\( \implies -17 = 25 + 14d \)
\( \implies -42 = 14d \implies d = -3 \) Thus, \( n = 15 \) and \( d = -3 \).
In simple words: By using the sum formula with the first and last terms, we find there are 15 terms. Then, we calculate the common difference is -3.

Exam Tip: Using the formula \( S_n = \frac{n}{2}(a + a_n) \) is the fastest way to find \( n \) when both boundaries are known.

Question 33. 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 first n terms is \( S_n = 3n^2 - 4n \). The nth term \( a_n \) is found by: \( a_n = S_n - S_{n-1} \) 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, find \( a_n \): \( a_n = (3n^2 - 4n) - (3n^2 - 10n + 7) \)
\( \implies a_n = 3n^2 - 4n - 3n^2 + 10n - 7 = 6n - 7 \) Thus, the nth term is \( 6n - 7 \).
In simple words: To find the formula for the nth term, we subtract the sum of \( n-1 \) terms from the sum of \( n \) terms. This simplifies to \( 6n - 7 \).

Exam Tip: You can verify your general term formula by checking if \( a_1 = S_1 \). Here, \( a_1 = 6(1) - 7 = -1 \) and \( S_1 = 3(1)^2 - 4(1) = -1 \), confirming accuracy.

Question 34. The sum of n terms of an A.P. is 3n2 + 5n. Find the A.P. Hence, find its 16th term
Answer: Given the sum of n terms: \( S_n = 3n^2 + 5n \) The nth term is: \( a_n = S_n - S_{n-1} \)
\( \implies a_n = (3n^2 + 5n) - [3(n - 1)^2 + 5(n - 1)] \)
\( \implies a_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5] \)
\( \implies a_n = 3n^2 + 5n - [3n^2 - n - 2] \)
\( \implies a_n = 6n + 2 \) To find the A.P., substitute \( n = 1, 2, 3 \): \( a_1 = 6(1) + 2 = 8 \) \( a_2 = 6(2) + 2 = 14 \) \( a_3 = 6(3) + 2 = 20 \) The A.P. is 8, 14, 20, ... Now, find the 16th term: \( a_{16} = 6(16) + 2 = 96 + 2 = 98 \)
In simple words: We find the general term formula is \( 6n + 2 \). This gives the sequence as 8, 14, 20... and tells us the 16th term is 98.

Exam Tip: When the question says "Hence", use the general formula for \( a_n \) to calculate the 16th term directly instead of using separate calculations.

Question 35. Find the sum of n terms of an A.P whose nth term is given by tn = 5 – 6n
Answer: The nth term is \( t_n = 5 - 6n \). The first term \( a \) is found by substituting \( n = 1 \): \( a = t_1 = 5 - 6(1) = -1 \) Using the sum formula: \( S_n = \frac{n}{2} (a + t_n) \)
\( \implies S_n = \frac{n}{2} [-1 + (5 - 6n)] \)
\( \implies S_n = \frac{n}{2} [4 - 6n] \)
\( \implies S_n = n(2 - 3n) = 2n - 3n^2 \) Thus, the sum is \( 2n - 3n^2 \).
In simple words: The first number is -1. Using the formula with the first and last terms, we calculate the sum of n terms as \( 2n - 3n^2 \).

Exam Tip: Factoring out common factors inside the brackets before multiplying by \( n \) makes the final expression simpler and easier to write.

Question 36. Find the sum of all natural numbers less than 100 which are divisible by 6
Answer: The natural numbers less than 100 divisible by 6 are: 6, 12, 18, ..., 96. This forms an A.P. with \( a = 6 \), \( d = 6 \), and \( a_n = 96 \). Find \( n \) using the general term formula: \( a_n = a + (n - 1)d \)
\( \implies 96 = 6 + (n - 1)6 \)
\( \implies 90 = (n - 1)6 \implies n - 1 = 15 \implies n = 16 \) Now, find the sum of these 16 terms: \( S_{16} = \frac{16}{2} (a + a_n) \)
\( \implies S_{16} = 8(6 + 96) \)
\( \implies S_{16} = 8(102) = 816 \) Thus, the sum is 816.
In simple words: The numbers divisible by 6 are 6, 12 up to 96. There are 16 such numbers, and adding them all up gives 816.

Exam Tip: Be sure to write out the first few and the last term of the sequence to show the examiner you have identified the bounds correctly.

Question 37. Find the sum of 3 digit numbers which are not divisible by 7
Answer: We can find this sum by subtracting the sum of three-digit numbers divisible by 7 from the total sum of all three-digit numbers. 1. Sum of all three-digit numbers (100 to 999): Here, first term \( a = 100 \), last term \( L = 999 \), and total terms \( n = 999 - 100 + 1 = 900 \). \( S_{\text{all}} = \frac{900}{2} (100 + 999) = 450(1099) = 494550 \) 2. Sum of three-digit numbers divisible by 7: First number divisible by 7 is 105 (since \( 7 \times 15 = 105 \)). Last number divisible by 7 is 994 (since \( 7 \times 142 = 994 \)). This forms an A.P.: 105, 112, ..., 994. Find the number of terms \( m \): \( 994 = 105 + (m - 1)7 \)
\( \implies 889 = (m - 1)7 \implies m - 1 = 127 \implies m = 128 \) Sum of these numbers: \( S_{\text{div } 7} = \frac{128}{2} (105 + 994) = 64(1099) = 70336 \) 3. Difference: \( S = S_{\text{all}} - S_{\text{div } 7} = 494550 - 70336 = 424214 \) Thus, the sum is 424214.
In simple words: We find the sum of all three-digit numbers (494,550) and subtract the sum of those divisible by 7 (70,336). The final answer is 424,214.

Exam Tip: Calculating the non-divisible numbers indirectly by subtraction is much faster and more accurate than trying to sum them directly.

Question 38. Find the sum of all the natural numbers upto 100, which are not divisible by 5
Answer: We can solve this by subtracting the sum of multiples of 5 from the sum of all natural numbers up to 100. 1. Sum of all natural numbers from 1 to 100: \( S_{\text{all}} = \frac{100(101)}{2} = 5050 \) 2. Sum of numbers divisible by 5 (5, 10, ..., 100): These form an A.P. with first term \( a = 5 \), last term \( L = 100 \), and number of terms \( n = 20 \). \( S_{\text{div } 5} = \frac{20}{2} (5 + 100) = 10(105) = 1050 \) 3. Difference: \( S = S_{\text{all}} - S_{\text{div } 5} = 5050 - 1050 = 4000 \) Thus, the sum of natural numbers up to 100 not divisible by 5 is 4000.
In simple words: The sum of all numbers 1 to 100 is 5,050. Subtracting the sum of multiples of 5 (1,050) leaves us with 4,000.

Exam Tip: Practice both the formula \( \frac{n(n+1)}{2} \) for natural numbers and general A.P. sum formulas to write fast answers.

Question 39. Find the sum of all three digit numbers which leave the remainder 3 when divided by 5
Answer: Three-digit numbers start at 100 and end at 999. The first three-digit number leaving a remainder of 3 when divided by 5 is 103. The last three-digit number leaving a remainder of 3 when divided by 5 is 998. These numbers form an A.P.: 103, 108, 113, ..., 998. Here, first term \( a = 103 \), common difference \( d = 5 \), and last term \( a_n = 998 \). Find the number of terms, \( n \): \( a_n = a + (n - 1)d \)
\( \implies 998 = 103 + (n - 1)5 \)
\( \implies 895 = (n - 1)5 \implies n - 1 = 179 \implies n = 180 \) Now, find the sum of these 180 terms: \( S_{180} = \frac{180}{2} (a + a_n) \)
\( \implies S_{180} = 90(103 + 998) \)
\( \implies S_{180} = 90(1101) = 99090 \) Thus, the sum is 99090.
In simple words: The numbers start at 103 and increase by 5 up to 998. There are 180 such numbers, and adding them up gives a total of 99,090.

Exam Tip: Be sure to write down the steps of calculating the first and last terms. Dividing 100 and 999 by 5 helps determine these boundaries.

Question 40. Find the sum of first seven multiples of 5
Answer: The first seven multiples of 5 are: 5, 10, 15, 20, 25, 30, and 35. This forms an A.P. with first term \( a = 5 \), last term \( a_n = 35 \), and number of terms \( n = 7 \). Using the sum formula: \( S_7 = \frac{7}{2} (a + a_n) \)
\( \implies S_7 = \frac{7}{2} (5 + 35) \)
\( \implies S_7 = \frac{7}{2} (40) = 7 \times 20 = 140 \) Thus, the sum is 140.
In simple words: Listing the first 7 multiples of 5 and adding them together using the formula gives 140.

Exam Tip: For simple series with very few terms, you can add them directly to cross-verify your calculated answer quickly.

Question 41. Find the sum of all natural numbers up to 100, which are not divisible by 5
Answer: We can solve this by subtracting the sum of numbers divisible by 5 from the sum of all natural numbers up to 100. 1. Sum of all natural numbers from 1 to 100: \( S_{\text{all}} = \frac{100(101)}{2} = 5050 \) 2. Sum of numbers divisible by 5 (5, 10, ..., 100): These form an A.P. with \( a = 5 \), \( L = 100 \), and \( n = 20 \). \( S_{\text{div } 5} = \frac{20}{2} (5 + 100) = 10(105) = 1050 \) 3. Difference: \( S = 5050 - 1050 = 4000 \) Thus, the sum is 4000.
In simple words: We take the total sum of all numbers up to 100 (5,050) and subtract the sum of multiples of 5 (1,050) to get 4,000.

Exam Tip: This is a repeat of Question 38 in the worksheet. Confirm your steps are exactly the same to maintain consistency.

Question 42. If 2 + 5 + 8 + …………………………+ x = 155, find x
Answer: The given series is 2 + 5 + 8 + ... + x = 155. This is an A.P. with first term \( a = 2 \), common difference \( d = 5 - 2 = 3 \), and sum \( S_n = 155 \). Using the sum formula: \( S_n = \frac{n}{2} [2a + (n - 1)d] = 155 \)
\( \implies \frac{n}{2} [2(2) + (n - 1)3] = 155 \)
\( \implies n[4 + 3n - 3] = 310 \)
\( \implies n[3n + 1] = 310 \)
\( \implies 3n^2 + n - 310 = 0 \) Solving this quadratic equation: \( 3n^2 - 30n + 31n - 310 = 0 \)
\( \implies 3n(n - 10) + 31(n - 10) = 0 \)
\( \implies (n - 10)(3n + 31) = 0 \) Since the number of terms \( n \) must be a positive integer: \( n = 10 \) The last term \( x \) is the 10th term (\( a_{10} \)): \( x = a_{10} = a + 9d = 2 + 9(3) = 2 + 27 = 29 \) Thus, the value of x is 29.
In simple words: We find that there are 10 terms in the series by solving a quadratic equation. The 10th term, which is x, is calculated to be 29.

Exam Tip: Be careful with factorisation when solving quadratic equations in A.P. problems. Only positive integers are acceptable values for the number of terms \( n \).

Question 43. Find the sum of the following A.P: 1 + 3 + 5 + …….. + 199.
Answer: The given A.P. is 1, 3, 5, ..., 199. Here, first term \( a = 1 \), common difference \( d = 2 \), and last term \( a_n = 199 \). Find the number of terms \( n \): \( a_n = a + (n - 1)d \)
\( \implies 199 = 1 + (n - 1)2 \)
\( \implies 198 = (n - 1)2 \implies n - 1 = 99 \implies n = 100 \) Now, find the sum of these 100 terms: \( S_{100} = \frac{100}{2} (a + a_n) \)
\( \implies S_{100} = 50(1 + 199) \)
\( \implies S_{100} = 50(200) = 10000 \) Thus, the sum is 10000.
In simple words: This is a list of the first 100 odd numbers. Adding them all together gives 10,000.

Exam Tip: The sum of the first \( n \) odd natural numbers is always \( n^2 \). This provides a very quick way to verify your answer for odd-number series.

Question 44. 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: Let \( a = 100 \) be the first term and \( d \) be the common difference. The sum of the first six terms is: \( S_6 = \frac{6}{2} [2a + 5d] = 3 [2(100) + 5d] = 600 + 15d \) The sum of the first twelve terms is: \( S_{12} = \frac{12}{2} [2a + 11d] = 6 [2(100) + 11d] = 1200 + 66d \) The sum of the next six terms (from 7th to 12th term) is: \( S_{\text{next } 6} = S_{12} - S_6 = (1200 + 66d) - (600 + 15d) = 600 + 51d \) According to the given condition: \( S_6 = 5 \times S_{\text{next } 6} \)
\( \implies 600 + 15d = 5(600 + 51d) \)
\( \implies 600 + 15d = 3000 + 255d \)
\( \implies 600 - 3000 = 255d - 15d \)
\( \implies -2400 = 240d \implies d = -10 \) Thus, the common difference is -10.
In simple words: We write equations for the first 6 terms and the next 6 terms. Solving the equation shows the numbers must go down by 10 each time.

Exam Tip: Always express the "sum of next k terms" as \( S_{2k} - S_k \). This is much cleaner than writing a new A.P. starting with the \( (k+1) \)-th term.

Question 45. Find the number of terms of the A.P, 63, 60, 57, ……….. So that their sum is 693
Answer: The given A.P. is 63, 60, 57, ... Here, first term \( a = 63 \) and common difference \( d = 60 - 63 = -3 \). The sum of terms is \( S_n = 693 \). Using the sum formula: \( S_n = \frac{n}{2} [2a + (n - 1)d] = 693 \)
\( \implies \frac{n}{2} [2(63) + (n - 1)(-3)] = 693 \)
\( \implies \frac{n}{2} [126 - 3n + 3] = 693 \)
\( \implies n(129 - 3n) = 1386 \)
\( \implies 129n - 3n^2 = 1386 \) Divide both sides by 3: \( 43n - n^2 = 462 \)
\( \implies n^2 - 43n + 462 = 0 \) Solving this quadratic equation: \( n^2 - 21n - 22n + 462 = 0 \)
\( \implies n(n - 21) - 22(n - 21) = 0 \)
\( \implies (n - 21)(n - 22) = 0 \) Thus, \( n = 21 \) or \( n = 22 \). Both values of n are valid because the 22nd term is \( a_{22} = 63 + 21(-3) = 0 \), meaning adding this extra term does not change the total sum.
In simple words: Either 21 or 22 terms can be taken to get a sum of 693. This is because the 22nd term in this decreasing sequence is exactly 0.

Exam Tip: If a quadratic equation yields two positive integer solutions, always explain why both values of \( n \) are mathematically valid.

Question 46. How many terms of the sequence 18, 16, 14, …………, should be taken so that their sum is 0
Answer: The given sequence is 18, 16, 14, ... Here, first term \( a = 18 \) and common difference \( d = -2 \). We require the sum of terms to be zero: \( S_n = 0 \)
\( \implies \frac{n}{2} [2a + (n - 1)d] = 0 \)
\( \implies \frac{n}{2} [2(18) + (n - 1)(-2)] = 0 \)
\( \implies \frac{n}{2} [36 - 2n + 2] = 0 \)
\( \implies \frac{n}{2} [38 - 2n] = 0 \) Since \( n \) represents the number of terms and cannot be zero, we solve: \( 38 - 2n = 0 \implies 2n = 38 \implies n = 19 \) Thus, 19 terms must be taken.
In simple words: Since the numbers drop below zero, the negative numbers will eventually cancel out the positive ones. This complete balance happens when we take exactly 19 terms.

Exam Tip: Since \( n \neq 0 \) for any real sequence of terms, you can safely divide both sides of the equation by \( n \) to simplify the algebra.

Question 47. 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 value of the first prize be \( a \). Since each prize is Rs. 40 less than the preceding one, the common difference is \( d = -40 \). The number of prizes is \( n = 7 \), and the total sum of the prizes is \( S_7 = 1400 \). Using the sum formula: \( S_7 = \frac{7}{2} [2a + (7 - 1)d] = 1400 \)
\( \implies \frac{7}{2} [2a + 6(-40)] = 1400 \)
\( \implies \frac{7}{2} [2a - 240] = 1400 \) Multiply both sides by 2 and divide by 7: \( 2a - 240 = \frac{2800}{7} \)
\( \implies 2a - 240 = 400 \)
\( \implies 2a = 640 \implies a = 320 \) The values of the prizes are: 1st prize: Rs. 320 2nd prize: \( 320 - 40 = \) Rs. 280 3rd prize: \( 280 - 40 = \) Rs. 240 4th prize: \( 240 - 40 = \) Rs. 200 5th prize: \( 200 - 40 = \) Rs. 160 6th prize: \( 160 - 40 = \) Rs. 120 7th prize: \( 120 - 40 = \) Rs. 80 Thus, the values of the prizes are Rs. 320, Rs. 280, Rs. 240, Rs. 200, Rs. 160, Rs. 120, and Rs. 80.
In simple words: We find that the first prize is Rs. 320. Each prize is Rs. 40 less, ending with the last prize at Rs. 80.

Exam Tip: Be sure to write out the value of all seven prizes separately at the end of your answer to fulfill the question requirements fully.

Question 48. 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 22nd term \( a_{22} = 149 \). First, find the first term \( a \): \( a_{22} = a + 21d = 149 \)
\( \implies a + 21(7) = 149 \)
\( \implies a + 147 = 149 \implies a = 2 \) Now, find the sum of the first 22 terms using the last term formula: \( S_{22} = \frac{22}{2} (a + a_{22}) \)
\( \implies S_{22} = 11(2 + 149) \)
\( \implies S_{22} = 11(151) = 1661 \) Thus, the sum is 1661.
In simple words: The 22nd term tells us the sequence starts at 2. Using this starting term, we calculate that the sum of the first 22 numbers is 1,661.

Exam Tip: Identifying and solving for the first term \( a \) is a necessary first step before you can use the sum formula.

Question 49. Find the sum of the following A.P: 3, 9/2, 6, 15/2……. To 25 terms
Answer: The given A.P. is \( 3, \frac{9}{2}, 6, \frac{15}{2}, \dots \) Here, first term \( a = 3 \), common difference \( d = \frac{9}{2} - 3 = \frac{3}{2} \), and number of terms \( n = 25 \). Using the sum formula: \( S_{25} = \frac{25}{2} [2a + (25 - 1)d] \)
\( \implies S_{25} = \frac{25}{2} [2(3) + 24\left(\frac{3}{2}\right)] \)
\( \implies S_{25} = \frac{25}{2} [6 + 36] \)
\( \implies S_{25} = \frac{25}{2} [42] = 25 \times 21 = 525 \) Thus, the sum of the first 25 terms is 525.
In simple words: By substituting the starting value of 3 and step size of 1.5 into the sum formula, we find the total of 25 terms is 525.

Exam Tip: Keep the common difference in fractional form rather than decimal form to avoid calculation errors during cancellation steps.

Question 50. 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 \( a \) be the first term and \( d \) be the common difference. We are given: \( \frac{S_p}{S_q} = \frac{p^2}{q^2} \)
\( \implies \frac{\frac{p}{2}[2a + (p - 1)d]}{\frac{q}{2}[2a + (q - 1)d]} = \frac{p^2}{q^2} \)
\( \implies \frac{p[2a + (p - 1)d]}{q[2a + (q - 1)d]} = \frac{p^2}{q^2} \) Dividing both sides by \( \frac{p}{q} \): \( \frac{2a + (p - 1)d}{2a + (q - 1)d} = \frac{p}{q} \) Cross-multiplying: \( q [2a + (p - 1)d] = p [2a + (q - 1)d] \)
\( \implies 2aq + pqd - qd = 2ap + pqd - pd \) Subtracting \( pqd \) 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 by \( (q - p) \): \( 2a = d \) Thus, the common difference is twice the first term. Hence proved.
In simple words: Expanding and cross-multiplying the sum formulas simplifies down to show that the step size 'd' is exactly double the starting number 'a'.

Exam Tip: Look out for terms like \( pqd \) on both sides of the equation and cancel them out early to simplify your proof.

Question 51. An auditorium has 50rows with 20 seats in the first row, 22 in the second, 24 in the third and so fourth. How many seats are In the auditorium?
Answer: The number of seats in the rows forms an A.P.: 20, 22, 24, ... Here, first term \( a = 20 \), common difference \( d = 2 \), and number of rows \( n = 50 \). Using the sum formula: \( S_{50} = \frac{50}{2} [2a + (50 - 1)d] \)
\( \implies S_{50} = 25 [2(20) + 49(2)] \)
\( \implies S_{50} = 25 [40 + 98] \)
\( \implies S_{50} = 25 [138] = 3450 \) Thus, there are 3450 seats in the auditorium.
In simple words: The seats in each row increase by 2. Adding the seats of all 50 rows using the progression sum formula gives 3,450 seats in total.

Exam Tip: Word problems are easily solved once you map the values given in the paragraph directly to \( a \), \( d \), and \( n \).

Question 52. The sum of the first five terms of an A.P is 25 and the sum of of its next five terms is – 75. Find the 10th term of the A .P
Answer: Let \( a \) be the first term and \( d \) be the common difference. The sum of the first five terms is: \( S_5 = 25 \implies \frac{5}{2} [2a + 4d] = 25 \implies 5(a + 2d) = 25 \implies a + 2d = 5 \) (Equation 1) The sum of the next five terms is -75, so the sum of the first ten terms is: \( S_{10} = S_5 + \text{sum of next five terms} = 25 + (-75) = -50 \) Using the sum formula for 10 terms: \( S_{10} = \frac{10}{2} [2a + 9d] = -50 \implies 5(2a + 9d) = -50 \implies 2a + 9d = -10 \) (Equation 2) Multiply Equation 1 by 2: \( 2a + 4d = 10 \) (Equation 3) Subtracting Equation 3 from Equation 2: \( 5d = -20 \implies d = -4 \) Substitute \( d = -4 \) in Equation 1: \( a + 2(-4) = 5 \implies a - 8 = 5 \implies a = 13 \) Now, find the 10th term: \( a_{10} = a + 9d = 13 + 9(-4) = 13 - 36 = -23 \) Thus, the 10th term is -23.
In simple words: Solving equations for the sums reveals that the sequence starts at 13 and decreases by 4 each step. This makes the 10th term -23.

Exam Tip: Remember that the "sum of next five terms" is not \( S_5 \). You must add it to the first sum to find the total sum \( S_{10} \).

Question 53. In an A.P the sum of first n terms 3n2 / 2 + 5n / 2 . Find its 25th term.
Answer: The sum of first n terms is: \( S_n = \frac{3n^2 + 5n}{2} \) We need to find the 25th term \( a_{25} \), which is: \( a_{25} = S_{25} - S_{24} \) First, calculate \( S_{25} \): \( S_{25} = \frac{3(25^2) + 5(25)}{2} = \frac{3(625) + 125}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000 \) Next, calculate \( S_{24} \): \( S_{24} = \frac{3(24^2) + 5(24)}{2} = \frac{3(576) + 120}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924 \) Now, find the 25th term: \( a_{25} = 1000 - 924 = 76 \) Thus, the 25th term is 76.
In simple words: Subtracting the sum of the first 24 terms (924) from the sum of the first 25 terms (1000) gives the value of the 25th term as 76.

Exam Tip: Using the formula \( a_n = S_n - S_{n-1} \) with specific numbers is much faster and less error-prone than finding the general expression for \( a_n \) first.

Question 54. The sum of first five multiples of 3 is
(a) 45
(b) 65
(c) 75
(d) 90
Answer: (a) 45
In simple words: The first five multiples of 3 are 3, 6, 9, 12, and 15. Adding them up gives a total of 45.

Exam Tip: For simple multiple-choice questions with small numbers, adding them up directly is the fastest way to get the correct option.

Question 55. First term of an A.P is -3 and common difference is -2, then fourth term is
(a) 3
(b) -3
(c) 4
(d) -9
Answer: (d) -9
In simple words: Starting at -3 and decreasing by 2 at each step, we reach -9 after three steps.

Exam Tip: Double-check sign calculations when adding negative numbers to negative values.

Question 56. If the pth term of an A.P is q and qth term of A.P is p, then its (p + q)th term is
(a) 0
(b) p + q
(c) p - q
(d) pq
Answer: (a) 0
In simple words: Solving the equations reveals that the step size is -1. Using this, the value at the (p + q)-th position works out to be exactly 0.

Exam Tip: This is a standard identity. Memorising that \( a_{p+q} = 0 \) when \( a_p = q \) and \( a_q = p \) saves valuable time in competitive exams.

Question 57. The sum of first 11 terms of an A.P whose middle term is 30, is
(a) 320
(b) 330
(c) 340
(d) none of these
Answer: (b) 330
In simple words: The sum of 11 terms is exactly 11 times the middle term. Since the middle term is 30, the total sum is 330.

Exam Tip: For any A.P. with an odd number of terms \( N \), the sum is always given by \( S_N = N \times \text{middle term} \). Use this shortcut to solve MCQs quickly.

Question 58. In an A.P, if d = - 2, n = 5 and an = 0, then the value of a is
(a) 10
(b) 5
(c) -8
(d) 8
Answer: (d) 8
In simple words: Since the 5th term is 0 and the sequence decreases by 2 at each step, working backwards to the first term gives us 8.

Exam Tip: Rearrange the standard term formula as \( a = a_n - (n - 1)d \) to find the starting term quickly.

Question 59. If the common difference of an A.P is 3, then a20 - a15 is
(a) 5
(b) 3
(c) 15
(d) 20
Answer: (c) 15
In simple words: The 20th term is exactly 5 steps of size 3 ahead of the 15th term. Therefore, the difference is 15.

Exam Tip: The difference between any two terms is always \( a_x - a_y = (x - y)d \). This makes calculating differences instant.

Question 60. The common difference of an A.P having its nth term (3n + 5) is
(a) 3
(b) 5
(c) 7
(d) none of these
Answer: (a) 3
In simple words: The first two terms are 8 and 11. The difference between them is 3.

Exam Tip: For any general term expressed as \( An + B \), the coefficient of \( n \) is always the common difference of the sequence.

Question 61. The next term of the A.P. √18, √50, √98, ……………,is
(a) √146
(b) √128
(c) √162
(d) √200
Answer: (c) √162
In simple words: Simplifying the square roots gives \( 3\sqrt{2} \), \( 5\sqrt{2} \), and \( 7\sqrt{2} \). The next term must be \( 9\sqrt{2} \), which is \( \sqrt{162} \).

Exam Tip: Always simplify square roots by extracting perfect squares to make the common difference visible.

Question 62. The sum of all natural numbers from 1 to 100 is
(a) 4050
(b) 5050
(c) 6050
(d) 7050
Answer: (b) 5050
In simple words: Adding all the numbers from 1 to 100 gives a total of 5,050.

Exam Tip: Use the standard formula \( \frac{n(n+1)}{2} \) for the sum of first n natural numbers to get the answer instantly.

Question 63. The 30th term of the A.P : 10, 7, 4, ………. Is
(a) 97
(b) -77
(c) -87
(d) 87
Answer: (b) -77
In simple words: Starting at 10 and decreasing by 3 at each step, we reach -77 at the 30th term.

Exam Tip: Take extra care with signs when adding negative values in multi-step arithmetic.

Question 64. Which term of the A.P 17, 34, 51, ……….. is 171
(a) 10
(b) 11
(c) 9
(d) term does not belong to A.P
Answer: (d) term does not belong to A.P
In simple words: Every term in this sequence is a multiple of 17. Since 171 is not divisible by 17, it cannot be a term of this sequence.

Exam Tip: If solving \( a + (n - 1)d = x \) results in a non-integer value for \( n \), the number \( x \) does not belong to the progression.

Question 65. 5 times 5th term of A .P is equal to 8 times 8th term the A.P then its 13th term is
(a) 5
(b) 8
(c) 12
(d) 0
Answer: (d) 0
In simple words: Simplifying the algebraic equation given in the question shows that the 13th term is exactly 0.

Exam Tip: If \( m \cdot a_m = n \cdot a_n \) for an A.P., then the \( (m+n) \)-th term is always 0.

Question 66. The value of a 30 - a 20 for the A.P. 2, 7, 12, 17, ………… is
(a) 100
(b) 10
(c) 50
(d) 20
Answer: (c) 50
In simple words: The 30th term is 10 steps ahead of the 20th term. Since the step size is 5, the total difference is 50.

Exam Tip: Solve the algebraic difference \( a_x - a_y = (x-y)d \) first instead of calculating the values of individual terms.