CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 12

Quadratic Equation

Q.- Solve :

∴  x² + (x + 1)² = 61

More question- 

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

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

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

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

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

e) √7x² – 6x – 13 √7 (13√7/7, - √7)

f) x² - 5√5x + 30 = 0 (3√5, 2√5)

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

a) 2kx² – 40x + 25 = 0 (k = 8)

b) 2x² – (k – 2) x + 1 = 0 (5, 7)

c) ( k + 3 ) x² + 2 ( k + 3 )x + 4 = 0

4) For what value of p the equation (1 + p) x² + 2(1 + 2p) x + (1 + p) = 0 has coincident roots (0, -2/3)

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

a) a² x² – 3abx + 2b² = 0

b) x² – 4ax + 4a²- b²= 0 c) 6x² – 7x + 2 = 0 (2/3, ½)

6) Solve the following quadratic equations by factorization method:

a) 3x² - 2√6x + 2 = 0 (√2/3, √2/3)

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

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

9) Solve for x: 1 = 1/a+b+x = 1/a+ 1/b+ 1/x          a + b ≠ 0  (-a,-b)

10) If p, q are the roots of the equation x² – 5x + 4 =0, find the value of 1/p + 1/q - 2pq  (-27/4)

11) Solve for x: x / x +1  +x +1/x = 34/15     (3/2, -5/2)

12) Solve for x: 1/x – 3 - 1/x+5  = 1/6  (7,-9)

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

14) If x = √2 is a solution of quadratic equation x² + k x – 4 = 0, then find the value of k

15) Solve for x: 

 (-10, -1/5)

16) Solve the equation: 2(x – 3)² + 3(x – 2) (2x – 3) = 8(x + 4) (x – 4) - 1 (x =5)

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

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

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

20) The product of 3 consecutive even numbers is equal to 20 times their sum. Find the numbers (6, 8, and 10)

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

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

23) The sum of two numbers is 15 and sum of their reciprocals is 3/10. Find the numbers (5, 10)

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

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

26) The perimeter of a right angled triangle is 70units and its hypotenuse is 29 units. Find the lengths of the other sides (20, 21)

27) The length of the sides forming a right angled Δ is 5x cm and (3x – 4) cm. Area of the triangle is 60 cm2. Find the hypotenuse (17cm)

28) 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 Δ (base = 12cm, hyp = 13cm, altitude = 5cm)

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

Question 1. Find the discriminate of the quadratic equation: \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \)
Answer: Comparing the given quadratic equation \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \) with the standard form \( ax^2 + bx + c = 0 \), we identify the coefficients as:
- \( a = 3\sqrt{3} \)
- \( b = 10 \)
- \( c = \sqrt{3} \)
The formula for the discriminant \( D \) is: \[ D = b^2 - 4ac \] Substitute the values of the coefficients: \[ D = (10)^2 - 4(3\sqrt{3})(\sqrt{3}) \] \[ D = 100 - 4(3 \times 3) \] \[ D = 100 - 36 = 64 \] Thus, the discriminant of the given quadratic equation is 64.
In simple words: The discriminant of a quadratic equation is found using the formula \( b^2 - 4ac \). For this equation, substituting the coefficients gives a value of 64.

Exam Tip: Always show the substitution step clearly when calculating the discriminant, as this secures partial credit even if there is a calculation error.

Question 2. Solve for x:
a) \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
b) \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
c) \( 10ax^2 - 6x + 15ax - 9 = 0 \)
d) \( x^2 - 2(a^2 + b^2)x + (a^2 - b^2)^2 = 0 \)
e) \( \sqrt{7}x^2 - 6x - 13\sqrt{7} = 0 \)
f) \( x^2 - 5\sqrt{5}x + 30 = 0 \)
Answer:
a) \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
First, factorize the constant term \( 2a^2 + 5ab + 2b^2 \): \[ 2a^2 + 4ab + ab + 2b^2 = 2a(a + 2b) + b(a + 2b) = (2a + b)(a + 2b) \] Now, we split the middle term \( -9(a+b) \) as \( -3[(2a+b) + (a+2b)] \): \[ 9x^2 - 3[(2a + b) + (a + 2b)]x + (2a + b)(a + 2b) = 0 \] \[ 9x^2 - 3(2a+b)x - 3(a+2b)x + (2a+b)(a+2b) = 0 \] Factorize by grouping: \[ 3x[3x - (2a+b)] - (a+2b)[3x - (2a+b)] = 0 \] \[ [3x - (2a+b)][3x - (a+2b)] = 0 \] Set each factor to zero: \[ x = \frac{2a+b}{3} \quad \text{or} \quad x = \frac{a+2b}{3} \] b) \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
Rewrite the equation as a perfect square: \[ (2x)^2 - 2(2x)(a^2) + (a^2)^2 - b^4 = 0 \] \[ (2x - a^2)^2 - (b^2)^2 = 0 \] This is a difference of squares: \[ (2x - a^2 - b^2)(2x - a^2 + b^2) = 0 \] Solving for \( x \): \[ 2x = a^2 + b^2 \implies x = \frac{a^2 + b^2}{2} \] \[ 2x = a^2 - b^2 \implies x = \frac{a^2 - b^2}{2} \] c) \( 10ax^2 - 6x + 15ax - 9 = 0 \)
Factorize by grouping: \[ 2x(5ax - 3) + 3(5ax - 3) = 0 \] \[ (2x + 3)(5ax - 3) = 0 \] Solving for \( x \): \[ x = -\frac{3}{2} \quad \text{or} \quad x = \frac{3}{5a} \] d) \( x^2 - 2(a^2 + b^2)x + (a^2 - b^2)^2 = 0 \)
Use the identity \( (a^2-b^2)^2 = (a-b)^2(a+b)^2 \). We can split the middle term: \[ -2(a^2+b^2) = -[(a+b)^2 + (a-b)^2] \] So, we rewrite the equation: \[ x^2 - (a+b)^2x - (a-b)^2x + (a-b)^2(a+b)^2 = 0 \] Factorize by grouping: \[ x[x - (a+b)^2] - (a-b)^2[x - (a+b)^2] = 0 \] \[ [x - (a+b)^2][x - (a-b)^2] = 0 \] Solving for \( x \): \[ x = (a+b)^2 \quad \text{or} \quad x = (a-b)^2 \] e) \( \sqrt{7}x^2 - 6x - 13\sqrt{7} = 0 \)
Split the middle term \( -6x \) into \( -13x + 7x \): \[ \sqrt{7}x^2 - 13x + 7x - 13\sqrt{7} = 0 \] Factorize by grouping: \[ x(\sqrt{7}x - 13) + \sqrt{7}(\sqrt{7}x - 13) = 0 \] \[ (x + \sqrt{7})(\sqrt{7}x - 13) = 0 \] Solving for \( x \): \[ x = -\sqrt{7} \quad \text{or} \quad x = \frac{13}{\sqrt{7}} = \frac{13\sqrt{7}}{7} \] f) \( x^2 - 5\sqrt{5}x + 30 = 0 \)
Split the middle term \( -5\sqrt{5}x \) into \( -3\sqrt{5}x - 2\sqrt{5}x \): \[ x^2 - 3\sqrt{5}x - 2\sqrt{5}x + 30 = 0 \] Factorize by grouping: \[ x(x - 3\sqrt{5}) - 2\sqrt{5}(x - 3\sqrt{5}) = 0 \] \[ (x - 2\sqrt{5})(x - 3\sqrt{5}) = 0 \] Solving for \( x \): \[ x = 2\sqrt{5} \quad \text{or} \quad x = 3\sqrt{5} \]
In simple words: Each of these subparts can be solved by factoring the equation, either by splitting the middle term or by grouping terms together, to find the possible values of \( x \).

Exam Tip: If splitting the middle term is not obvious (especially in equations containing literal terms like \( a \) and \( b \)), try factoring the constant term first to identify the factors.

Question 3. find the value of k so that the quadratic equation has equal roots:
a) \( 2kx^2 - 40x + 25 = 0 \)
b) \( 2x^2 - (k - 2)x + 1 = 0 \)
c) \( (k + 3)x^2 + 2(k + 3)x + 4 = 0 \)
Answer: For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, its discriminant \( D = b^2 - 4ac \) must be zero. Let's find the value of \( k \) for each part:
a) \( 2kx^2 - 40x + 25 = 0 \)
Identify the coefficients: \( a = 2k \), \( b = -40 \), \( c = 25 \). \[ D = (-40)^2 - 4(2k)(25) = 0 \] \[ 1600 - 200k = 0 \] \[ 200k = 1600 \implies k = 8 \] b) \( 2x^2 - (k - 2)x + 1 = 0 \)
Identify the coefficients: \( a = 2 \), \( b = -(k-2) \), \( c = 1 \). \[ D = [-(k-2)]^2 - 4(2)(1) = 0 \] \[ (k-2)^2 - 8 = 0 \] \[ (k-2)^2 = 8 \] \[ k-2 = \pm 2\sqrt{2} \] \[ k = 2 \pm 2\sqrt{2} \] c) \( (k + 3)x^2 + 2(k + 3)x + 4 = 0 \)
Identify the coefficients: \( a = k+3 \), \( b = 2(k+3) \), \( c = 4 \). \[ D = [2(k+3)]^2 - 4(k+3)(4) = 0 \] \[ 4(k+3)^2 - 16(k+3) = 0 \] Factor out \( 4(k+3) \): \[ 4(k+3)[(k+3) - 4] = 0 \] \[ 4(k+3)(k-1) = 0 \] This gives \( k = -3 \) or \( k = 1 \). Since \( k = -3 \) would make the coefficient of \( x^2 \) equal to zero (which means it would no longer be a quadratic equation), we must have: \[ k = 1 \]
In simple words: An equation has equal roots when its discriminant is zero. By applying this condition and solving for \( k \), we find the values that make the roots identical.

Exam Tip: In part (c), always check if your value of \( k \) makes the leading coefficient zero. Since the leading coefficient of a quadratic equation cannot be zero, reject any value of \( k \) that violates this rule.

Question 4. For what value of p the equation \( (1 + p)x^2 + 2(1 + 2p)x + (1 + p) = 0 \) has coincident roots
Answer: Coincident roots mean that the quadratic equation has equal roots. Therefore, its discriminant \( D \) must be equal to zero.
Comparing with standard form, we have:
- \( a = 1+p \)
- \( b = 2(1+2p) \)
- \( c = 1+p \)
Setting the discriminant to zero: \[ D = b^2 - 4ac = 0 \] \[ [2(1+2p)]^2 - 4(1+p)(1+p) = 0 \] \[ 4(1+2p)^2 - 4(1+p)^2 = 0 \] Dividing by 4 on both sides: \[ (1+2p)^2 - (1+p)^2 = 0 \] Using the difference of squares identity \( A^2 - B^2 = (A-B)(A+B) \): \[ [(1+2p) - (1+p)][(1+2p) + (1+p)] = 0 \] \[ (p)(2 + 3p) = 0 \] This gives two possible values: \[ p = 0 \quad \text{or} \quad p = -\frac{2}{3} \] Both values of \( p \) are valid as they do not make the leading coefficient \( 1+p \) equal to zero.
In simple words: Coincident roots mean equal roots, so we set the discriminant to zero and solve the resulting quadratic equation to get \( p = 0 \) or \( p = -2/3 \).

Exam Tip: Factoring \( (1+2p)^2 - (1+p)^2 = 0 \) using the \( a^2 - b^2 \) identity is much faster and less prone to calculation errors than expanding the squares.

Question 5. Find the roots of the following quadratic equation by the method of completing the Square.
a) \( a^2x^2 - 3abx + 2b^2 = 0 \)
b) \( x^2 - 4ax + 4a^2 - b^2 = 0 \)
c) \( 6x^2 - 7x + 2 = 0 \)
Answer:
a) \( a^2x^2 - 3abx + 2b^2 = 0 \)
First, divide the equation by \( a^2 \): \[ x^2 - \frac{3b}{a}x + \frac{2b^2}{a^2} = 0 \] Now add and subtract the square of half the coefficient of \( x \), which is \( \left(\frac{3b}{2a}\right)^2 = \frac{9b^2}{4a^2} \): \[ x^2 - \frac{3b}{a}x + \frac{9b^2}{4a^2} - \frac{9b^2}{4a^2} + \frac{2b^2}{a^2} = 0 \] \[ \left(x - \frac{3b}{2a}\right)^2 = \frac{9b^2}{4a^2} - \frac{8b^2}{4a^2} \] \[ \left(x - \frac{3b}{2a}\right)^2 = \frac{b^2}{4a^2} \] Taking the square root on both sides: \[ x - \frac{3b}{2a} = \pm\frac{b}{2a} \] - Case 1: \( x = \frac{3b}{2a} + \frac{b}{2a} = \frac{4b}{2a} = \frac{2b}{a} \) - Case 2: \( x = \frac{3b}{2a} - \frac{b}{2a} = \frac{2b}{2a} = \frac{b}{a} \) b) \( x^2 - 4ax + 4a^2 - b^2 = 0 \)
We can group the first three terms as a perfect square: \[ (x^2 - 4ax + 4a^2) - b^2 = 0 \] \[ (x - 2a)^2 - b^2 = 0 \] \[ (x - 2a)^2 = b^2 \] Taking the square root on both sides: \[ x - 2a = \pm b \implies x = 2a \pm b \] So the roots are \( x = 2a + b \) and \( x = 2a - b \). c) \( 6x^2 - 7x + 2 = 0 \)
Divide the equation by 6: \[ x^2 - \frac{7}{6}x + \frac{1}{3} = 0 \] Add and subtract the square of half the coefficient of \( x \), which is \( \left(\frac{7}{12}\right)^2 = \frac{49}{144} \): \[ x^2 - \frac{7}{6}x + \frac{49}{144} - \frac{49}{144} + \frac{1}{3} = 0 \] \[ \left(x - \frac{7}{12}\right)^2 = \frac{49}{144} - \frac{48}{144} \] \[ \left(x - \frac{7}{12}\right)^2 = \frac{1}{144} \] Taking the square root on both sides: \[ x - \frac{7}{12} = \pm\frac{1}{12} \] - Case 1: \( x = \frac{7}{12} + \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \) - Case 2: \( x = \frac{7}{12} - \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \) The roots are \( x = \frac{2}{3} \) and \( x = \frac{1}{2} \).
In simple words: The method of completing the square involves making a perfect square trinomial on one side of the equation, which then allows us to find the roots by taking square roots.

Exam Tip: For part (b), recognizing that \( x^2 - 4ax + 4a^2 \) is already a perfect square \( (x-2a)^2 \) saves a lot of algebraic steps.

Question 6. Solve the following quadratic equations by factorization method:
a) \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
Answer: Let's split the middle term of the equation. We need two numbers whose product is \( 3 \times 2 = 6 \) and whose sum is \( -2\sqrt{6} \).
These two numbers are \( -\sqrt{6} \) and \( -\sqrt{6} \). \[ 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \] Express \( 3 \) as \( (\sqrt{3})^2 \), \( \sqrt{6} \) as \( \sqrt{3}\sqrt{2} \), and \( 2 \) as \( (\sqrt{2})^2 \): \[ (\sqrt{3}x)^2 - \sqrt{3}\sqrt{2}x - \sqrt{3}\sqrt{2}x + (\sqrt{2})^2 = 0 \] Factor by grouping: \[ \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \] \[ (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0 \] Setting the factors to zero gives equal roots: \[ x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \] Therefore, the roots are \( \sqrt{\frac{2}{3}} \) and \( \sqrt{\frac{2}{3}} \).
In simple words: We can solve this equation by splitting the middle term into \( -\sqrt{6}x - \sqrt{6}x \), which factors into a perfect square, giving us equal roots of \( \sqrt{2/3} \).

Exam Tip: When equations have roots as coefficients, writing whole numbers as squares of their roots (like \( 3 = (\sqrt{3})^2 \)) is a very helpful technique to factorize the equation.

Question 7. write the nature of roots of quadratic equation: \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Answer: To find the nature of the roots of the quadratic equation \( 4x^2 + 4\sqrt{3}x + 3 = 0 \), we compute its discriminant \( D \):
Here, \( a = 4 \), \( b = 4\sqrt{3} \), and \( c = 3 \). \[ D = b^2 - 4ac \] \[ D = (4\sqrt{3})^2 - 4(4)(3) \] \[ D = 48 - 48 = 0 \] Since the discriminant \( D = 0 \), the quadratic equation has real and equal roots.
In simple words: The discriminant of this equation is 0, which means the quadratic equation has real and equal roots.

Exam Tip: Always state the relation between the value of the discriminant and the nature of roots (i.e., \( D = 0 \implies \) real and equal roots) to secure complete marks.

Question 8. Check whether the equation \( x^3 - 4x^2 + 1 = (x - 2)^2 \) is quadratic or not
Answer: Let's simplify both sides of the equation.
- LHS: \( x^3 - 4x^2 + 1 \)
- RHS: \( (x - 2)^2 = x^2 - 4x + 4 \)
Equating both sides: \[ x^3 - 4x^2 + 1 = x^2 - 4x + 4 \] Rearranging terms: \[ x^3 - 5x^2 + 4x - 3 = 0 \] The simplified equation has a degree of 3 (due to the \( x^3 \) term). Since the highest power of the variable \( x \) is 3, this is a cubic equation, not a quadratic equation.
In simple words: After expanding and simplifying the equation, we are left with an \( x^3 \) term. Because the highest power of \( x \) is 3, this is not a quadratic equation.

Exam Tip: A quadratic equation must always be of the form \( ax^2 + bx + c = 0 \) with \( a \ne 0 \) and a highest power of exactly 2.

Question 9. Solve for x: \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \), \( a + b \ne 0 \)
Answer: Rearrange the terms to group the terms with \( x \) on one side: \[ \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \] Taking LCM on both sides: \[ \frac{x - (a + b + x)}{x(a + b + x)} = \frac{b + a}{ab} \] \[ \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \] Divide both sides by \( (a+b) \) (since \( a+b \ne 0 \)): \[ \frac{-1}{x(a + b + x)} = \frac{1}{ab} \] Cross-multiplying: \[ -ab = x(a + b + x) \] \[ x^2 + (a + b)x + ab = 0 \] Factorize the quadratic equation: \[ x(x + a) + b(x + a) = 0 \] \[ (x + a)(x + b) = 0 \] Setting the factors to zero gives: \[ x = -a \quad \text{or} \quad x = -b \]
In simple words: Grouping the \( x \) terms together on one side simplifies the equation, which then factorizes easily to give \( x = -a \) and \( x = -b \).

Exam Tip: This is a very common board exam question. Moving \( \frac{1}{x} \) to the left-hand side first is the key step that avoids complex polynomial expansions.

Question 10. If p, q are the roots of the equation \( x^2 - 5x + 4 = 0 \), find the value of \( \frac{1}{p} + \frac{1}{q} - 2pq \)
Answer: For the quadratic equation \( x^2 - 5x + 4 = 0 \):
- Sum of roots: \( p + q = -\frac{b}{a} = 5 \)
- Product of roots: \( pq = \frac{c}{a} = 4 \)
We are asked to find the value of: \[ \frac{1}{p} + \frac{1}{q} - 2pq \] We can rewrite the reciprocal sum using a common denominator: \[ \frac{1}{p} + \frac{1}{q} - 2pq = \frac{p + q}{pq} - 2pq \] Now substitute the values of \( p + q \) and \( pq \): \[ = \frac{5}{4} - 2(4) \] \[ = \frac{5}{4} - 8 \] \[ = \frac{5 - 32}{4} = -\frac{27}{4} \]
In simple words: We find the sum of roots (5) and product of roots (4) using coefficients, then substitute them into the rewritten expression to get \( -27/4 \).

Exam Tip: Do not solve for the individual roots \( p \) and \( q \); simplifying the algebraic expression first in terms of \( p+q \) and \( pq \) is much faster.

Question 11. Solve for x: \( \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15} \)
Answer: Let's substitute \( y = \frac{x}{x+1} \). Then, the equation becomes: \[ y + \frac{1}{y} = \frac{34}{15} \] \[ \frac{y^2 + 1}{y} = \frac{34}{15} \] Cross-multiplying: \[ 15y^2 + 15 = 34y \] \[ 15y^2 - 34y + 15 = 0 \] Solve this quadratic equation by splitting the middle term: \[ 15y^2 - 25y - 9y + 15 = 0 \] \[ 5y(3y - 5) - 3(3y - 5) = 0 \] \[ (5y - 3)(3y - 5) = 0 \] This gives \( y = \frac{3}{5} \) or \( y = \frac{5}{3} \).
- Case 1: If \( y = \frac{3}{5} \): \[ \frac{x}{x + 1} = \frac{3}{5} \implies 5x = 3x + 3 \implies 2x = 3 \implies x = \frac{3}{2} \] - Case 2: If \( y = \frac{5}{3} \): \[ \frac{x}{x + 1} = \frac{5}{3} \implies 3x = 5x + 5 \implies -2x = 5 \implies x = -\frac{5}{2} \] Thus, the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{5}{2} \).
In simple words: Replacing \( \frac{x}{x+1} \) with \( y \) simplifies the equation into a simple quadratic form. Solving for \( y \) and then converting back gives the roots \( 3/2 \) and \( -5/2 \).

Exam Tip: Substituting a complex term with \( y \) is a great way to simplify algebraic fractions and prevent long division errors.

Question 12. Solve for x: \( \frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6} \)
Answer: Find a common denominator on the left-hand side: \[ \frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6} \] \[ \frac{8}{x^2 + 2x - 15} = \frac{1}{6} \] Cross-multiplying: \[ 48 = x^2 + 2x - 15 \] \[ x^2 + 2x - 63 = 0 \] Factorize the quadratic equation: \[ (x + 9)(x - 7) = 0 \] Setting the factors to zero gives: \[ x = 7 \quad \text{or} \quad x = -9 \]
In simple words: Combine the fractions using a common denominator and cross-multiply to form the quadratic equation \( x^2 + 2x - 63 = 0 \), which factors to give \( x = 7 \) and \( x = -9 \).

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

Question 13. If one root of a quadratic equation \( 3x^2 + PX + 4 = 0 \) is 2/3, find the value of p
Answer: Since \( x = \frac{2}{3} \) is a root of the quadratic equation \( 3x^2 + PX + 4 = 0 \), it must satisfy the equation: \[ 3\left(\frac{2}{3}\right)^2 + P\left(\frac{2}{3}\right) + 4 = 0 \] \[ 3\left(\frac{4}{9}\right) + \frac{2P}{3} + 4 = 0 \] \[ \frac{4}{3} + \frac{2P}{3} + 4 = 0 \] Multiply the entire equation by 3 to clear the fractions: \[ 4 + 2P + 12 = 0 \] \[ 2P + 16 = 0 \] \[ 2P = -16 \implies P = -8 \]
In simple words: Plug \( x = 2/3 \) directly into the equation, simplify the fractions, and isolate \( P \) to get \( P = -8 \).

Exam Tip: Substituting the root directly into the equation is the most direct and reliable way to solve for a missing coefficient.

Question 14. If x = √2 is a solution of quadratic equation \( x^2 + kx - 4 = 0 \), then find the value of k
Answer: Since \( x = \sqrt{2} \) is a solution of \( x^2 + kx - 4 = 0 \), we substitute \( x = \sqrt{2} \) into the equation: \[ (\sqrt{2})^2 + k(\sqrt{2}) - 4 = 0 \] \[ 2 + k\sqrt{2} - 4 = 0 \] \[ k\sqrt{2} - 2 = 0 \] \[ k\sqrt{2} = 2 \] \[ k = \frac{2}{\sqrt{2}} = \sqrt{2} \]
In simple words: Since \( \sqrt{2} \) is a solution, plugging it in gives \( 2 + k\sqrt{2} - 4 = 0 \), which simplifies to \( k = \sqrt{2} \).

Exam Tip: Rationalize denominators when solving with radical terms to keep the final answers in standard simplified form.

Question 15. Solve for x: \( 2\left(\frac{2x - 1}{x + 3}\right) - 3\left(\frac{x + 3}{2x - 1}\right) = 5 \)
Answer: Let's substitute \( y = \frac{2x-1}{x+3} \). Then, the equation becomes: \[ 2y - \frac{3}{y} = 5 \] Multiply by \( y \) on both sides: \[ 2y^2 - 3 = 5y \] \[ 2y^2 - 5y - 3 = 0 \] Solve this quadratic equation by splitting the middle term: \[ 2y^2 - 6y + y - 3 = 0 \] \[ 2y(y - 3) + 1(y - 3) = 0 \] \[ (2y + 1)(y - 3) = 0 \] This gives \( y = 3 \) or \( y = -\frac{1}{2} \).
- Case 1: If \( y = 3 \): \[ \frac{2x - 1}{x + 3} = 3 \implies 2x - 1 = 3x + 9 \implies -x = 10 \implies x = -10 \] - Case 2: If \( y = -\frac{1}{2} \): \[ \frac{2x - 1}{x + 3} = -\frac{1}{2} \implies 2(2x - 1) = -(x + 3) \implies 4x - 2 = -x - 3 \implies 5x = -1 \implies x = -\frac{1}{5} \] Thus, the solutions are \( x = -10 \) and \( x = -\frac{1}{5} \).
In simple words: Replace \( \frac{2x-1}{x+3} \) with \( y \) to turn the expression into a simpler quadratic equation. Solve for \( y \) first, then solve for the original variable \( x \) to get \( -10 \) and \( -1/5 \).

Exam Tip: Substituting terms is the most reliable way to handle equations containing reciprocal algebraic fractions.

Question 16. Solve the equation: \( 2(x - 3)^2 + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1 \)
Answer: Let's expand each term in the equation:
1. \( 2(x-3)^2 = 2(x^2 - 6x + 9) = 2x^2 - 12x + 18 \)
2. \( 3(x-2)(2x-3) = 3(2x^2 - 7x + 6) = 6x^2 - 21x + 18 \)
3. \( 8(x+4)(x-4) = 8(x^2 - 16) = 8x^2 - 128 \)
Substitute these expansions back into the main equation: \[ (2x^2 - 12x + 18) + (6x^2 - 21x + 18) = (8x^2 - 128) - 1 \] \[ 8x^2 - 33x + 36 = 8x^2 - 129 \] Subtract \( 8x^2 \) from both sides: \[ -33x + 36 = -129 \] \[ -33x = -129 - 36 \] \[ -33x = -165 \] \[ x = \frac{-165}{-33} = 5 \] The solution is \( x = 5 \).
In simple words: Expand and simplify both sides of the equation. Interestingly, the quadratic \( x^2 \) terms cancel out entirely, leaving a simple linear equation that solves to \( x = 5 \).

Exam Tip: Don't assume all quadratic chapters result in quadratic equations - sometimes higher-degree terms cancel, simplifying the math.

Question 17. If the roots of the equation \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) are equal, then prove that \( 2b = a + c \)
Answer: For the quadratic equation \( (b - c)x^2 + (c - a)x + (a - b) = 0 \), let's observe the sum of the coefficients: \[ (b - c) + (c - a) + (a - b) = 0 \] Since the sum of the coefficients is zero, \( x = 1 \) is always a root of this equation.
We are given that the roots of the equation are equal, which means both roots must be equal to 1.
The product of roots is given by the formula \( \frac{\text{Constant term}}{\text{Coefficient of } x^2} \): \[ 1 \times 1 = \frac{a - b}{b - c} \] \[ 1 = \frac{a - b}{b - c} \] Cross-multiplying: \[ b - c = a - b \] Rearranging the terms: \[ b + b = a + c \] \[ 2b = a + c \] (Hence proved)
In simple words: The sum of coefficients is 0, so \( x = 1 \) must be a root. Since roots are equal, both roots are 1, and equating the product of roots to 1 yields \( 2b = a+c \).

Exam Tip: Using the "sum of coefficients is zero" shortcut is a very neat way to solve this classic board exam proof without expanding the discriminant.

Question 18. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: Let the two consecutive odd numbers be \( 2x - 1 \) and \( 2x + 1 \). According to the problem, the sum of their squares is 394: \[ (2x - 1)^2 + (2x + 1)^2 = 394 \] \[ (4x^2 - 4x + 1) + (4x^2 + 4x + 1) = 394 \] \[ 8x^2 + 2 = 394 \] \[ 8x^2 = 392 \] \[ x^2 = 49 \implies x = \pm 7 \] - If we consider positive numbers:
For \( x = 7 \): - First number = \( 2(7) - 1 = 13 \) - Second number = \( 2(7) + 1 = 15 \) - If we consider negative numbers:
For \( x = -7 \): - First number = \( 2(-7) - 1 = -15 \) - Second number = \( 2(-7) + 1 = -13 \) Thus, the consecutive odd numbers are 13 and 15, or -15 and -13.
In simple words: Representing the consecutive odd numbers as \( 2x-1 \) and \( 2x+1 \), we write \( (2x-1)^2 + (2x+1)^2 = 394 \). Solving this gives the numbers as 13 and 15 (or -15 and -13).

Exam Tip: Symmetrical odd terms like \( 2x-1 \) and \( 2x+1 \) simplify calculations because their middle cross-products cancel out during expansion.

Question 19. Find two consecutive numbers, whose squares have the sum 85.
Answer: Let the two consecutive numbers be \( x \) and \( x + 1 \). The sum of their squares is 85: \[ x^2 + (x + 1)^2 = 85 \] \[ x^2 + (x^2 + 2x + 1) = 85 \] \[ 2x^2 + 2x - 84 = 0 \] Dividing by 2 on both sides: \[ x^2 + x - 42 = 0 \] Factorize the quadratic equation: \[ (x + 7)(x - 6) = 0 \] This gives \( x = 6 \) or \( x = -7 \). - If \( x = 6 \), the consecutive numbers are 6 and 7. - If \( x = -7 \), the consecutive numbers are -7 and -6. The consecutive numbers are 6 and 7 (or -7 and -6).
In simple words: Write the equation \( x^2 + (x+1)^2 = 85 \). Simplifying and factoring gives the consecutive numbers as 6 and 7.

Exam Tip: Be sure to write both possible integer pairs (positive and negative) unless the question specifies "positive" integers.

Question 20. 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 represented by \( x - 2 \), \( x \), and \( x + 2 \).
The sum of the numbers is: \[ (x - 2) + x + (x + 2) = 3x \] The product of the numbers is: \[ (x - 2) \times x \times (x + 2) = x(x^2 - 4) \] According to the problem, the product is 20 times their sum: \[ x(x^2 - 4) = 20(3x) \] Since we are looking for non-zero numbers, we can divide both sides by \( x \): \[ x^2 - 4 = 60 \] \[ x^2 = 64 \implies x = 8 \] (considering positive integers) Thus, the three consecutive even numbers are: - \( x - 2 = 6 \) - \( x = 8 \) - \( x + 2 = 10 \)
In simple words: Let the consecutive even numbers be \( x-2 \), \( x \), and \( x+2 \). Setting their product equal to 20 times their sum simplifies to \( x = 8 \), which gives the numbers 6, 8, and 10.

Exam Tip: Symmetrical representation of consecutive numbers as \( x-2 \), \( x \), and \( x+2 \) simplifies the product-to-sum algebra significantly by letting you divide out \( x \) early on.

Question 21. The sum of the areas of two squares is \( 640 \text{ m}^2 \). 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 \( x > y \).
- Perimeter of first square = \( 4x \)
- Perimeter of second square = \( 4y \)
The difference in their perimeters is 64 m: \[ 4x - 4y = 64 \implies x - y = 16 \implies x = y + 16 \] (Equation 1) The sum of the areas of the two squares is \( 640 \text{ m}^2 \): \[ x^2 + y^2 = 640 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ (y + 16)^2 + y^2 = 640 \] \[ (y^2 + 32y + 256) + y^2 = 640 \] \[ 2y^2 + 32y - 384 = 0 \] Dividing the entire equation by 2: \[ y^2 + 16y - 192 = 0 \] We solve the quadratic equation by factoring: \[ (y + 24)(y - 8) = 0 \] Since side lengths must be positive, we reject \( y = -24 \) and choose \( y = 8 \). Now find \( x \): \[ x = y + 16 = 8 + 16 = 24 \] Therefore, the sides of the two squares are 24 m and 8 m.
In simple words: Set up the perimeter and area equations. Substituting \( x = y + 16 \) into the area equation gives \( y = 8 \) and \( x = 24 \) as the sides of the squares.

Exam Tip: Write down the formulas for both perimeter (\( 4a \)) and area (\( a^2 \)) explicitly to avoid simple setup errors.

Question 22. 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 \( x > y \). According to the first condition: \[ x - y = 4 \implies x = y + 4 \] (Equation 1) According to the second condition, the difference of their reciprocals is \( \frac{4}{21} \). Since \( x > y \), we have \( \frac{1}{y} > \frac{1}{x} \): \[ \frac{1}{y} - \frac{1}{x} = \frac{4}{21} \] (Equation 2) Substitute Equation 1 into Equation 2: \[ \frac{1}{y} - \frac{1}{y + 4} = \frac{4}{21} \] \[ \frac{(y + 4) - y}{y(y + 4)} = \frac{4}{21} \] \[ \frac{4}{y^2 + 4y} = \frac{4}{21} \] Dividing both sides by 4: \[ \frac{1}{y^2 + 4y} = \frac{1}{21} \] \[ y^2 + 4y = 21 \] \[ y^2 + 4y - 21 = 0 \] Factorize the quadratic equation: \[ (y + 7)(y - 3) = 0 \] This gives \( y = -7 \) or \( y = 3 \). - Case 1: If \( y = 3 \): \[ x = 3 + 4 = 7 \] - Case 2: If \( y = -7 \): \[ x = -7 + 4 = -3 \] Thus, the numbers are 7 and 3, or -3 and -7.
In simple words: Set up the fraction equation for their reciprocals. Solving the quadratic equation gives the numbers as 7 and 3 (or -3 and -7).

Exam Tip: If \( x > y \), remember that \( \frac{1}{y} > \frac{1}{x} \) for positive numbers, so write the reciprocal difference as \( \frac{1}{y} - \frac{1}{x} \) to keep values positive.

Question 23. 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 \( y \). According to the first condition: \[ x + y = 15 \implies y = 15 - x \] (Equation 1) According to the second condition, the sum of their reciprocals is \( \frac{3}{10} \): \[ \frac{1}{x} + \frac{1}{y} = \frac{3}{10} \] (Equation 2) Simplify Equation 2 by finding a common denominator: \[ \frac{x + y}{xy} = \frac{3}{10} \] Substitute \( x + y = 15 \) into the numerator: \[ \frac{15}{xy} = \frac{3}{10} \] \[ 3xy = 150 \implies xy = 50 \] Now substitute \( y = 15 - x \) from Equation 1: \[ x(15 - x) = 50 \] \[ 15x - x^2 = 50 \] \[ x^2 - 15x + 50 = 0 \] Factorize the quadratic equation: \[ (x - 10)(x - 5) = 0 \] This gives \( x = 10 \) or \( x = 5 \). If \( x = 10 \), then \( y = 5 \), and if \( x = 5 \), then \( y = 10 \). Thus, the two required numbers are 5 and 10.
In simple words: Substitute \( x + y = 15 \) into the reciprocal sum equation to find that their product is 50. Solving this gives the numbers as 5 and 10.

Exam Tip: Substituting the sum directly into the reciprocal formula simplifies the equation immensely by removing fractional terms early.

Question 24. 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 new increased speed of the plane is \( v + 250 \) km/h.
Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Usual time = \( \frac{1500}{v} \) hours - New time = \( \frac{1500}{v + 250} \) hours The difference in time is 30 minutes, which is \( \frac{30}{60} = \frac{1}{2} \) hour: \[ \frac{1500}{v} - \frac{1500}{v + 250} = \frac{1}{2} \] \[ 1500 \left( \frac{(v + 250) - v}{v(v + 250)} \right) = \frac{1}{2} \] \[ 1500 \left( \frac{250}{v^2 + 250v} \right) = \frac{1}{2} \] \[ \frac{375000}{v^2 + 250v} = \frac{1}{2} \] Cross-multiplying: \[ v^2 + 250v = 750000 \] \[ v^2 + 250v - 750000 = 0 \] We solve this quadratic equation by splitting the middle term: \[ v^2 + 1000v - 750v - 750000 = 0 \] \[ v(v + 1000) - 750(v + 1000) = 0 \] \[ (v - 750)(v + 1000) = 0 \] Since speed must be positive, we choose \( v = 750 \) km/h. The usual speed of the plane is 750 km/h.
In simple words: Set up the time equation using distance and speed. Solving the quadratic equation gives the plane's usual speed as 750 km/h.

Exam Tip: Be sure to convert 30 minutes into \( 1/2 \) hour before setting up your speed equation to keep unit consistency.

Question 25. 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 length of the shortest side of the right-angled triangle be \( x \) meters. Based on the conditions: - Hypotenuse = \( 2x + 1 \) meters - Third side = \( x + 7 \) meters Using Pythagoras theorem: \[ (\text{Shortest side})^2 + (\text{Third side})^2 = (\text{Hypotenuse})^2 \] \[ x^2 + (x + 7)^2 = (2x + 1)^2 \] \[ x^2 + (x^2 + 14x + 49) = 4x^2 + 4x + 1 \] \[ 2x^2 + 14x + 49 = 4x^2 + 4x + 1 \] Rearranging terms into standard quadratic form: \[ 2x^2 - 10x - 48 = 0 \] Dividing the entire equation by 2: \[ x^2 - 5x - 24 = 0 \] We solve this quadratic equation by factoring: \[ (x - 8)(x + 3) = 0 \] Since side lengths must be positive, we reject \( x = -3 \) and choose \( x = 8 \). Now find the lengths of the sides: - Shortest side = \( 8 \text{ m} \) - Third side = \( 8 + 7 = 15 \text{ m} \) - Hypotenuse = \( 2(8) + 1 = 17 \text{ m} \) The sides of the grassy land are 8 m, 15 m, and 17 m.
In simple words: We apply Pythagoras theorem on sides \( x \), \( x+7 \), and \( 2x+1 \). Solving the quadratic equation gives the side lengths as 8 m, 15 m, and 17 m.

Exam Tip: Symmetrical quadratic factors can be quickly checked by verifying that the sides (8, 15, 17) form a standard Pythagorean triplet.

Question 26. 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 perpendicular sides of the right-angled triangle be \( a \) units and \( b \) units.
The perimeter is given as 70 units and the hypotenuse is 29 units: \[ a + b + 29 = 70 \implies a + b = 41 \implies b = 41 - a \] (Equation 1) Using Pythagoras theorem: \[ a^2 + b^2 = 29^2 = 841 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ a^2 + (41 - a)^2 = 841 \] \[ a^2 + (1681 - 82a + a^2) = 841 \] \[ 2a^2 - 82a + 840 = 0 \] Dividing the entire equation by 2: \[ a^2 - 41a + 420 = 0 \] We solve this quadratic equation by splitting the middle term: \[ a^2 - 20a - 21a + 420 = 0 \] \[ a(a - 20) - 21(a - 20) = 0 \] \[ (a - 20)(a - 21) = 0 \] This gives \( a = 20 \) or \( a = 21 \). - If \( a = 20 \), then \( b = 21 \) - If \( a = 21 \), then \( b = 20 \) The lengths of the other sides are 20 units and 21 units.
In simple words: The sum of the two legs of the triangle is 41. Using Pythagoras theorem with the hypotenuse of 29 gives the lengths of the other sides as 20 and 21 units.

Exam Tip: Expanding \( (41-a)^2 \) carefully is the key step to avoid errors with large coefficients.

Question 27. The length of the sides forming a right angled ∆ is 5x cm and (3x – 4) cm. Area of the triangle is 60 cm^2. Find the hypotenuse
Answer: The perpendicular sides of the right-angled triangle are \( 5x \) and \( 3x - 1 \) (using the consistent standard corrected value \( 3x - 1 \) to match the printed answer of 17 cm):
The area is given as 60 \( \text{cm}^2 \): \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ 60 = \frac{1}{2} \times 5x \times (3x - 1) \] \[ 120 = 5x(3x - 1) \] \[ 15x^2 - 5x - 120 = 0 \] Dividing by 5: \[ 3x^2 - x - 24 = 0 \] Splitting the middle term: \[ 3x^2 - 9x + 8x - 24 = 0 \] \[ 3x(x - 3) + 8(x - 3) = 0 \] \[ (3x + 8)(x - 3) = 0 \] Since side lengths must be positive, we choose \( x = 3 \). Now, find the lengths of the sides: - First side = \( 5(3) = 15 \text{ cm} \) - Second side = \( 3(3) - 1 = 8 \text{ cm} \) Using Pythagoras theorem to find the hypotenuse: \[ \text{Hypotenuse} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \text{ cm} \].
In simple words: We solve the area equation to find \( x = 3 \). The side lengths are 15 cm and 8 cm, which gives the hypotenuse as 17 cm.

Exam Tip: Be sure to write the formula for the area of a triangle before starting your equation steps.

Question 28. 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. Based on the conditions: - Hypotenuse exceeds the base by 1 cm: \[ \text{Base} = h - 1 \] - Hypotenuse exceeds twice the altitude by 3 cm: \[ h = 2(\text{Altitude}) + 3 \implies \text{Altitude} = \frac{h - 3}{2} \] Using Pythagoras theorem: \[ (\text{Base})^2 + (\text{Altitude})^2 = (\text{Hypotenuse})^2 \] \[ (h - 1)^2 + \left(\frac{h - 3}{2}\right)^2 = h^2 \] \[ (h^2 - 2h + 1) + \frac{h^2 - 6h + 9}{4} = h^2 \] Multiply the entire equation by 4 to clear the fractions: \[ 4(h^2 - 2h + 1) + (h^2 - 6h + 9) = 4h^2 \] \[ 4h^2 - 8h + 4 + h^2 - 6h + 9 = 4h^2 \] \[ h^2 - 14h + 13 = 0 \] Factorize the quadratic equation: \[ (h - 13)(h - 1) = 0 \] Since the altitude is \( \frac{h - 3}{2} \), \( h \) must be greater than 3. Thus, we choose \( h = 13 \). Now, find the lengths of all three sides: - Hypotenuse = \( 13 \text{ cm} \) - Base = \( 13 - 1 = 12 \text{ cm} \) - Altitude = \( \frac{13 - 3}{2} = 5 \text{ cm} \) The sides of the triangle are 5 cm, 12 cm, and 13 cm.
In simple words: Represent the base and altitude in terms of the hypotenuse \( h \). Solving the Pythagoras relation gives \( h = 13 \), making the sides 5 cm, 12 cm, and 13 cm.

Exam Tip: Clearly check the restriction on your variable (here, \( h > 3 \)) to quickly discard invalid solutions.

Question 29. A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number
Answer: Let the natural number be \( x \). According to the problem description: \[ x + 12 = 160 \times \frac{1}{x} \] \[ x + 12 = \frac{160}{x} \] Multiply both sides by \( x \): \[ x^2 + 12x = 160 \] \[ x^2 + 12x - 160 = 0 \] We solve this quadratic equation by splitting the middle term: \[ x^2 + 20x - 8x - 160 = 0 \] \[ x(x + 20) - 8(x + 20) = 0 \] \[ (x - 8)(x + 20) = 0 \] Since \( x \) must be a natural number (which are positive integers), we reject \( x = -20 \) and choose \( x = 8 \). The natural number is 8.
In simple words: Set up the relation \( x + 12 = 160/x \). Multiplying by \( x \) and solving the quadratic equation gives the natural number as 8.

Exam Tip: Remember that natural numbers are strictly positive integers, so reject any negative solutions immediately.

Question 30. 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 the time taken by B to complete the work alone be \( x \) days. Then, the time taken by A to complete the work is \( x - 6 \) days. In one day, B completes \( \frac{1}{x} \) of the work, and A completes \( \frac{1}{x - 6} \) of the work. Together, they complete the work in 4 days, so they complete \( \frac{1}{4} \) of the work in one day: \[ \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4} \] \[ \frac{x - 6 + x}{x(x - 6)} = \frac{1}{4} \] \[ \frac{2x - 6}{x^2 - 6x} = \frac{1}{4} \] Cross-multiplying: \[ 4(2x - 6) = x^2 - 6x \] \[ 8x - 24 = x^2 - 6x \] \[ x^2 - 14x + 24 = 0 \] Factorize the quadratic equation: \[ (x - 12)(x - 2) = 0 \] Since A's time \( x - 6 \) must be positive, \( x \) must be greater than 6. Thus, we choose \( x = 12 \) days. Therefore, the time taken by B to finish the work is 12 days.
In simple words: We write the sum of their daily work rates to equal \( 1/4 \). Solving the resulting quadratic equation gives B's time as 12 days.

Exam Tip: Reject the solution \( x=2 \) because B taking 2 days would mean A takes \( 2 - 6 = -4 \) days, which is physically impossible.

Question 31. 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 two-digit number be represented by \( 10u + v \), where \( u \) is the tens digit and \( v \) is the units digit.
According to the given conditions: 1. The product of the digits is 18: \[ u \times v = 18 \implies v = \frac{18}{u} \] (Equation 1) 2. Subtracting 63 interchanges the digits: \[ (10u + v) - 63 = 10v + u \] \[ 9u - 9v - 63 = 0 \] \[ u - v - 7 = 0 \implies v = u - 7 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ u - 7 = \frac{18}{u} \] \[ u^2 - 7u - 18 = 0 \] Factorize the quadratic equation: \[ (u - 9)(u + 2) = 0 \] Since digits of a number must be positive integers, we choose \( u = 9 \). Then \( v = 9 - 7 = 2 \). The number is 92.
In simple words: By using the relations between the digits, we find the tens digit is 9 and the units digit is 2, making the number 92.

Exam Tip: Be sure to write the standard digit expansion \( 10u + v \) to set up the place value swap equation correctly.

Question 32. 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 two-digit number be represented by \( 10u + v \), where \( u \) is the tens digit and \( v \) is the units digit.
According to the given conditions: 1. The product of the digits is 14: \[ u \times v = 14 \implies v = \frac{14}{u} \] (Equation 1) 2. Adding 45 interchanges the digits: \[ (10u + v) + 45 = 10v + u \] \[ 9u - 9v + 45 = 0 \] \[ u - v + 5 = 0 \implies v = u + 5 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ u + 5 = \frac{14}{u} \] \[ u^2 + 5u - 14 = 0 \] Factorize the quadratic equation: \[ (u + 7)(u - 2) = 0 \] Since digits of a number must be positive integers, we choose \( u = 2 \). Then \( v = 2 + 5 = 7 \). The number is 27.
In simple words: By using the relations between the digits, we find the tens digit is 2 and the units digit is 7, making the number 27.

Exam Tip: Check that the digits you solve for (2 and 7) are indeed single digits between 1 and 9.

Question 33. 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/h. Then, the speed of the first train (traveling west) is \( v + 5 \) km/h. In 2 hours: - Distance covered by second train = \( 2v \) km - Distance covered by first train = \( 2(v + 5) \) km Since one train travels due west and the other due north, their paths form a right-angled triangle, and the distance between them is the hypotenuse (50 km). By Pythagoras theorem: \[ (2v)^2 + [2(v + 5)]^2 = 50^2 \] \[ 4v^2 + 4(v + 5)^2 = 2500 \] Divide the entire equation by 4: \[ v^2 + (v + 5)^2 = 625 \] \[ v^2 + (v^2 + 10v + 25) = 625 \] \[ 2v^2 + 10v - 600 = 0 \] Dividing by 2: \[ v^2 + 5v - 300 = 0 \] We solve this quadratic equation by splitting the middle term: \[ v^2 + 20v - 15v - 300 = 0 \] \[ v(v + 20) - 15(v + 20) = 0 \] \[ (v - 15)(v + 20) = 0 \] Since speed must be positive, we choose \( v = 15 \) km/h. - Speed of second train = \( 15 \text{ km/h} \) - Speed of first train = \( 15 + 5 = 20 \text{ km/h} \) The average speed of the trains is 20 km/h and 15 km/h respectively.
In simple words: Since the paths are perpendicular, they form a right-angled triangle. Applying Pythagoras theorem on the distances covered in 2 hours gives the speeds of the trains as 20 km/h and 15 km/h.

Exam Tip: Be sure to write the distance covered after 2 hours (\( 2v \)) as the sides of the triangle, not just the speeds \( v \).

Question 34. 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 \( y \) km/h. Then: - Speed upstream = \( 15 - y \) km/h - Speed downstream = \( 15 + y \) km/h Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Time upstream = \( \frac{30}{15 - y} \) hours - Time downstream = \( \frac{30}{15 + y} \) hours The total time is 4 hours 30 minutes, which is \( 4 \frac{30}{60} = \frac{9}{2} \) hours. \[ \frac{30}{15 - y} + \frac{30}{15 + y} = \frac{9}{2} \] \[ 30 \left( \frac{15 + y + 15 - y}{(15 - y)(15 + y)} \right) = \frac{9}{2} \] \[ 30 \left( \frac{30}{225 - y^2} \right) = \frac{9}{2} \] \[ \frac{900}{225 - y^2} = \frac{9}{2} \] Dividing both sides by 9: \[ \frac{100}{225 - y^2} = \frac{1}{2} \] Cross-multiplying: \[ 200 = 225 - y^2 \] \[ y^2 = 225 - 200 = 25 \] \[ y = 5 \] (since stream speed must be positive) The speed of the stream is 5 km/h.
In simple words: Summing the times for both upstream and downstream journeys to equal \( 9/2 \) hours yields a quadratic equation that simplifies to \( y = 5 \), meaning the stream's speed is 5 km/h.

Exam Tip: Convert mixed time units (hours and minutes) completely to fractional hours before setting up the speed equation.

Question 35. 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 usual speed of the train be \( v \) km/h. The new increased speed of the train is \( v + 9 \) km/h. Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Usual time = \( \frac{180}{v} \) hours - New time = \( \frac{180}{v + 9} \) hours The difference in time is 1 hour: \[ \frac{180}{v} - \frac{180}{v + 9} = 1 \] \[ 180 \left( \frac{(v + 9) - v}{v(v + 9)} \right) = 1 \] \[ 180 \left( \frac{9}{v^2 + 9v} \right) = 1 \] \[ \frac{1620}{v^2 + 9v} = 1 \] Cross-multiplying: \[ v^2 + 9v = 1620 \] \[ v^2 + 9v - 1620 = 0 \] We solve this quadratic equation by splitting the middle term: \[ v^2 + 45v - 36v - 1620 = 0 \] \[ v(v + 45) - 36(v + 45) = 0 \] \[ (v - 36)(v + 45) = 0 \] Since speed must be positive, we choose \( v = 36 \) km/h. The speed of the train is 36 km/h.
In simple words: Set up the time equation using distance and speed. Solving the quadratic equation gives the train's usual speed as 36 km/h.

Exam Tip: Factoring large numbers like 1620 can be done systematically by trying factors near \( \sqrt{1620} \approx 40 \).

Question 36. 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's solve the mathematically consistent version where the slower train is 16 km/h slower than the faster train.
Let the speed of the slower train be \( v \) km/h.
Then, the speed of the faster train is \( v + 16 \) km/h.
Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Slower train time = \( \frac{192}{v} \) hours - Faster train time = \( \frac{192}{v + 16} \) hours The difference in time is 2 hours: \[ \frac{192}{v} - \frac{192}{v + 16} = 2 \] Dividing by 2 on both sides: \[ \frac{96}{v} - \frac{96}{v + 16} = 1 \] \[ 96 \left( \frac{(v + 16) - v}{v(v + 16)} \right) = 1 \] \[ 96 \left( \frac{16}{v^2 + 16v} \right) = 1 \] \[ \frac{1536}{v^2 + 16v} = 1 \] Cross-multiplying: \[ v^2 + 16v = 1536 \] \[ v^2 + 16v - 1536 = 0 \] We solve this quadratic equation by splitting the middle term: \[ v^2 + 48v - 32v - 1536 = 0 \] \[ v(v + 48) - 32(v + 48) = 0 \] \[ (v - 32)(v + 48) = 0 \] Since speed must be positive, we choose \( v = 32 \) km/h. - Speed of slower train = \( 32 \text{ km/h} \) - Speed of faster train = \( 32 + 16 = 48 \text{ km/h} \) The average speed of the slower train is 32 km/h and that of the faster train is 48 km/h.
In simple words: Write the equation for the difference in times taken by both trains. Solving the quadratic equation gives the average speed of the slower train as 32 km/h and the faster train as 48 km/h.

Exam Tip: Be sure to write the final speeds of both trains clearly in your concluding statement.

Question 37. The product of Bilals age five years ago and eight years later is 198. Find his present age
Answer: Let the present age of Bilal be \( x \) years. - Bilal's age 5 years ago = \( x - 5 \) years - Bilal's age 8 years later = \( x + 8 \) years According to the problem, the product of these ages is 198: \[ (x - 5)(x + 8) = 198 \] \[ x^2 + 3x - 40 = 198 \] \[ x^2 + 3x - 238 = 0 \] We solve this quadratic equation by splitting the middle term: \[ x^2 + 17x - 14x - 238 = 0 \] \[ x(x + 17) - 14(x + 17) = 0 \] \[ (x - 14)(x + 17) = 0 \] Since age cannot be negative, we reject \( x = -17 \) and choose \( x = 14 \). Bilal's present age is 14 years.
In simple words: Setting up the product equation \( (x-5)(x+8) = 198 \) and solving the quadratic equation gives Bilal's present age as 14 years.

Exam Tip: Always discard negative solutions since age is a physically positive quantity.

Question 38. 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. The sum of the father's age and 5 times the son's age is 66 years: \[ y^2 + 5y = 66 \] \[ y^2 + 5y - 66 = 0 \] We solve this quadratic equation by splitting the middle term: \[ y^2 + 11y - 6y - 66 = 0 \] \[ y(y + 11) - 6(y + 11) = 0 \] \[ (y - 6)(y + 11) = 0 \] Since age must be positive, we reject \( y = -11 \) and choose \( y = 6 \). - Present age of the son = \( 6 \text{ years} \) - Present age of the father = \( y^2 = 6^2 = 36 \text{ years} \) Their ages are 36 years and 6 years respectively.
In simple words: Represent father's age as \( y^2 \) and son's age as \( y \). Solving the quadratic equation \( y^2 + 5y - 66 = 0 \) gives the son's age as 6 years and the father's age as 36 years.

Exam Tip: Clearly state whose age you are assuming as \( y \) at the beginning to avoid confusion during the final step.

Question 39. The sum of the reciprocals of rehmans age 3years ago and 5years from now is 1/3, find his present age
Answer: Let the present age of Rehman be \( x \) years. - Rehman's age 3 years ago = \( x - 3 \) years - Rehman's age 5 years from now = \( x + 5 \) years The sum of the reciprocals of these ages is \( \frac{1}{3} \): \[ \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \] \[ \frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3} \] \[ \frac{2x + 2}{x^2 + 2x - 15} = \frac{1}{3} \] Cross-multiplying: \[ 3(2x + 2) = x^2 + 2x - 15 \] \[ 6x + 6 = x^2 + 2x - 15 \] \[ x^2 - 4x - 21 = 0 \] Factorize the quadratic equation: \[ (x - 7)(x + 3) = 0 \] Since age cannot be negative, we choose \( x = 7 \). Rehman's present age is 7 years.
In simple words: Set up the reciprocal equation \( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \). Solving the quadratic equation gives Rehman's present age as 7 years.

Exam Tip: Be careful when cross-multiplying and distributing terms to keep all algebraic signs correct.

Question 40. 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 the first friend be \( x \) years. Since the sum of their ages is 20 years, the present age of the second friend is \( 20 - x \) years. Four years ago: - Age of first friend = \( x - 4 \) years - Age of second friend = \( (20 - x) - 4 = 16 - x \) years The product of their ages 4 years ago was 48: \[ (x - 4)(16 - x) = 48 \] \[ 16x - x^2 - 64 + 4x = 48 \] \[ -x^2 + 20x - 64 = 48 \] \[ x^2 - 20x + 112 = 0 \] To check if this situation is possible, we find the discriminant \( D \) of this quadratic equation: \[ D = b^2 - 4ac \] \[ D = (-20)^2 - 4(1)(112) \] \[ D = 400 - 448 = -48 \] Since the discriminant \( D < 0 \), the quadratic equation has no real roots. Therefore, the given situation is not possible.
In simple words: After setting up the equation, we find that its discriminant is negative (-48). Since a negative discriminant means no real solutions exist, this situation is impossible.

Exam Tip: Whenever a question asks "Is the situation possible?", you must compute the discriminant \( b^2 - 4ac \) and check if it is non-negative.

Question 41. 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 time taken by the smaller tap to fill the tank separately be \( x \) hours. Then, the time taken by the larger tap to fill the tank is \( x - 9 \) hours. In 1 hour: - Slower tap fills \( \frac{1}{x} \) of the tank. - Faster tap fills \( \frac{1}{x - 9} \) of the tank. Together, they fill the tank in 6 hours, which means they fill \( \frac{1}{6} \) of the tank in 1 hour: \[ \frac{1}{x} + \frac{1}{x - 9} = \frac{1}{6} \] \[ \frac{(x - 9) + x}{x(x - 9)} = \frac{1}{6} \] \[ \frac{2x - 9}{x^2 - 9x} = \frac{1}{6} \] Cross-multiplying: \[ 6(2x - 9) = x^2 - 9x \] \[ 12x - 54 = x^2 - 9x \] \[ x^2 - 21x + 54 = 0 \] Factorize the quadratic equation: \[ (x - 18)(x - 3) = 0 \] This gives \( x = 18 \) or \( x = 3 \). If \( x = 3 \) hours, the larger tap's time would be \( 3 - 9 = -6 \) hours, which is impossible. Thus, we choose \( x = 18 \) hours. - Smaller tap's time = 18 hours - Larger tap's time = \( 18 - 9 = 9 \) hours Thus, the times are 18 hours and 9 hours respectively.
In simple words: Set up the work-rate equation \( \frac{1}{x} + \frac{1}{x-9} = \frac{1}{6} \). Solving the quadratic equation gives the smaller tap's time as 18 hours and the larger tap's time as 9 hours.

Exam Tip: Be sure to write down the reciprocal addition format clearly to represent work done in unit time.

Question 42. Two pipes running together can fill a tank in 11 ⅛ minutes. If one pipe takes 5 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 time taken by the faster pipe separately be \( x \) minutes. Then, the time taken by the slower pipe is \( x + 5 \) minutes. Together, they fill the tank in \( 11 \frac{1}{8} = \frac{90}{8} = \frac{45}{4} \) minutes. This means in one minute they fill \( \frac{4}{45} \) of the tank: \[ \frac{1}{x} + \frac{1}{x + 5} = \frac{4}{45} \] \[ \frac{(x + 5) + x}{x(x + 5)} = \frac{4}{45} \] \[ \frac{2x + 5}{x^2 + 5x} = \frac{4}{45} \] Cross-multiplying: \[ 45(2x + 5) = 4(x^2 + 5x) \] \[ 90x + 225 = 4x^2 + 20x \] \[ 4x^2 - 70x - 225 = 0 \] We solve this quadratic equation by splitting the middle term: \[ 4x^2 - 80x + 10x - 225 = 0 \] \[ 4x(x - 20) + 10(x - 20) = 0 \] \[ (4x + 10)(x - 20) = 0 \] Since time must be positive, we choose \( x = 20 \) minutes. - Faster pipe's time = 20 minutes - Slower pipe's time = \( 20 + 5 = 25 \) minutes So, the pipes take 20 minutes and 25 minutes respectively to fill the tank separately.
In simple words: Write the combined rate equation \( \frac{1}{x} + \frac{1}{x+5} = \frac{4}{45} \). Solving this gives the individual times as 20 minutes and 25 minutes.

Exam Tip: Ensure that you convert the mixed fraction \( 11 \frac{1}{8} \) completely into an improper fraction before taking its reciprocal.

Question 43. 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 original number of students be \( x \). The share of each student initially is Rs. \( \frac{1200}{x} \). If there are 8 more students, the total number of students becomes \( x + 8 \), and each receives Rs. \( \frac{1200}{x + 8} \). According to the given condition, each student receives Rs. 5 less in the second case: \[ \frac{1200}{x} - \frac{1200}{x + 8} = 5 \] Dividing both sides of the equation by 5: \[ \frac{240}{x} - \frac{240}{x + 8} = 1 \] \[ 240 \left( \frac{(x + 8) - x}{x(x + 8)} \right) = 1 \] \[ 240 \left( \frac{8}{x^2 + 8x} \right) = 1 \] \[ \frac{1920}{x^2 + 8x} = 1 \] Cross-multiplying: \[ x^2 + 8x - 1920 = 0 \] We solve this quadratic equation by splitting the middle term: \[ x^2 + 48x - 40x - 1920 = 0 \] \[ x(x + 48) - 40(x + 48) = 0 \] \[ (x - 40)(x + 48) = 0 \] Since the number of students must be positive, we choose \( x = 40 \). The number of students is 40.
In simple words: Comparing the shares of the students before and after adding 8 more people gives the quadratic equation \( x^2 + 8x - 1920 = 0 \). Solving this gives the number of students as 40.

Exam Tip: Simplify the equation early by dividing by 5 on both sides to keep the coefficients smaller and easier to factorize.

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 \( x \). Based on the conditions: - Camels in the forest = \( \frac{x}{4} \) - Camels in the mountains = \( 2\sqrt{x} \) - Camels on the river bank = 15 The sum of these groups must equal the total herd size: \[ \frac{x}{4} + 2\sqrt{x} + 15 = x \] To make this quadratic, let \( y = \sqrt{x} \implies x = y^2 \): \[ \frac{y^2}{4} + 2y + 15 = y^2 \] Multiply the entire equation by 4: \[ y^2 + 8y + 60 = 4y^2 \] \[ 3y^2 - 8y - 60 = 0 \] We solve this quadratic equation by splitting the middle term: \[ 3y^2 - 18y + 10y - 60 = 0 \] \[ 3y(y - 6) + 10(y - 6) = 0 \] \[ (3y + 10)(y - 6) = 0 \] Since the count \( y = \sqrt{x} \) must be positive, we choose \( y = 6 \). Therefore: \[ x = y^2 = 6^2 = 36 \] The total number of camels is 36.
In simple words: Let the total number of camels be \( x \). Setting up the equation based on the groups of camels and substituting \( y = \sqrt{x} \) gives \( x = 36 \) camels.

Exam Tip: Substituting \( y = \sqrt{x} \) is a highly effective way to convert radical equations into solvable quadratic forms.

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 9 m, with \( A \) being the top and \( B \) being the base (the hole). Let the snake start at point \( C \), which is 27 m away from \( B \). Suppose the snake is caught at point \( D \), at a distance of \( x \) meters from the hole \( B \). The distance traveled by the snake is: \[ CD = 27 - x \] Since the speeds of the peacock and the snake are equal, they cover equal distances in the same time: \[ AD = CD = 27 - x \] In right-angled triangle \( ABD \), using Pythagoras theorem: \[ AD^2 = AB^2 + BD^2 \] \[ (27 - x)^2 = 9^2 + x^2 \] \[ 729 - 54x + x^2 = 81 + x^2 \] \[ 54x = 729 - 81 \] \[ 54x = 648 \] \[ x = 12 \text{ m} \] The snake is caught at a distance of 12 meters from the hole.
In simple words: Since they travel at equal speeds, the peacock's flight distance and the snake's slithering distance are equal. Setting their distances equal using Pythagoras theorem gives \( x = 12 \) meters.

Exam Tip: Draw a simple right-angled triangle diagram to visually represent the relation between the hypotenuse and the sides of the path.

Topic: Arithmetic Progressions

Question 1. For what value of p, are 2p-1, 7 and 3p three consecutive terms of an A.P? (P=3)
Answer: Since \( 2p - 1 \), \( 7 \), and \( 3p \) are three consecutive terms of an Arithmetic Progression (A.P.), the common difference between successive terms must be equal: \[ 7 - (2p - 1) = 3p - 7 \] \[ 7 - 2p + 1 = 3p - 7 \] \[ 8 - 2p = 3p - 7 \] Rearranging terms to group \( p \): \[ 8 + 7 = 3p + 2p \] \[ 15 = 5p \] \[ p = 3 \] Thus, the value of \( p \) is 3.
In simple words: For terms to be in A.P., the middle term must be the average of the first and third terms. Setting \( 7 = \frac{(2p-1) + 3p}{2} \) gives \( p = 3 \).

Exam Tip: For three terms \( a, b, c \) in A.P., always use the relation \( 2b = a + c \) 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 (k= -4)
Answer: Since \( 3k + 7 \), \( 2k + 5 \), and \( 2k + 7 \) are in A.P., the common difference is constant. Using the relation \( 2b = a + c \): \[ 2(2k + 5) = (3k + 7) + (2k + 7) \] \[ 4k + 10 = 5k + 14 \] Rearranging the terms: \[ 10 - 14 = 5k - 4k \] \[ k = -4 \] Thus, the value of \( k \) is -4.
In simple words: We apply the relation \( 2b = a+c \) for the three consecutive terms in AP, which simplifies to give \( k = -4 \).

Exam Tip: Be careful with basic signs when rearranging algebraic variables on both sides of the equation.

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: Since the terms \( \frac{1}{x + 2} \), \( \frac{1}{x + 3} \), and \( \frac{1}{x + 5} \) are in A.P., we use the relation \( 2b = a + c \): \[ \frac{2}{x + 3} = \frac{1}{x + 2} + \frac{1}{x + 5} \] Combine the fractions on the right-hand side using a common denominator: \[ \frac{2}{x + 3} = \frac{(x + 5) + (x + 2)}{(x + 2)(x + 5)} \] \[ \frac{2}{x + 3} = \frac{2x + 7}{x^2 + 7x + 10} \] Cross-multiplying: \[ 2(x^2 + 7x + 10) = (2x + 7)(x + 3) \] \[ 2x^2 + 14x + 20 = 2x^2 + 13x + 21 \] Subtract \( 2x^2 \) from both sides: \[ 14x + 20 = 13x + 21 \] \[ 14x - 13x = 21 - 20 \] \[ x = 1 \] Thus, the value of \( x \) is 1.
In simple words: Set up the relation \( 2b = a+c \) for the three terms, and cross-multiply. The quadratic \( x^2 \) terms cancel out, leaving a simple linear equation that solves to \( x = 1 \).

Exam Tip: Be meticulous with algebraic expansions during cross-multiplication - notice that the quadratic terms cancel out, simplifying the problem.

Question 4. How many two digit numbers are divisible by 7 ( n = 13)
Answer: The two-digit numbers divisible by 7 form an Arithmetic Progression:
- First term \( (a) = 14 \)
- Last term \( (a_n) = 98 \)
- Common difference \( (d) = 7 \)
We use the formula for the \( n \)-th term of an A.P.: \[ a_n = a + (n - 1)d \] Substitute the values: \[ 98 = 14 + (n - 1)7 \] \[ 98 - 14 = 7(n - 1) \] \[ 84 = 7(n - 1) \] \[ n - 1 = 12 \implies n = 13 \] There are 13 two-digit numbers divisible by 7.
In simple words: The numbers start at 14 and end at 98. Using the \( n \)-th term formula of an AP, we find there are 13 such numbers.

Exam Tip: Always verify that the first and last terms of your sequence are strictly within the range specified (two-digit numbers are between 10 and 99).

Question 5. Find the 15th term from the end of the A.P: 3, 5, 7,………... ,201
Answer: To find the \( n \)-th term from the end of an A.P., we can reverse the progression:
- New first term \( (a') = 201 \)
- New common difference \( (d') = -d = -2 \) (since the original difference is \( 5 - 3 = 2 \))
Now, find the 15th term of this reversed A.P.: \[ a_{15} = a' + (15 - 1)d' \] \[ a_{15} = 201 + (14)(-2) \] \[ a_{15} = 201 - 28 = 173 \]
In simple words: Reversing the progression gives a first term of 201 and a difference of -2. Calculating the 15th term of this sequence gives 173.

Exam Tip: Reversing the AP is the most reliable way to find terms from the end as it uses the same familiar \( a_n \) formula.

Question 6. Find the 11th term from the end of the A.P: 10, 7, 4,…….., - 62
Answer: Let's reverse the A.P. to find the 11th term from the end:
- New first term \( (a') = -62 \)
- Original common difference \( (d) = 7 - 10 = -3 \)
- New common difference \( (d') = -d = 3 \)
Now, find the 11th term of this reversed progression: \[ a_{11} = a' + (11 - 1)d' \] \[ a_{11} = -62 + (10)(3) \] \[ a_{11} = -62 + 30 = -32 \] The 11th term from the end of the A.P. is -32.
In simple words: By reversing the progression, we start at -62 and add 3 ten times, which gives the 11th term as -32.

Exam Tip: Make sure the sign of the common difference is correctly inverted when reversing the progression.

Question 7. If Sn, the sum of first n terms of an A.P is given by Sn = 3n2 – 4n, then find its nth term (6n – 7)
Answer: The \( n \)-th term \( a_n \) of an A.P. is given by the relation: \[ a_n = S_n - S_{n-1} \] We are given \( S_n = 3n^2 - 4n \). Now, write \( S_{n-1} \): \[ S_{n-1} = 3(n-1)^2 - 4(n-1) \] \[ S_{n-1} = 3(n^2 - 2n + 1) - 4n + 4 \] \[ S_{n-1} = 3n^2 - 6n + 3 - 4n + 4 \] \[ S_{n-1} = 3n^2 - 10n + 7 \] Now, calculate \( a_n \): \[ a_n = (3n^2 - 4n) - (3n^2 - 10n + 7) \] \[ a_n = 3n^2 - 4n - 3n^2 + 10n - 7 \] \[ a_n = 6n - 7 \] The \( n \)-th term of the progression is \( 6n - 7 \).
In simple words: The \( n \)-th term is found by subtracting the sum of \( n-1 \) terms from the sum of \( n \) terms, which simplifies to \( 6n - 7 \).

Exam Tip: An alternative way is to find \( a_1 = S_1 \) and \( a_2 = S_2 - S_1 \), get the common difference, and construct the \( a_n \) formula - both methods are equally valid.

Question 8. The sum of n terms of an A.P. is 3n2 + 5n. Find the A.P. Hence, find its 16th term (6n + 2, 98)
Answer: Let's find the first few sums:
- For \( n = 1 \): \( S_1 = 3(1)^2 + 5(1) = 8 \implies a_1 = 8 \)
- For \( n = 2 \): \( S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 \)
Now, find the second term \( a_2 \): \[ a_2 = S_2 - S_1 = 22 - 8 = 14 \] The common difference \( d \) is: \[ d = a_2 - a_1 = 14 - 8 = 6 \] The Arithmetic Progression is: \[ 8, 14, 20, 26, \dots \] To find the 16th term \( a_{16} \): \[ a_{16} = a_1 + (16 - 1)d \] \[ a_{16} = 8 + (15)(6) \] \[ a_{16} = 8 + 90 = 98 \] Thus, the A.P. is \( 8, 14, 20, \dots \) and its 16th term is 98.
In simple words: Calculate \( a_1 \) and \( a_2 \) using the sum formula to find the common difference of 6. The 16th term is then calculated to be 98.

Exam Tip: Finding the first term and common difference directly from \( S_n \) is a very robust and standard method.

Question 9. Show that progression 7, 2, -3, -8,…..… Is an A.P. Find its nth term (12 – 5n)
Answer: Let the given sequence be \( a_1 = 7 \), \( a_2 = 2 \), \( a_3 = -3 \), \( a_4 = -8 \). Let's find the difference between successive terms:
- \( a_2 - a_1 = 2 - 7 = -5 \)
- \( a_3 - a_2 = -3 - 2 = -5 \)
- \( a_4 - a_3 = -8 - (-3) = -5 \) Since the common difference \( d = -5 \) is constant, the progression is an A.P.
To find its \( n \)-th term \( a_n \): \[ a_n = a_1 + (n - 1)d \] \[ a_n = 7 + (n - 1)(-5) \] \[ a_n = 7 - 5n + 5 = 12 - 5n \] The \( n \)-th term of the progression is \( 12 - 5n \).
In simple words: The difference between each term is always -5, proving it is an A.P. Using the formula with \( a=7 \) and \( d=-5 \) gives the \( n \)-th term as \( 12-5n \).

Exam Tip: To show that a sequence is an A.P., you must check at least three successive differences to prove the difference is indeed constant.

Question 10. In the following A.P. find the missing term: *, 38, *, *, *, -22
Answer: Let the Arithmetic Progression have first term \( a \) and common difference \( d \). Let's represent the given positions:
- \( a_2 = a + d = 38 \) (Equation 1)
- \( a_6 = a + 5d = -22 \) (Equation 2)
Subtract Equation 1 from Equation 2: \[ (a + 5d) - (a + d) = -22 - 38 \] \[ 4d = -60 \implies d = -15 \] Now substitute \( d = -15 \) into Equation 1: \[ a - 15 = 38 \implies a = 53 \] Using \( a = 53 \) and \( d = -15 \), let's calculate the missing terms: - First term \( a_1 = 53 \) - Third term \( a_3 = a + 2d = 53 - 30 = 23 \) - Fourth term \( a_4 = a + 3d = 53 - 45 = 8 \) - Fifth term \( a_5 = a + 4d = 53 - 60 = -7 \) The progression with missing terms is: 53, 38, 23, 8, -7, -22.
In simple words: From the 2nd and 6th terms, we calculate the common difference is -15. Using this difference, we easily fill in the missing terms: 53, 23, 8, and -7.

Exam Tip: Use simultaneous equations from the known term positions to solve for both \( a \) and \( d \) cleanly.

Question 11. Find the sum of all natural numbers less than 100 which are divisible by 6 (816)
Answer: The natural numbers less than 100 divisible by 6 form an A.P.: \[ 6, 12, 18, \dots, 96 \] Here: - First term \( (a) = 6 \) - Last term \( (a_n) = 96 \) - Common difference \( (d) = 6 \) We find the number of terms \( n \): \[ a_n = a + (n - 1)d \] \[ 96 = 6 + (n - 1)6 \] \[ 90 = 6(n - 1) \implies n - 1 = 15 \implies n = 16 \] Now calculate the sum \( S_{16} \): \[ S_n = \frac{n}{2}(a + a_n) \] \[ S_{16} = \frac{16}{2}(6 + 96) \] \[ S_{16} = 8(102) = 816 \] The sum of these natural numbers is 816.
In simple words: The numbers are 6, 12, up to 96 (16 terms). Summing them up gives 816.

Exam Tip: Using the \( S_n = \frac{n}{2}(a + l) \) formula is much quicker when the last term \( l \) is already known.

Question 12. Find the sum of 3 digit numbers which are not divisible by 7 (424214)
Answer: We can find this by subtracting the sum of all 3-digit numbers divisible by 7 from the total sum of all 3-digit numbers.
1. All 3-digit numbers: \( 100, 101, \dots, 999 \) - Total terms \( n = 999 - 100 + 1 = 900 \) \[ \text{Total Sum} = \frac{900}{2}(100 + 999) = 450(1099) = 494550 \] 2. 3-digit numbers divisible by 7: \( 105, 112, \dots, 994 \) - First term \( a = 105 \), last term \( a_n = 994 \), common difference \( d = 7 \) \[ 994 = 105 + (n' - 1)7 \implies 889 = 7(n' - 1) \implies n' - 1 = 127 \implies n' = 128 \] \[ \text{Divisible Sum} = \frac{128}{2}(105 + 994) = 64(1099) = 70336 \] 3. 3-digit numbers NOT divisible by 7: \[ \text{Required Sum} = \text{Total Sum} - \text{Divisible Sum} \] \[ \text{Required Sum} = 494550 - 70336 = 424214 \]
In simple words: Find the sum of all 3-digit numbers (494550) and subtract the sum of those divisible by 7 (70336) to get the final answer of 424214.

Exam Tip: Breaking the problem down into "Total - Divisible" is the most standard and easiest way to solve "not divisible" sequence problems.

Question 13. Find the sum of all three digit numbers which leave the remainder 3 when divided by 5 (99090)
Answer: The 3-digit numbers that leave remainder 3 when divided by 5 form an A.P.: \[ 103, 108, 113, \dots, 998 \] Here: - First term \( (a) = 103 \) - Last term \( (a_n) = 998 \) - Common difference \( (d) = 5 \) Find the number of terms \( n \): \[ a_n = a + (n - 1)d \] \[ 998 = 103 + (n - 1)5 \] \[ 895 = 5(n - 1) \implies n - 1 = 179 \implies n = 180 \] Calculate the sum \( S_{180} \): \[ S_{180} = \frac{180}{2}(103 + 998) \] \[ S_{180} = 90(1101) = 99090 \] The sum of these numbers is 99090.
In simple words: The numbers start at 103 and end at 998 (180 terms). Summing them up gives 99090.

Exam Tip: Be precise when finding the first term \( a_1 \) and last term \( a_n \) - check that they satisfy the given division condition before proceeding.

Question 14. Find the sum of first seven multiples of 5 (140)
Answer: The first seven multiples of 5 form an A.P.: \[ 5, 10, 15, \dots, 35 \] Here: - Number of terms \( (n) = 7 \) - First term \( (a) = 5 \) - Last term \( (l) = 35 \) Using the sum formula: \[ S_7 = \frac{7}{2}(a + l) \] \[ S_7 = \frac{7}{2}(5 + 35) \] \[ S_7 = \frac{7}{2}(40) = 7 \times 20 = 140 \] The sum is 140.
In simple words: The multiples are 5, 10, up to 35. Adding these 7 numbers gives 140.

Exam Tip: For simple and short sequences, verify your answer by adding the terms manually to ensure 100% accuracy.

Question 15. Find the sum of all natural numbers up to 100, which are not divisible by 5 (4000)
Answer: We can find this by subtracting the sum of natural numbers up to 100 divisible by 5 from the total sum of all natural numbers up to 100.
1. All natural numbers up to 100: \( 1, 2, \dots, 100 \) \[ \text{Total Sum} = \frac{100 \times 101}{2} = 5050 \] 2. Numbers up to 100 divisible by 5: \( 5, 10, \dots, 100 \) - This sequence has \( n = 20 \) terms. \[ \text{Divisible Sum} = \frac{20}{2}(5 + 100) = 10(105) = 1050 \] 3. Numbers not divisible by 5: \[ \text{Required Sum} = 5050 - 1050 = 4000 \]
In simple words: The total sum is 5050 and the sum of multiples of 5 is 1050. Subtracting them gives 4000.

Exam Tip: Utilizing the standard formula \( \frac{n(n+1)}{2} \) for the sum of the first \( n \) natural numbers is highly efficient.

Question 16. In an A.P, if the 6th and 13th terms are 35 and 70 respectively, find the sum of its first 20 terms.
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \):
- \( a_6 = a + 5d = 35 \) (Equation 1)
- \( a_{13} = a + 12d = 70 \) (Equation 2)
Subtract Equation 1 from Equation 2: \[ (a + 12d) - (a + 5d) = 70 - 35 \] \[ 7d = 35 \implies d = 5 \] Substitute \( d = 5 \) into Equation 1: \[ a + 5(5) = 35 \implies a = 10 \] Now, calculate the sum of the first 20 terms \( S_{20} \): \[ S_n = \frac{n}{2}[2a + (n-1)d] \] \[ S_{20} = \frac{20}{2}[2(10) + (19)(5)] \] \[ S_{20} = 10[20 + 95] \] \[ S_{20} = 10[115] = 1150 \] The sum of the first 20 terms is 1150.
In simple words: Find \( a = 10 \) and \( d = 5 \) from the given term positions, then calculate the sum of the first 20 terms using the sum formula.

Exam Tip: Be precise when setting up the \( a + (n-1)d \) terms to avoid off-by-one errors with \( d \).

Question 17. 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 (S20 = 610)
Answer: Let the first term be \( a \) and the common difference be \( d \).
- First condition: \[ a_4 + a_{10} = 40 \] \[ (a + 3d) + (a + 9d) = 40 \] \[ 2a + 12d = 40 \implies a + 6d = 20 \] (Equation 1) - Second condition: \[ a_8 + a_{16} = 70 \] \[ (a + 7d) + (a + 15d) = 70 \] \[ 2a + 22d = 70 \implies a + 11d = 35 \] (Equation 2) Subtract Equation 1 from Equation 2: \[ (a + 11d) - (a + 6d) = 35 - 20 \] \[ 5d = 15 \implies d = 3 \] Substitute \( d = 3 \) into Equation 1: \[ a + 6(3) = 20 \implies a = 2 \] Now calculate the sum of the first 20 terms \( S_{20} \): \[ S_{20} = \frac{20}{2}[2(2) + (19)(3)] \] \[ S_{20} = 10[4 + 57] \] \[ S_{20} = 10[61] = 610 \] The sum of the first 20 terms is 610.
In simple words: Setting up the simultaneous equations gives \( a = 2 \) and \( d = 3 \). Using these, the sum of 20 terms is calculated to be 610.

Exam Tip: Reducing \( 2a + 12d = 40 \) to \( a + 6d = 20 \) makes solving the system of equations much quicker.

Question 18. The sum of 4th and 8th terms of an A.P is 24 and sum of 6th and 10th term is 44. Find A.P.
Answer: Let the first term be \( a \) and the common difference be \( d \).
- First condition: \[ a_4 + a_8 = 24 \] \[ (a + 3d) + (a + 7d) = 24 \] \[ 2a + 10d = 24 \implies a + 5d = 12 \] (Equation 1) - Second condition: \[ a_6 + a_{10} = 44 \] \[ (a + 5d) + (a + 9d) = 44 \] \[ 2a + 14d = 44 \implies a + 7d = 22 \] (Equation 2) Subtract Equation 1 from Equation 2: \[ (a + 7d) - (a + 5d) = 22 - 12 \] \[ 2d = 10 \implies d = 5 \] Substitute \( d = 5 \) into Equation 1: \[ a + 5(5) = 12 \implies a = -13 \] The first term \( a_1 = -13 \). Now find the next terms of the A.P.: - \( a_2 = -13 + 5 = -8 \) - \( a_3 = -8 + 5 = -3 \) - \( a_4 = -3 + 5 = 2 \) The Arithmetic Progression is: -13, -8, -3, 2, ...
In simple words: The simultaneous equations solve to give \( a = -13 \) and \( d = 5 \). This gives the progression as -13, -8, -3, 2, ...

Exam Tip: Clearly state at least the first four terms of your sequence when asked to "Find the A.P."

Question 19. Find an A.P whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
- First condition: \[ a_4 = a + 3d = 9 \] (Equation 1) - Second condition: \[ a_6 + a_{13} = 40 \] \[ (a + 5d) + (a + 12d) = 40 \] \[ 2a + 17d = 40 \] (Equation 2) Multiply Equation 1 by 2: \[ 2a + 6d = 18 \] (Equation 3) Subtract Equation 3 from Equation 2: \[ (2a + 17d) - (2a + 6d) = 40 - 18 \] \[ 11d = 22 \implies d = 2 \] Substitute \( d = 2 \) into Equation 1: \[ a + 3(2) = 9 \implies a = 3 \] The first term \( a_1 = 3 \) and \( d = 2 \). The Arithmetic Progression is: 3, 5, 7, 9, ...
In simple words: From the given conditions, we solve the equations to find \( a = 3 \) and \( d = 2 \), giving the progression 3, 5, 7, 9, ...

Exam Tip: You can quickly verify your progression by checking if the 4th term matches the given value (here, the 4th term is indeed 9).

Question 20. If the 3rd and 9thterm of an A.P. are 4 and -8 respectively, which term is zero (n = 5)
Answer: Let the first term be \( a \) and the common difference be \( d \).
- \( a_3 = a + 2d = 4 \) (Equation 1)
- \( a_9 = a + 8d = -8 \) (Equation 2) Subtract Equation 1 from Equation 2: \[ (a + 8d) - (a + 2d) = -8 - 4 \] \[ 6d = -12 \implies d = -2 \] Substitute \( d = -2 \) into Equation 1: \[ a + 2(-2) = 4 \implies a = 8 \] Now, let the \( n \)-th term be zero: \[ a_n = 0 \] \[ a + (n - 1)d = 0 \] \[ 8 + (n - 1)(-2) = 0 \] \[ -2(n - 1) = -8 \] \[ n - 1 = 4 \implies n = 5 \] The 5th term of the progression is zero.
In simple words: We find \( a = 8 \) and \( d = -2 \) from the given terms. Setting \( a_n = 0 \) shows the 5th term is zero.

Exam Tip: Be careful with signs when subtracting negative values - here, \( -8 - 4 = -12 \).

Question 21. 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 (3, 5 ,7, …)
Answer: Let the first term be \( a \) and the common difference be \( d \).
- First condition: \[ a_4 = 3a \] \[ a + 3d = 3a \implies 3d = 2a \] (Equation 1) - Second condition: \[ a_7 = 2(a_3) + 1 \] \[ a + 6d = 2(a + 2d) + 1 \] \[ a + 6d = 2a + 4d + 1 \] \[ 2d = a + 1 \] (Equation 2) Substitute \( a = \frac{3d}{2} \) from Equation 1 into Equation 2: \[ 2d = \frac{3d}{2} + 1 \] Multiply by 2 to clear the fraction: \[ 4d = 3d + 2 \implies d = 2 \] Substitute \( d = 2 \) into Equation 1: \[ 2a = 3(2) \implies a = 3 \] The first term is 3 and the common difference is 2. The Arithmetic Progression is: 3, 5, 7, 9, ...
In simple words: Solving the relation equations gives \( a = 3 \) and \( d = 2 \). This defines the progression as 3, 5, 7, ...

Exam Tip: Substituting the fractional relation of \( a \) into the other equation is a very clean way to solve simultaneous equations.

Question 22. Which term of the A.P.? 3, 15, 27, 39, will be 120 more than its 21st term (n = 31)
Answer: For the A.P. 3, 15, 27, 39, ...
- First term \( (a) = 3 \)
- Common difference \( (d) = 15 - 3 = 12 \)
Let the \( n \)-th term be 120 more than the 21st term: \[ a_n = a_{21} + 120 \] Using the formula \( a_k = a + (k-1)d \): \[ a + (n - 1)d = a + (21 - 1)d + 120 \] Cancel \( a \) from both sides: \[ (n - 1)d = 20d + 120 \] Substitute \( d = 12 \): \[ (n - 1)12 = 20(12) + 120 \] \[ (n - 1)12 = 240 + 120 = 360 \] \[ n - 1 = \frac{360}{12} = 30 \implies n = 31 \] The 31st term of the A.P. is 120 more than its 21st term.
In simple words: Cancelling \( a \) early on simplifies the equation to \( (n-1)12 = 360 \), which gives \( n = 31 \) directly.

Exam Tip: Cancelling the first term \( a \) from both sides of the equation before substituting values is a great shortcut to save time.

Question 23. In an A.P., the first term is 25, nth term is -17 and sum to first n terms is 60.Find n and d the common difference.
Answer: We are given:
- First term \( (a) = 25 \)
- \( n \)-th term \( (a_n) = -17 \)
- Sum \( (S_n) = 60 \)
Using the sum formula: \[ S_n = \frac{n}{2}(a + a_n) \] \[ 60 = \frac{n}{2}(25 - 17) \] \[ 60 = \frac{n}{2}(8) \] \[ 60 = 4n \implies n = 15 \] Now, use the \( n \)-th term formula with \( n = 15 \) to find \( d \): \[ a_{15} = a + 14d \] \[ -17 = 25 + 14d \] \[ -17 - 25 = 14d \] \[ -42 = 14d \implies d = -3 \] Thus, \( n = 15 \) and \( d = -3 \).
In simple words: Using the sum formula gives \( n = 15 \). Substituting this back into the \( n \)-th term formula gives the common difference \( d = -3 \).

Exam Tip: Always use the simpler sum formula \( S_n = \frac{n}{2}(a+l) \) when both the first and last terms are given.

Question 24. which term of the sequence 114, 109, 104… is the first negative term? (n =24)
Answer: For the given sequence:
- First term \( (a) = 114 \)
- Common difference \( (d) = 109 - 114 = -5 \)
We want to find the first term where \( a_n < 0 \): \[ a + (n - 1)d < 0 \] \[ 114 + (n - 1)(-5) < 0 \] \[ 114 - 5n + 5 < 0 \] \[ 119 < 5n \] \[ n > \frac{119}{5} = 23.8 \] The smallest integer greater than 23.8 is 24.
Thus, the 24th term is the first negative term.
In simple words: Set up the inequality \( a_n < 0 \). Solving this shows that \( n \) must be greater than 23.8, so the 24th term is the first negative term.

Exam Tip: Be careful with inequality direction when dividing by a negative number, or rearrange terms to keep coefficients positive as shown above.

Question 25. Which term of the sequence 121, 117, 113… is the first negative term? (32)
Answer: For the given sequence:
- First term \( (a) = 121 \)
- Common difference \( (d) = 117 - 121 = -4 \)
We set up the inequality for \( a_n < 0 \): \[ a + (n - 1)d < 0 \] \[ 121 + (n - 1)(-4) < 0 \] \[ 121 - 4n + 4 < 0 \] \[ 125 < 4n \] \[ n > \frac{125}{4} = 31.25 \] The smallest integer greater than 31.25 is 32.
Thus, the 32nd term is the first negative term of the sequence.
In simple words: The numbers decrease by 4. Setting the \( n \)-th term to be less than 0 shows that \( n \) must be greater than 31.25, meaning the 32nd term is the first negative term.

Exam Tip: Clearly show the division step to justify why you round up to the next integer for the final value of \( n \).

Question 26. If the 4th term of an A.P is twice the 8th term, prove that the 10th term is twice the 11th term
Answer: Let the first term be \( a \) and the common difference be \( d \).
The given condition is: \[ a_4 = 2(a_8) \] \[ a + 3d = 2(a + 7d) \] \[ a + 3d = 2a + 14d \] \[ a = -11d \] (Equation 1) Now, let's write the expressions for the 10th and 11th terms: - \( a_{10} = a + 9d \) - \( a_{11} = a + 10d \) Substitute \( a = -11d \) into these: - \( a_{10} = -11d + 9d = -2d \) - \( a_{11} = -11d + 10d = -d \) Comparing the two terms: \[ a_{10} = -2d = 2(-d) = 2(a_{11}) \] Thus, the 10th term is twice the 11th term. (Hence proved)
In simple words: The given condition simplifies to \( a = -11d \). Substituting this into the formulas for the 10th and 11th terms shows that the 10th term (\( -2d \)) is indeed twice the 11th term (\( -d \)).

Exam Tip: Expressing all term formulas in terms of a single variable (here, \( d \)) makes comparison proofs very straightforward.

Question 27. If 2 + 5 + 8 + …………………………+ x = 155, find x (n = 10, x = a10=29)
Answer: This is a sum of an A.P. sequence:
- First term \( (a) = 2 \)
- Common difference \( (d) = 5 - 2 = 3 \)
- Sum \( (S_n) = 155 \)
We use the sum formula to find the number of terms \( n \): \[ S_n = \frac{n}{2}[2a + (n-1)d] \] \[ 155 = \frac{n}{2}[2(2) + (n-1)3] \] \[ 310 = n[4 + 3n - 3] \] \[ 310 = n[3n + 1] \] \[ 3n^2 + n - 310 = 0 \] We solve this quadratic equation: \[ 3n^2 - 30n + 31n - 310 = 0 \] \[ 3n(n - 10) + 31(n - 10) = 0 \] \[ (3n + 31)(n - 10) = 0 \] Since \( n \) must be a positive integer, we choose \( n = 10 \). Now, find the 10th term, which is \( x \): \[ x = a_{10} = a + 9d \] \[ x = 2 + 9(3) = 2 + 27 = 29 \] Thus, \( x = 29 \).
In simple words: Solving the sum equation gives \( n = 10 \). Calculating the 10th term of the sequence gives the final value of \( x = 29 \).

Exam Tip: Be careful to solve for \( x \) (the last term) rather than stopping at the number of terms \( n \).

Question 28. Find the sum of the following A.P: 1 + 3 + 5 + …….. + 199. (10000)
Answer: This is the sum of consecutive odd numbers:
- First term \( (a) = 1 \)
- Last term \( (a_n) = 199 \)
- Common difference \( (d) = 2 \)
Find the number of terms \( n \): \[ a_n = a + (n - 1)d \] \[ 199 = 1 + (n - 1)2 \] \[ 198 = 2(n - 1) \implies n - 1 = 99 \implies n = 100 \] Now, calculate the sum \( S_{100} \): \[ S_{100} = \frac{100}{2}(1 + 199) \] \[ S_{100} = 50(200) = 10000 \]
In simple words: The sequence has 100 terms of odd numbers. Summing them up gives 10000.

Exam Tip: Keep in mind that the sum of the first \( n \) odd numbers is always \( n^2 \) - this serves as a great shortcut to verify your work.

Question 29. For A.P. a1, a2, a3… if a4/a7 = 2/3 , find a6/a8
Answer: We are given: \[ \frac{a_4}{a_7} = \frac{2}{3} \] \[ \frac{a + 3d}{a + 6d} = \frac{2}{3} \] Cross-multiplying: \[ 3(a + 3d) = 2(a + 6d) \] \[ 3a + 9d = 2a + 12d \] \[ a = 3d \] (Equation 1) Now, find the ratio \( \frac{a_6}{a_8} \): \[ \frac{a_6}{a_8} = \frac{a + 5d}{a + 7d} \] Substitute \( a = 3d \) into the expression: \[ \frac{a_6}{a_8} = \frac{3d + 5d}{3d + 7d} \] \[ \frac{a_6}{a_8} = \frac{8d}{10d} = \frac{4}{5} \] The ratio \( a_6 / a_8 \) is \( 4/5 \).
In simple words: The given ratio simplifies to \( a = 3d \). Substituting this into the formula for \( a_6/a_8 \) simplifies the ratio to \( 4/5 \).

Exam Tip: Substituting the relation \( a = 3d \) eliminates all variables from the fraction, leaving a simple numerical ratio.

Question 30. 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 (d= - 10)
Answer: Let the first term be \( a = 100 \) and the common difference be \( d \).
According to the given condition: \[ S_6 = 5 \times (S_{12} - S_6) \] \[ S_6 = 5S_{12} - 5S_6 \] \[ 6S_6 = 5S_{12} \] Now, write the formulas for \( S_6 \) and \( S_{12} \): - \( S_6 = \frac{6}{2}[2(100) + 5d] = 3[200 + 5d] = 600 + 15d \) - \( S_{12} = \frac{12}{2}[2(100) + 11d] = 6[200 + 11d] = 1200 + 66d \) Substitute these back: \[ 6(600 + 15d) = 5(1200 + 66d) \] \[ 3600 + 90d = 6000 + 330d \] \[ 3600 - 6000 = 330d - 90d \] \[ -2400 = 240d \implies d = -10 \] The common difference of the AP is -10.
In simple words: The relation \( 6S_6 = 5S_{12} \) is derived from the condition. Substituting the formulas and solving for \( d \) gives the common difference as -10.

Exam Tip: Representing "the sum of the next 6 terms" as \( S_{12} - S_6 \) is the cleanest way to set up the equation.

Question 31. The angles of a triangle are in A.P, the last being half the greatest. Find the angles. (40˚, 60˚, 80˚)
Answer: Let the three angles of the triangle in A.P. be \( a - d \), \( a \), and \( a + d \). The sum of angles in a triangle is \( 180^\circ \): \[ (a - d) + a + (a + d) = 180^\circ \] \[ 3a = 180^\circ \implies a = 60^\circ \] So, the angles are \( 60^\circ - d \), \( 60^\circ \), and \( 60^\circ + d \). According to the condition, the smallest angle is half the greatest angle: \[ 60^\circ - d = \frac{60^\circ + d}{2} \] \[ 2(60^\circ - d) = 60^\circ + d \] \[ 120^\circ - 2d = 60^\circ + d \] \[ 120^\circ - 60^\circ = 3d \] \[ 60^\circ = 3d \implies d = 20^\circ \] Now find the three angles: - \( a - d = 60^\circ - 20^\circ = 40^\circ \) - \( a = 60^\circ \) - \( a + d = 60^\circ + 20^\circ = 80^\circ \) The angles of the triangle are 40˚, 60˚, and 80˚.
In simple words: Symmetrical representation of angles gives \( a = 60^\circ \). Using the "smallest is half of largest" condition gives \( d = 20^\circ \), which defines the angles as 40˚, 60˚, and 80˚.

Exam Tip: Always use the symmetrical representation \( a-d, a, a+d \) for three terms in A.P., as this instantly solves for \( a \) using the sum.

Question 32. The sum of 3 numbers in A.P is 3 and their product is -35. Find the numbers (7, 1, and -5)
Answer: Let the three numbers in A.P. be \( a - d \), \( a \), and \( a + d \). The sum of the numbers is 3: \[ (a - d) + a + (a + d) = 3 \] \[ 3a = 3 \implies a = 1 \] The product of the numbers is -35: \[ (a - d) \times a \times (a + d) = -35 \] Substituting \( a = 1 \): \[ (1 - d) \times 1 \times (1 + d) = -35 \] \[ 1 - d^2 = -35 \] \[ d^2 = 36 \implies d = \pm 6 \] - If \( d = 6 \), the numbers are \( 1 - 6 = -5 \), \( 1 \), and \( 1 + 6 = 7 \). - If \( d = -6 \), the numbers are \( 1 - (-6) = 7 \), \( 1 \), and \( 1 + (-6) = -5 \). In both cases, the three numbers are -5, 1, and 7.
In simple words: Representing the terms symmetrically gives \( a = 1 \). Using the product condition yields \( d^2 = 36 \), which gives the numbers as -5, 1, and 7.

Exam Tip: Be sure to write out the sequence of numbers for both positive and negative values of \( d \); they yield the same set of numbers but in different order.

Question 33. Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two Is 154. Find the integers (3, 5, 7……….).
Answer: Note: The printed answer in parenthesis is \( (3, 5, 7\dots) \), which is a sequence of odd numbers. However, the question says "Three consecutive positive integers". Let's solve using the mathematically consistent standard version where they are consecutive positive integers:
Let the three consecutive positive integers be \( x \), \( x + 1 \), and \( x + 2 \).
According to the condition: \[ x^2 + (x + 1)(x + 2) = 154 \] \[ x^2 + (x^2 + 3x + 2) = 154 \] \[ 2x^2 + 3x - 152 = 0 \] We solve this quadratic equation by splitting the middle term: \[ 2x^2 - 16x + 19x - 152 = 0 \] \[ 2x(x - 8) + 19(x - 8) = 0 \] \[ (2x + 19)(x - 8) = 0 \] Since \( x \) must be a positive integer, we choose \( x = 8 \). The three consecutive positive integers are: - \( x = 8 \) - \( x + 1 = 9 \) - \( x + 2 = 10 \) If the question is instead "consecutive positive odd integers" \( 2k-1, 2k+1, 2k+3 \): \[ (2k-1)^2 + (2k+1)(2k+3) = 154 \implies 8k^2 + 8k + 4 = 154 \implies 8k^2 + 8k - 150 = 0 \] (not clean integer roots). Thus, the three consecutive positive integers are 8, 9, and 10.
In simple words: Let the three consecutive positive integers be \( x \), \( x+1 \), and \( x+2 \). Setting up the equation gives \( x = 8 \), so the integers are 8, 9, and 10.

Exam Tip: Always verify that your final integers satisfy the original word problem description to ensure there are no setup errors.

Question 34. Find the sum of n terms of an A.P whose nth term is given by tn = 5 – 6n (2n – 3n2)
Answer: The \( n \)-th term is \( t_n = 5 - 6n \). Let's find the first few terms:
- For \( n = 1 \): \( a_1 = t_1 = 5 - 6(1) = -1 \)
- For \( n = 2 \): \( a_2 = t_2 = 5 - 6(2) = -7 \) The common difference \( d \) is: \[ d = a_2 - a_1 = -7 - (-1) = -6 \] Now calculate the sum of \( n \) terms \( S_n \): \[ S_n = \frac{n}{2}[2a_1 + (n-1)d] \] \[ S_n = \frac{n}{2}[2(-1) + (n-1)(-6)] \] \[ S_n = \frac{n}{2}[-2 - 6n + 6] \] \[ S_n = \frac{n}{2}[4 - 6n] \] \[ S_n = n(2 - 3n) = 2n - 3n^2 \] The sum of the first \( n \) terms is \( 2n - 3n^2 \).
In simple words: Find \( a_1 = -1 \) and \( d = -6 \) from the \( t_n \) formula. Substituting these into the sum formula gives \( 2n - 3n^2 \).

Exam Tip: Alternatively, you can use \( S_n = \frac{n}{2}(a_1 + t_n) \), which is much faster: \( S_n = \frac{n}{2}(-1 + 5 - 6n) = \frac{n}{2}(4 - 6n) = 2n - 3n^2 \).

Question 35. Find the middle term of A.P: 1, 8, 15, ……………, 505 (253)
Answer: Let's find the number of terms \( n \) in the A.P.:
- First term \( (a) = 1 \)
- Common difference \( (d) = 8 - 1 = 7 \)
- Last term \( (a_n) = 505 \) \[ 505 = 1 + (n - 1)7 \] \[ 504 = 7(n - 1) \implies n - 1 = 72 \implies n = 73 \] Since \( n = 73 \) is odd, there is a single middle term at the position: \[ \text{Middle Position} = \frac{n + 1}{2} = \frac{74}{2} = 37\text{th term} \] Now calculate the 37th term \( a_{37} \): \[ a_{37} = a + 36d \] \[ a_{37} = 1 + 36(7) \] \[ a_{37} = 1 + 252 = 253 \] The middle term of the A.P. is 253.
In simple words: The sequence has 73 terms. The middle term is the 37th term, which is calculated to be 253.

Exam Tip: If the total number of terms \( n \) is even, there will be two middle terms, but since \( n = 73 \) is odd here, we have exactly one middle term.

Question 36. Find the number of terms of the A.P, 63, 60, 57, ……….. So that their sum is 693 (n = 22, 21)
Answer: For the given progression:
- First term \( (a) = 63 \)
- Common difference \( (d) = 60 - 63 = -3 \)
- Sum \( (S_n) = 693 \) Using the sum formula: \[ S_n = \frac{n}{2}[2a + (n-1)d] \] \[ 693 = \frac{n}{2}[2(63) + (n-1)(-3)] \] \[ 1386 = n[126 - 3n + 3] \] \[ 1386 = n[129 - 3n] \] Divide the entire equation by 3: \[ 462 = n[43 - n] \] \[ n^2 - 43n + 462 = 0 \] We solve this quadratic equation by splitting the middle term: \[ n^2 - 21n - 22n + 462 = 0 \] \[ n(n - 21) - 22(n - 21) = 0 \] \[ (n - 21)(n - 22) = 0 \] This gives two possible integer values: \[ n = 21 \quad \text{or} \quad n = 22 \] Both values are positive integers, which means both 21 terms and 22 terms can sum to 693.
In simple words: Solving the quadratic equation gives \( n = 21 \) or \( n = 22 \). Both are valid because the 22nd term is zero, so adding it doesn't change the sum.

Exam Tip: Explain why there are two answers: since the 22nd term of the progression is \( 63 + 21(-3) = 0 \), adding the 22nd term does not change the sum of the sequence.

Question 37. How many terms of the sequence 18, 16, 14, …………, should be taken so that their sum is 0 (n= 19)
Answer: For the given sequence:
- First term \( (a) = 18 \)
- Common difference \( (d) = 16 - 18 = -2 \)
- Sum \( (S_n) = 0 \) Using the sum formula: \[ S_n = \frac{n}{2}[2a + (n-1)d] \] \[ 0 = \frac{n}{2}[2(18) + (n-1)(-2)] \] Since \( n \ne 0 \), we can divide both sides by \( \frac{n}{2} \): \[ 36 - 2n + 2 = 0 \] \[ 38 - 2n = 0 \] \[ 2n = 38 \implies n = 19 \] Therefore, 19 terms should be taken so that their sum is 0.
In simple words: Setting the sum to zero and solving the formula gives \( n = 19 \). The sum is zero because the negative terms eventually cancel out the positive terms.

Exam Tip: Since \( n \) represents the number of terms, it must always be a non-zero positive integer.

Question 38. 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. (320, 280, 240, 200, 160, 120, 80)
Answer: Let the value of the first prize be Rs. \( a \). Since each prize is Rs. 40 less than the preceding one:
- Common difference \( (d) = -40 \)
- Number of prizes \( (n) = 7 \)
- Total sum \( (S_7) = 1400 \) Using the sum formula: \[ S_7 = \frac{7}{2}[2a + 6d] \] \[ 1400 = \frac{7}{2}[2a + 6(-40)] \] Multiply both sides by \( \frac{2}{7} \): \[ 400 = 2a - 240 \] \[ 2a = 640 \implies a = 320 \] The first prize is Rs. 320. Now calculate the remaining prizes by subtracting Rs. 40 successively: - First prize = Rs. 320 - Second prize = Rs. 280 - Third prize = Rs. 240 - Fourth prize = Rs. 200 - Fifth prize = Rs. 160 - Sixth prize = Rs. 120 - Seventh prize = Rs. 80 The values of the prizes are Rs. 320, Rs. 280, Rs. 240, Rs. 200, Rs. 160, Rs. 120, and Rs. 80.
In simple words: Setting up the sum equation gives the first prize as Rs. 320. Subtracting Rs. 40 successively gives the other prizes.

Exam Tip: List all 7 values clearly in your final answer to ensure you satisfy the full requirements of the question.

Question 39. Verify that a + b, (a + 1) + b, (a + 1) + (b + 1) ……….. Is an A.P. and then write its next term (a+2) + (b+1)
Answer: Let the terms of the sequence be:
- \( T_1 = a + b \)
- \( T_2 = (a + 1) + b = a + b + 1 \)
- \( T_3 = (a + 1) + (b + 1) = a + b + 2 \) Let's find the difference between successive terms: - \( T_2 - T_1 = (a + b + 1) - (a + b) = 1 \) - \( T_3 - T_2 = (a + b + 2) - (a + b + 1) = 1 \) Since the difference between successive terms is constant (\( d = 1 \)), the sequence is an A.P.
The next term \( T_4 \) of the progression is: \[ T_4 = T_3 + 1 = a + b + 3 \] We can write this in standard grouped form: \[ T_4 = (a + 2) + (b + 1) \] Thus, the sequence is an A.P. and its next term is \( (a+2) + (b+1) \).
In simple words: Each term is exactly 1 larger than the previous one, showing it is an A.P. with \( d = 1 \). The next term in the grouped form is \( (a+2) + (b+1) \).

Exam Tip: Showing that \( T_2 - T_1 = T_3 - T_2 = \text{constant} \) is the standard way to verify that any sequence is an A.P.

Question 40. Determine the A.P whose 3rd term is 16 and 7th term exceeds the 5th term by 12 (4, 6, 10, 16 …)
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
- First condition: \[ a_3 = a + 2d = 16 \] (Equation 1) - Second condition: \[ a_7 = a_5 + 12 \] \[ a + 6d = (a + 4d) + 12 \] Cancel \( a \) from both sides: \[ 6d = 4d + 12 \] \[ 2d = 12 \implies d = 6 \] Substitute \( d = 6 \) into Equation 1: \[ a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4 \] The first term \( a_1 = 4 \) and \( d = 6 \). The Arithmetic Progression is: 4, 10, 16, 22, ...
In simple words: The second condition simplifies to \( d = 6 \). Substituting this into the first condition gives \( a = 4 \), defining the progression as 4, 10, 16, 22, ...

Exam Tip: Notice that the terms printed in parenthesis in the original worksheet `(4, 6, 10, 16 …)` has a typo (it has 6 instead of 10), but our derived progression \( 4, 10, 16, 22, \dots \) is mathematically exact.

Question 41. 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 (n = 7 )
Answer: Let's find the formula for the \( n \)-th term of both progressions:
- For the first A.P. \( 9, 7, 5, \dots \):
First term \( a_1 = 9 \) and common difference \( d_1 = 7 - 9 = -2 \). \[ T_{n1} = a_1 + (n - 1)d_1 = 9 + (n - 1)(-2) = 11 - 2n \] - For the second A.P. \( 15, 12, 9, \dots \):
First term \( a_2 = 15 \) and common difference \( d_2 = 12 - 15 = -3 \). \[ T_{n2} = a_2 + (n - 1)d_2 = 15 + (n - 1)(-3) = 18 - 3n \] Since the \( n \)-th terms of both progressions are equal: \[ 11 - 2n = 18 - 3n \] \[ 3n - 2n = 18 - 11 \] \[ n = 7 \] Thus, the value of \( n \) is 7.
In simple words: Write the general formula for the \( n \)-th term of both progressions and set them equal to each other. Solving this simple equation gives \( n = 7 \).

Exam Tip: Be careful with the signs of the common differences, which are both negative because the terms of both progressions are decreasing.

Question 42. Find the sum of first 22 terms of an A.P. in which d = 7 and 22nd term is 149 (1661)
Answer: We are given:
- Common difference \( d = 7 \)
- 22nd term \( T_{22} = 149 \)
We can find the first term \( a \) using the formula for the \( n \)-th term: \[ T_{22} = a + 21d \] \[ 149 = a + 21(7) \] \[ 149 = a + 147 \] \[ a = 2 \] Now, calculate the sum of the first 22 terms (\( S_{22} \)) using the formula: \[ S_n = \frac{n}{2}(a + T_n) \] \[ S_{22} = \frac{22}{2}(2 + 149) \] \[ S_{22} = 11(151) = 1661 \] The sum of the first 22 terms is 1661.
In simple words: Use the 22nd term to find the first term \( a = 2 \). Then apply the sum formula with the first and last terms to get 1661.

Exam Tip: Using the simpler formula \( S_n = \frac{n}{2}(a + l) \) when the last term is known is faster and avoids unnecessary calculations.

Question 43. Find the sum of the following A.P: 3, 9/2, 6, 15/2……. To 25 terms (525)
Answer: For the given progression:
- First term \( a = 3 \)
- Common difference \( d = \frac{9}{2} - 3 = \frac{3}{2} \)
- Number of terms \( n = 25 \) Using the sum formula: \[ S_n = \frac{n}{2}[2a + (n - 1)d] \] \[ S_{25} = \frac{25}{2}\left[2(3) + (25 - 1)\left(\frac{3}{2}\right)\right] \] \[ S_{25} = \frac{25}{2}\left[6 + 24\left(\frac{3}{2}\right)\right] \] \[ S_{25} = \frac{25}{2}[6 + 36] \] \[ S_{25} = \frac{25}{2}(42) = 25 \times 21 = 525 \] The sum of the first 25 terms is 525.
In simple words: The common difference here is \( 1.5 \) (or \( 3/2 \)). Substituting \( a = 3 \) and \( d = 3/2 \) into the sum formula gives \( 525 \).

Exam Tip: Keep common differences in fraction form (\( 3/2 \)) rather than decimals to simplify your arithmetic during multiplication steps.

Question 44. The ratio of the sum to p terms and q terms of an A.P. is p2 : q2. Prove that the common difference of the A.P.is twice the first term
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \). We are given: \[ \frac{S_p}{S_q} = \frac{p^2}{q^2} \] Using the sum formula: \[ \frac{\frac{p}{2}[2a + (p - 1)d]}{\frac{q}{2}[2a + (q - 1)d]} = \frac{p^2}{q^2} \] Cancel \( \frac{1}{2} \) and one \( p \) and \( q \) from both sides: \[ \frac{2a + (p - 1)d}{2a + (q - 1)d} = \frac{p}{q} \] Cross-multiplying: \[ q[2a + (p - 1)d] = p[2a + (q - 1)d] \] \[ 2aq + pqd - qd = 2ap + pqd - pd \] Cancel the common term \( pqd \) from both sides: \[ 2aq - qd = 2ap - pd \] \[ 2aq - 2ap = qd - pd \] \[ 2a(q - p) = d(q - p) \] Dividing both sides by \( (q - p) \) (since \( p \ne q \)): \[ 2a = d \implies d = 2a \] Thus, the common difference is twice the first term. (Hence proved)
In simple words: Write the ratio of sums and simplify it. Cross-multiplying the remaining terms cancels out the mixed variables, directly proving \( d = 2a \).

Exam Tip: Cancelling the common factors \( p \) and \( q \) from both sides early in the step simplifies the algebraic equations significantly.

Question 45. If 6 times the sixth term of an A.P is equal to 15 times the fifteenth term, find its 21st term (0)
Answer: Let the first term be \( a \) and the common difference be \( d \). Based on the given condition: \[ 6(T_6) = 15(T_{15}) \] Using the \( n \)-th term formula: \[ 6(a + 5d) = 15(a + 14d) \] Dividing both sides by 3: \[ 2(a + 5d) = 5(a + 14d) \] \[ 2a + 10d = 5a + 70d \] \[ 5a - 2a = 10d - 70d \] \[ 3a = -60d \implies a = -20d \] Now, we calculate the 21st term (\( T_{21} \)): \[ T_{21} = a + 20d \] Substituting \( a = -20d \): \[ T_{21} = -20d + 20d = 0 \] Thus, the 21st term of the A.P. is 0.
In simple words: Simplify the given expression to show that \( a = -20d \). Since the 21st term formula is \( a + 20d \), substituting \( a \) gives exactly 0.

Exam Tip: When a problem asks to find a specific term and gives a relation of two other terms, look to express \( a \) in terms of \( d \) first.

Question 46. 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? (3450)
Answer: The number of seats in successive rows forms an Arithmetic Progression:
- First row (\( a \)) = 20 seats
- Common difference (\( d \)) = 2 seats
- Number of rows (\( n \)) = 50 rows Using the sum formula to find the total number of seats (\( S_{50} \)): \[ S_n = \frac{n}{2}[2a + (n - 1)d] \] \[ S_{50} = \frac{50}{2}[2(20) + (50 - 1)2] \] \[ S_{50} = 25[40 + 98] \] \[ S_{50} = 25[138] = 3450 \] Therefore, there are 3450 seats in the auditorium.
In simple words: The seating arrangement forms an AP starting at 20 and increasing by 2 for each of the 50 rows. Summing this progression yields 3450 total seats.

Exam Tip: Always identify the terms \( a \), \( d \), and \( n \) from the word problem clearly before applying the sum formula.

Question 47. I n 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 \( T_n = 218 \) First, let's find the total number of terms \( n \): \[ T_n = a + (n - 1)d \] \[ 218 = 8 + (n - 1)7 \] \[ 210 = 7(n - 1) \implies n - 1 = 30 \implies n = 31 \] Since \( n = 31 \) is odd, there is a single middle term at position: \[ \text{Middle Position} = \frac{n + 1}{2} = \frac{32}{2} = 16\text{th term} \] Now, calculate the 16th term (\( T_{16} \)): \[ T_{16} = a + 15d \] \[ T_{16} = 8 + 15(7) = 8 + 105 = 113 \] Thus, the middle term of the A.P. is 113.
In simple words: The sequence has 31 terms. The middle term is the 16th term, which we calculate to be 113.

Exam Tip: When \( n \) is odd, there is exactly one middle term at position \( \frac{n+1}{2} \).

Question 48. 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 the first term be \( a \) and the common difference be \( d \).
- Sum of first 5 terms (\( S_5 \)) = 25: \[ \frac{5}{2}[2a + 4d] = 25 \implies 5(a + 2d) = 25 \implies a + 2d = 5 \] (Equation 1) - Sum of next 5 terms is \( -75 \).
This means the sum of the first 10 terms (\( S_{10} \)) is: \[ S_{10} = S_5 + \text{sum of next 5 terms} = 25 + (-75) = -50 \] Using the sum formula for \( S_{10} \): \[ \frac{10}{2}[2a + 9d] = -50 \implies 5(2a + 9d) = -50 \implies 2a + 9d = -10 \] (Equation 2) Multiply Equation 1 by 2: \[ 2a + 4d = 10 \] (Equation 3) Subtract Equation 3 from Equation 2: \[ 5d = -20 \implies d = -4 \] Substitute \( d = -4 \) into Equation 1: \[ a + 2(-4) = 5 \implies a - 8 = 5 \implies a = 13 \] Now, find the 10th term (\( T_{10} \)): \[ T_{10} = a + 9d = 13 + 9(-4) = 13 - 36 = -23 \] The 10th term of the A.P. is -23.
In simple words: The sum of the first 5 terms is 25, and the sum of the first 10 terms is \( 25 - 75 = -50 \). Solving these equations gives \( a = 13 \) and \( d = -4 \), which gives the 10th term as -23.

Exam Tip: Representing "the sum of the next 5 terms" as \( S_{10} - S_5 \) avoids dealing with complex starting terms for the second interval.

Question 49. In an A.P the sum of first n terms is \( \frac{3n^2}{2} + \frac{5n}{2} \). Find its 25th term. (76)
Answer: The sum of first \( n \) terms is given by: \[ S_n = \frac{3n^2}{2} + \frac{5n}{2} \] We can find the 25th term (\( T_{25} \)) using the relation: \[ T_n = S_n - S_{n-1} \] Let's calculate \( S_{25} \) and \( S_{24} \): - For \( n = 25 \): \[ S_{25} = \frac{3(25)^2}{2} + \frac{5(25)}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000 \] - For \( n = 24 \): \[ S_{24} = \frac{3(24)^2}{2} + \frac{5(24)}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924 \] Now, calculate \( T_{25} \): \[ T_{25} = S_{25} - S_{24} = 1000 - 924 = 76 \] The 25th term of the A.P. is 76.
In simple words: Find the sum of the first 25 terms (1000) and subtract the sum of the first 24 terms (924) to get the 25th term as 76.

Exam Tip: Subtracting \( S_{n-1} \) from \( S_n \) is a very quick and standard way to find any specific term from a given sum formula.