Question

(ii) Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd
and the 4th is 105. The terms are in ascending order.​

Answer 1

ANSWER:

Given:

• Sum of 4 consecutive terms of an AP = 36
• Product of 2nd and 4th term = 105

To Find:

• The 4 numbers

Assumption:

• Let the terms be a-3d, a-d, a+d, a+3d.

Solution:

⇒ Sum of the terms = 36

So, on placing the terms,

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 36

⇒ a - 3d + a - d + a + d + a + 3d = 36

⇒ a + a + a + a - 3d - d + d + 3d = 36

All the ds get cancelled. So,

⇒ 4a = 36

⇒ a = 9 ------------(1)

Now, we know that:

Product of 2nd and 3rd term = 105

⇒ (a - d)*(a + 3d) = 105

⇒ a² + 3ad - ad - 3d² = 105

From (1),

⇒ (9)² + 2(9)d - 3d² = 105

⇒ 81 + 18d - 3d² = 105

Transposing 81 to RHS,

⇒ 18d - 3d² = 105 - 81

⇒ 18d - 3d² = 24

Transposing 24 to LHS,

⇒ -3d² + 18d -24 = 0

⇒ -(3d² - 18d + 24) = 0

⇒ 3d² - 18d + 24 = 0

Taking 3 common and cancelling it,

⇒ d² - 6d + 8 = 0

On splitting the middle term,

⇒ d² - 2d - 4d + 8 = 0

⇒ d(d - 2) - 4(d - 2) = 0

⇒ (d - 2)*(d - 4) = 0

So,

⇒ d = 2 OR d = 4 ----------(2)

The terms when d = 2:

• a - 3d = 9 - 3(2) = 9 - 6 = 3
• a - d = 9 - 2 = 7
• a + d = 9 + 2 = 11
• a + 3d = 9 + 3(2) = 9 + 6 = 15

The terms when d = 4:

• a - 3d = 9 - 3(4) = 9 - 12 = -3
• a - d = 9 - 4 = 5
• a + d = 9 + 4 = 13
• a + 3d = 9 + 3(4) = 9 + 12 = 21

Hence, the terms are 3, 7, 11, 15 OR -3, 5, 13, 21.

Verification:

1. Sum of terms:

⇒ 3 + 7 + 11 + 15 & -3 + 5 + 13 + 21

⇒ 36

2. Product of the 2nd and the 4th term:

⇒ 7 * 15 & 5 * 21

⇒ 105

Hence verified!!!