Question

The sum of four consecutive numbers in an A.P. is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers.

Answer 1

Let the 4 numbers which are in AP be

a – 3d, a – d, a + d, a + 3d   ...(i)

Given, the sum of these terms = 32

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8

Given, the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15

`((a - 3d)(a + 3d))/((a - d)(a + d)) = 7/15`

`(a^2 - 9d^2)/(a^2 - d^2) = 7/15`

15a² – 135d² = 7a² – 7d²

8a² = 128d²

128d² = 512   ...(∵ a = 8)

d² = 4

d = ± 2

Putting the value d = ± 2 and a = 8 in (i), we get the required number as 2, 6, 10, 14.

Answer 2

Let the four consecutive terms of A.P. be (a – 3d), (a – d), (a + d) and (a + 3d)

By given conditions,

a – 3d + a – d + a + d + a + 3d = 32

⇒ 4a = 32

⇒ a = 8

And `((a - 3d)(a + 3d))/((a - d)(a + d)) = 7/15`

⇒ `(a^2 - 9d^2)/(a^2 - d^2) = 7/15`

⇒ `((8)^2 - 9d^2)/((8)^2 - d^2) = 7/15`

⇒ `(64 - 9d^2)/(64 - d^2) = 7/15`

⇒ 960 – 135d² = 448 – 7d²

⇒ 7d² – 135d² = 448 – 960

⇒ –128d² = –512

⇒ d² = 4

⇒ d = ± 2

Hence, the number are 2, 6, 10 and 14 or 14, 10, 6 and 2.