Question

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

HINT: Let these parts be (a – 3d), (a – d), (a + d) and (a + 3d).

Answer 1

Let the four parts in AP be (a – 3d), (a – d), (a + d) and (a + 3d).

Then, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8   ...(1)

Also,

(a – 3d) (a + 3d) : (a – d) (a + d) = 7 : 15

⇒ `((8 - 3d) (8 + 3d))/((8 - d)(8 + d)) = 7/15`   ...[From (1)]

⇒ `(64 - 9d^2)/(64 - d^2) = 7/15`

⇒ 15(64 – 9d²) = 7(64 – d²)

⇒ 960 – 135d² = 448 – 7d²

⇒ 135d² – 7d² = 960 – 448

⇒ 128d² = 512

⇒ d² = 4

⇒ d = ±2

When a = 8 and d = 2,

a – 3d = 8 – 3 × 2 = 8 – 6 = 2

a – d = 8 – 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3 × 2 = 8 + 6 = 14

When a = 8 and d = –2,

a – 3d = 8 – 3 × (–2) = 8 + 6 = 14

a – d = 8 – (–2) = 8 + 2 = 10

a + d = 8 – 2 = 6

a + 3d = 8 + 3 × (–2) = 8 – 6 = 2

Hence, the four parts are 2, 6, 10 and 14.