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.

Question

Philoid · Accepted Answer

Let 32 be divided into parts as (a - 3d), (a - d), (a + d) and (a + 3d).

Now (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8

Now, we are given that product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

i.e. [(a - 3d) × (a + 3 d)] : [(a - d) × (a + d)] = 7 : 15

⇒ =

⇒ 15[(a - 3d) × (a + 3 d)] = 7[(a - d) × (a + d)]

⇒ 15[a² - 9d²] = 7 [a² - d²]

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 8a² - 128d² = 0

⇒ 8a² = 128d²

Putting the value of a, we get,

512 = 128 d²

⇒ d² = 4

⇒ d = 2

∴ If d = 2, then the numbers are 2, 6, 10, 14.

If d = - 1, then the numbers are 14, 10, 6, 2.