The sum of four consecutive numbers in an AP 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.

Let the numbers in A.P be a-3d, a-d, a+d, a+3d
Given: Sum of four consecutive numbers = 32
∴ a-3d + a-d + a+d + a+3d = 32
⇒ 4a = 32
⇒ a = 32/4 = 8
Now, A.T.Q,
(a-3d)(a+3d)/(a-d)(a+d) = 7/15
⇒ (a²-9d²)/(a²-d²) = 7/15
⇒ 15(a²-9d²) = 7(a²-d²)
⇒ 15(64-9d²) = 7(64-d²) [∵ a=8]
⇒ 960 - 135d² = 448 - 7d²
⇒ 960 - 448 = 135d² - 7d²
⇒ 512 = 128d²
⇒ d² = 512/128 = 4
⇒ d = ±2
∴ Either, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2