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

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.

Answer

Let the four consecutive terms of the A.P. be a - 3d, a - d, a + d, a + 3d.

By given conditions 

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

4a = 32 

a = 8 

$ \frac {(a - 3d)(a+3d)}{(a - d)(a+d)} = \frac{7}{15} $

8a² = 128d²

d² = 4

$ d = \pm{2} $

Numbers are 2, 6, 10, 14 or 14, 10, 6, 2.