The sum of four consecutive numbers in an AP is 32 and the ratio of the product

• Arithmetic Progression (C10)

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 numbers in AP be;

(a - 3d), (a - d), (a + d) and (a + 3d)

So, according to the question

a - 3d + a - d + a + d + a + 3d = 32

4a = 32

a = 8

Now,

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

15(a² - 9d²) = 7(a² - d²)

15a² - 135d² = 7a² - 7d²

8a² = 128d²

Putting the value of a = 8, we get

d² = 4

d = ±2

So, the four consecutive numbers are 2, 6, 10, 14 or 14, 10, 6, 2