Question

the sum of four consecutive numbers in an ap is 32 and the ratio of the product of the first and last term to the product of two middle term is 7 is to 15 find the numbers​

Answer 1

2, 6, 10, 14 a+2d a+d a-d a-2d are the numbers General form

Answer 2

Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

So,

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

4a=32

a=8

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

15(a2-9d2) = 7(a2-d2)

15a2-135d2 = 7a2-7d2

8a2-128d2 = 0

d2 = 4, ±2

Therefore d=4 or d=±2

So, when a=8 and d=2, the numbers are 2,6,10,14.

When a=8,d=-2 the numbers are 14,10,6,2