Question

The sum of four consecutive number s in an AP is 32 and the ratio of the product of the first and the last term to product of two middle terms is7:15. Find the numbers

Answer 1

Hence,

a=8

d=2

The terms are

1. a-3d. = 8 - 6 = 2

2. a-d = 8 - 2=6

3. a+d = 8 + 2 =10

4 . a+3d = 8 + 6 = 14

Answer 2

four consecutive terms of an ap : a - 3d, a-d,a+d and a+3d
A.T.Q
a-3d +a-d +a+d+a+3d =32
4a=32
a=8 ......(1)
(a-3d) (a+3d):(a-d)(a+d)=7:15
a^2-9d^2/a^2-d^2=7/15
64-9d^2/64-d^2=7/15
(from (1) put a=8)
15(64-9d^2) =7(64-d^2)
960-135d^2=448-7d^2
960-448=135d^2-7d^2
512 =128d^2
d^2=512/128
d^2=4
d=+-2