Question

The sum of 4 consecutive terms which are in an ap is 32 and the ratio of the product of the first and the last term of the product of 2 middle terms is 7:15. Find the number

Answer 1

a-3d+a-d+a+d+a+3d=32
4a=32
a=8
a2-9d2/a2-d2=7/15
128d2=512
d=+-4

Answer 2

Answer:

Let the four consecutive term of the AP be ,

a -  3 d , a - d , a + d , a + 3 d  

A.T.Q.

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

4 a = 32

a = 8

and  

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

15  ( a²  - 9 d² ) = 7 ( a²  -  d² )

8 a²  =  128 d² [ a = 8 ]  

d = ± 2

Therefore , numbers are 2 , 6 , 10 , 14 or 14 , 10 , 6 , 2 .