Question

the sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. find the numbers.

Answer 1

The answer is given below :

Let, the four consecutive numbers are
(a - 3d), (a - d), (a + d) and (a + 3d).

Given that,

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

=> 4a = 32

=> a = 8

So, the numbers are

(8 - 3d), (8 - d), (8 + d) and (8 + 3d).

Given that :

(8 - 3d)(8 + 3d) : (8 - d)(8 + d) = 7 : 15

=> (64 - 9d²) : (64 - d²) = 7 : 15

=> (64 - 9d²)/(64 - d²) = 7/15

=> 960 - 135d² = 448 - 7d²

=> 128d² = 512

=> d² = 4

So, d = ± 2.

So, the numbers are :

2, 6, 10, 14

or,

14, 10, 6, 2.

Therefore, the four consecutive numbers are

2, 6, 10, 14.

Thank you for your question.

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 .