The sum of four consecutive terms which are in . . .

Question : The sum of four consecutive terms which are in an arithmetic progression 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 four terms of the AP.

Doubt by Bhavya 

Solution : 

Let the four consecutive terms of the AP be 

a₁=a-3d

a₂=a-d

a₃=a+d

a₄=a+3d

ATQ

a₁+a₂+a₃+a₄=32

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

4a+0=32

4a=32

a=32/4

a=8 — (1)

Also

a₁×a₄:a₂×a₃=7:15 (Given)
[a₁×a₄]/[a₂×a₃]=7/15

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

[a²-9d²]/[a²-d²]=7/15 [∵(a+b)(a-b)=a²-b²]

[(8)²-9d²]/[(8)²-d²]=7/15 [Using eq (1)]

[64-9d²]/[64-d²]=7/15

15[64-9d²]=7[64-d²]

960-135d²=448-7d²
960-448=135d²-7d²

512=128d²
512/128=d²

4=d²

d²=4

d=±√4

d=±2

When a=8 and d=2

a₁=a-3d 
= 8-3(2) 
= 8-6 = 2

a₂=a-d
=8-2

=6

a₃=a+d
=8+2
=10

a₄=a+3d

=8+3(2)

=8+6

=14

When a=8 and d=-2

a₁=a-3d
= 8-3(-2)
= 8+6 = 14

a₂=a-d
=8-(-2)

=10

a₃=a+d
=8+(-2)
=6

a₄=a+3d

=8+3(-2)

=8-6

=2

Hence, the required four terms of AP are 2,6,8,14 or 14,8,6,2