Divide 56 in four parts in AP such that the ratio of the product of their . . .

Question : Divide 56 in four parts in AP such that the ratio of the product of their extremes to the product of means is 5:6.

Doubt by Vaishnavi

Solution : 

Let the four parts which are in AP are
a₁=a-3d
a₂=a-d
a₃=a+d
a₄=a+3d

ATQ

a₁+a₂+a₃+a₄=56

a-3d+a-d+a+d+a+3d=56

4a=56

a=56/4

a=14 — (1)

(a₁×a₄):(a₂×a₃) = 5:6

(a₁×a₄)/(a₂×a₃) = 5/6

(a-3d)(a+3d)/(a-d)(a+d)=5/6
(a²-9d²)/(a²-d²)=5/6
(a²-9d²)/(a²-d²)=5/6
6(a²-9d²)/=5×(a²-d²)
6a²-54d²=5a²-5d²
6a²-5a²=54d²-5d²
a²=49d²
14²=49d² [From equation (1)]
196/49=d²
4=d²
d=±√4

d=±2

When d=+2, a=14

a₁=a-3d = 14-3(2) = 14-6 = 8

a₂=a-d = 14-2 = 12
a₃=a+d = 14+2 = 16
a₄=a+3d = 14+3(2) = 14+6 = 20

When d=-2, a=14

a₁=a-3d = 14-3(-2) = 14+6 = 20

a₂=a-d = 14-(-2) = 14+2 = 16
a₃=a+d = 14+(-2) = 14-2 = 12
a₄=a+3d = 14+3(-2) = 14-6 = 8

Hence, the required four parts are 8,12,16,20 or 20,16,12,8