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
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
(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
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
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
Similar Question
Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7:15.