Doubt by Shaurya
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=32
a-3d+a-d+a+d+a+3d=32
4a=32
a=32/4
a=8 — (1)
(a1×a4):(a2×a3) = 7:15
(a1×a4)/(a2×a3) = 7/15
(a-3d)(a+3d)/(a-d)(a+d)=7/15
(a²-9d²)/(a²-d²)=7/15
(a²-9d²)/(a²-d²)=7/15
15(a²-9d²)/=7×(a²-d²)
15a²-135d²=7a²-7d²
15a²-7a²=135d²-7d²
8a²=128d²
8(8)²/128=d² [From equation (1)]
8×64/128=d²
8/2=d²
4=d²
d²=4
d=±√4
(a²-9d²)/(a²-d²)=7/15
(a²-9d²)/(a²-d²)=7/15
15(a²-9d²)/=7×(a²-d²)
15a²-135d²=7a²-7d²
15a²-7a²=135d²-7d²
8a²=128d²
8(8)²/128=d² [From equation (1)]
8×64/128=d²
8/2=d²
4=d²
d²=4
d=±√4
d=±2
When d=+2, a=8
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
a3=a+d = 8+2 = 10
a4=a+3d = 8+3(2) = 8+6 = 14
When d=-2, a=8
a1=a-3d = 8-3(-2) = 8+6 = 14
a2=a-d = 8-(-2) = 8+2 = 10
a3=a+d = 8+(-2) = 8-2 = 6
a4=a+3d = 8+3(-2) = 8-6 = 2
a3=a+d = 8+(-2) = 8-2 = 6
a4=a+3d = 8+3(-2) = 8-6 = 2
Hence, the required four parts are 2, 6, 10, 14 or 14, 10, 6, 2
Similar 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.