Question

divide 32 into 4 parts which are in AP such that the product of the extremes is to the product of means as 7 : 15

Answer 1

Let 4 terms in A.P be
(a-3d),(a-d),(a+d),(a+3d)
sum=4
a-3d+a+3d+a-d+a+d=32
4a=32
a=8

acc to ques,
((a-3d)*(a+3d))/((a-d)*(a+d))=7/15
(a²-9d²)/(a²-d²)=7/15
(64-9d²) /(64-d²)=7/15
15(64-9d²)=7(64-d²)
960-135d²=448-7d²
960-448=135d²-7d²
512=128d²
256=64d²
d=±16/8=±2
four numbers:for d=2
a-3d=8-6=2
a-d=8-2=6
a+d=8+2=10
a+3d=16
2,6,10,16