Question

Divide $32$ into $4$ parts which are in an $AP$ such that the ratio of the product of the extremes to the products of means is $7:15$.

Answer 1

Let the numbers be: $a-3d,a-d,a+d,a+3d$

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

$4a=32$

$a=8$

$\frac{(8-3d)(8+3d)}{(8-d)(8+d)}=\frac{7}{15}$

$\frac{64+24d-24d-9d^2}{64+8d-8d-d^2}=\frac{7}{15}$

$\frac{64-9d^2}{64-d^2}=\frac{7}{15}$

$15(64-9d^2)=7(64-d^2)$

Solving the above equation, we get,

$d=2$

The numbers are: $a-3d,a-d,a+d,a+3d=2,6,10,14$