Question

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

Answer 1

Let the four parts be $(a-3d),(a-d),(a+d)$ and $(a+3d)$.

Then, Sum of the numbers $=32$
$⟹(a-3d)+(a-d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8$
It is given that

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{7}{15}$
$⟹\frac{a^2-9d^2}{a^2-d^2}=\frac{7}{15}$
$⟹\frac{64-9d^2}{64-d^2}=\frac{7}{15}⟹128d^2=512⟹d^2=4⟹d=\pm 2$

Thus, the four parts are $a-3d,a-d,a+d$ and $3d$, i.e. $2,6,10,14$.