Question

Divide $96$ into four parts which are in A.P. and the ratio between product of their means to product of their extremes is $15:7$.

Answer 1

Let $4$ parts be $a-3d,a-d,a+d,a+3d$

$\Rightarrow$$(a-3d)+(a-d)+(a+d)+(a+3d)=96$

$\Rightarrow$$4a=96$

$\Rightarrow$$a=24$

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

$\Rightarrow$$7a^2-7d^2=15a^2-135d^2$

$\Rightarrow$$128d^2=8a^2$

$\Rightarrow$$d=\frac{a}{4}=6$

$\Rightarrow$$4$ parts are $6,18,30,42$