Question

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

Answer 1

Let the four parts are $a-3d,$ $a-d,$ $a+d$ and $a+3d$

where common difference is $2d$

Now their sum is $96$

$a-3d+a-d+a+d+a+3d=96$

$4a=96$

$a=24$

Also it is given that,

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

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

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

$128d^2=8a^2=8\times (24{)}^2$

$d^2=36$

$d=6$

So, common difference$d=6$