# Divide 44 into 4 parts following an AP such that the ratio of the product of extremes to the product of the means is 5:14.

$\begin{array}{1 1} 2,10,18,26 \\ 1,7,15,19 \\ 1,8, 15,21 \\ 2,8,14,20 \end{array}$

Explanation : Let the 4 terms in AP = $a-3d\;,a-d\;,a+d\;,a+3d$
$sum=44$
$a+3d+a+d+a-d+a-3d=44$
$4a=44\quad\;a=11$
$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=5/14$
$14\;(a^2-9d^2)=5(a^2-d^2)$
$a=11\quad,1694-126d^2=605-5d^2$
$121d^2=1089$
$d^2=\frac{1089}{121}=9\quad\;d=\pm3$
$terms\; are\;11-9\;,11-3\;,11+3\;,11+9$
$=2,8,14,20.$