Question

Divide $56$ into $4$ parts which are in AP such that the ratio of products of extremes to the products of means is $5:6.$

Answer 1

Let the 4 numbers which are in AP be $(a-3d),(a-d),(a+d),(a+3d)$
Their sum, $(a-3d)+(a-d)+(a+d)+(a+3d)$ = $4a$

$4a=56$ [$\because$ $56$ is divided into $4$ numbers which are in AP]
$\Rightarrow$ $a=14$

It is given that the ratio of products of extremes to the products of means is $5:6.$

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=5:6$

$\Rightarrow 6(a^2-9d^2)=5(a^2-d^2)$

$\Rightarrow a^2=54d^2-5d^2$

$\Rightarrow {14}^2=49d^2$ [$\because a=14$]

$\Rightarrow d^2=\frac{{14}^2}{7^2}$

$\Rightarrow d=\pm \frac{14}{7}=\pm 2$

for $d=2$ [Taking positive value of $d,$ as the numbers are positive]

$\Rightarrow a-3d=14-6=8$

$\Rightarrow a-d=14-2=12$

$\Rightarrow a+d=14+2=16$

$\Rightarrow a+3d=14+6=20$

Numbers are $8,12,16,20$