Question

$28$ is divided into four parts in A.P. so that ratio of the product of first and third with the product of second and fourth is $8:15$,then the sum of these parts is

Answer 1

Assume numbers are $a-3d,a-d,a+d,a+3d$
Here sum $=4a=28\therefore a=7.$
Also
$\frac{(a-3d)(a+d)}{(a-d)(a+3d)}=\frac{8}{15}$
simplifying

$15[a^2-3d^2-2ad]=8[a^2-3d^2+2ad]$

or $7[a^2-3d^2]=46ad$ or
$7(49-3d^2)=46\times 7.d$ or
$49-3d^2=46d$
or $3d^2+46d-49=0(d-1)(3d+49)=0\therefore d=1.$
$\therefore$ Required numbers are $4,6,8,10$