Question

Divide $56$ in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is $5:6$.

Answer 1

Let the four number in $AP$ be

$a-3d,a-d,a+d,a+3d$

The sum total of number $=56$

then

$a-3d+a-d+a+d+a+3d=56$

$\Rightarrow 4a=56$

$\Rightarrow a=\mathrm{14............}\left(1\right)$

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

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

$\Rightarrow 6a^2-54d^2=5a^2-5d^2$

$\Rightarrow a^2=49d^2$

$\Rightarrow a=\pm 7d$

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

$\therefore d=\pm 2$

Putting $d=2$, we get the numbers $(14-6),(14-2),(14+2),(14+6)$

The four number in $AP$ are

$8,12,16,20$