Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5:6

Answer 1

Let's assume the four parts be

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

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

or $4a=56$

or $a=14$

Give,

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

or, $\frac{a^2-9d^2}{a^2-d^2}=\frac{5}{6}$

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

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

$\Rightarrow d^2=\frac{14\times 14}{49}$

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

$\Rightarrow d=\pm 2$

If $d=\pm 2$ then four parts will be

$14-6,14-2,14+2,14+6$

$8,12,16,20$

If $d=-2$ then four parts will be

$14+6$ $14+2$ $14-2$ $14-6$

$20,16,12,8$

So, $4$ numbers are $8,12,16,20.$

Hence, the answer is $8,12,16,20.$