Question

Divide $20$ into four parts which are in A.P., and such that the product of the first and fourth is to the product of the second and third in the ratio of $2$ to $3$.

Answer 1

Let the numbers that are in A.P are

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

Given sum $=20$

$\Rightarrow a-3d+a-d+a+d+a+3d=20\Rightarrow 4a=20\Rightarrow a=5$

Also $\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{2}{3}$
$\Rightarrow \frac{(5-3d)(5+3d)}{(5-d)(5+d)}=\frac{2}{3}\Rightarrow \frac{25-{9d}^2}{25-d^2}=\frac{2}{3}\Rightarrow 75-27d^2=50-2d^2\Rightarrow 25=25d^2\Rightarrow d=\pm 1$

So two sequences can be formed with $a=5$ and $d=\pm 1$ that are

$2,4,6,8$ and $8,6,4,2$