If the roots of $x^5-40x^4+P{x}^3+Q{x}^2+Rx+S=0$  are in G.P. and sum of their reciprocals is 10, then $\left|S\right|$  is:

The given roots of the equation $x^5-40x^4+P{x}^3+Q{x}^3+Rx+S=0$ 

are in G.P.

Let roots to be $a,ar,a{r}^2,a{r}^3,a{r}^4$

Sum of the roots : $a+ar+a{r}^2+a{r}^3+a{r}^4=40$  and

Given that, sum of their reciprocals : $\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}+\frac{1}{a{r}^4}$ $=\frac{1+r+r^2+r^3+r^4}{a{r}^4}=\frac{40}{a^2{r}^4}=10$

$\Rightarrow$ $a^2{r}^4=4$  or $a{r}^2=\pm 2$

Now $S=$ product of roots $=a.ar.a{r}^2.a{r}^3.a{r}^4$

$=a^5{r}^{10}$

$={\left(a{r}^2\right)}^5$

$=\pm 32$

$\therefore$  $\left|S\right|=32$