Question

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

• A. 40
• B. 32
• C. 8
• D. 2

Answer 1

Answer: (B)

$x^5-40x^4+P{x}^3+Q{x}^2+Rx+S=0$ ....(1)
$a+ar+a{r}^2+a{r}^3+a{r}^4=40$
$$\begin{gathered} \frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}+\frac{1}{a{r}^4}=10 \\ <br/>\frac{r^4+r^3+r^2+r+1}{a{r}^4}=10 \end{gathered}$$ ....(2)
$\text{ From(1) and }(2)$
$a^2{r}^4=4\Rightarrow a^2=\pm 2$
$-S=a\cdot ar\cdot a{r}^2\cdot a{r}^{3}a{r}^4=a^5{r}^{10}={\left(a{r}^2\right)}^5$
$\therefore S=\pm 32\Rightarrow |S|=32$