Question

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

• A. $32$
• B. - $\frac{1}{32}$
• C. $-32$
• D. $\frac{1}{32}$

Answer 1

Answer: (A), (C)

$32$
$-32$

Let the roots be $a,ar,a{r}^2,a{r}^3$ and $a{r}^4$.

Using sum of roots = $-\frac{a_1}{a_o}$, where $a_o{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+......................+a_n=0$ is the polynomial

Then $a+ar+a{r}^2+a{r}^3+a{r}^4=40$
i.e $a[\frac{r^5-1}{r-1}]=40$
$\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}+\frac{1}{a{r}^4}=10$
i.e $\frac{1}{a^2{r}^4}\frac{a(r^5-1)}{(r-1)}=10$ i.e $a^2{r}^4=4$
i.e $a{r}^2=+2$ or $-2$
$S=-a^5{r}^{10}=-(\pm 2{)}^5=\pm 32$