Question

If the roots of the equation $x ^5-40 x ^4+ Px ^3+ Qx ^2+ Rx + S =0$ 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)

$x^5-40 x^4+P x^3+Q x^2+R x+S=0$ ....(1)
$a+a r+a r^2+a r^3+a r^4=40 $
$\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+\frac{1}{a r^3}+\frac{1}{a r^4}=10 \\
\frac{r^4+r^3+r^2+r+1}{a r^4}=10 $ ....(2)
$\text { From(1) and }(2)$
$a^2 r^4=4 \Rightarrow a^2= \pm 2$
$-S=a \cdot a r \cdot a r^2 \cdot a r^3 a r^4=a^5 r^{10}=\left(a r^2\right)^5$
$\therefore S= \pm 32 \Rightarrow|S|=32$