Question

If the roots of the equation $x^5-80x^4+p{x}^3+q{x}^2+rx-s=0$ are in G. P and all the terms are positive, such that the sum of its reciprocals is $20$, then find the value of s

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

Answer: (C)

$32$

Let the roots be $a,ax,a{x}^2,a{x}^3,a{x}^4$.
$a+ax+a{x}^2+a{x}^3+a{x}^4=80$ .......(1)
$\left(1/a\right)+\left(1/ax\right)+\left(1/a{x}^2\right)+\left(1/a{x}^3\right)+\left(1/a{x}^4\right)=20$.........(2)
$\Rightarrow \frac{a(x^4+x^3+x^2+x+1)}{a^2{x}^4}=20$
solving equation (1) and (2)
$a{x}^2=2$
s=multiplication of all the roots$={\left(a{x}^2\right)}^5=32$