Question

All the roots of the equation $x^5-40x^4+A{x}^3+B{x}^2+Cx+D=0$ are positive and are in geometric progression with common ratio $r$. If sum of the reciprocal of roots is $10,$ then

• A. mid term of geometric progression is $2$
• B. $D=32$
• C. $\frac{1}{r}+r=\frac{-1\pm \sqrt{85}}{2}$
• D. $\frac{1}{r}+r=\frac{\sqrt{85}-1}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A mid term of geometric progression is $2$
D $\frac{1}{r}+r=\frac{\sqrt{85}-1}{2}$

Let roots are $\frac{a}{r^2},\frac{a}{r},a,ar,a{r}^2.$

$\Rightarrow a(\frac{1}{r^2}+\frac{1}{r}+1+r+r^2)=40\cdots \left(1\right)$
According to the question,
$\frac{1}{a}(r^2+r+1+\frac{1}{r}+\frac{1}{r^2})=10\cdots \left(2\right)$

From equations $\left(1\right)$ and $\left(2\right),$ we get
$a^2=4$
$\Rightarrow a=2$ as roots are positive

$-D=$ product of roots
$\Rightarrow -D=a^5$
$\Rightarrow D=-32$

Now, from equation $\left(1\right),$ we have
$(\frac{1}{r^2}+\frac{1}{r}+1+r+r^2)=20$
$\Rightarrow (\frac{1}{r^2}+r^2)+(r+\frac{1}{r})+1=20$
$\Rightarrow {(r+\frac{1}{r})}^2+(r+\frac{1}{r})-21=0$
$\Rightarrow \frac{1}{r}+r=\frac{-1\pm \sqrt{85}}{2}$
Since, roots are positive.
$\therefore r+\frac{1}{r}=\frac{\sqrt{85}-1}{2}$