Question

If the roots of $10x^3-c{x}^2-54x-27=0$ are in harmonic progression, then the value of $c$ is

• A. $9$
• B. $14$
• C. $17$
• D. $27$

Answer 1

The correct option is A $9$

Given roots of equation $10x^3-c{x}^2-54x-27=0$ are in H.P.

Replacing $x$ by $\frac{1}{x}$, then we get

$27x^3+54x^2+cx-10=0$ .....(i)

Now, roots of equation (i) are in A.P.

So, let the roots be $\alpha -\beta ,\alpha ,\alpha +\beta$, then

$\alpha -\beta +\alpha +\alpha +\beta =-\frac{54}{27}=-2$

$\Rightarrow \alpha =-\frac{2}{3}$

$\because \alpha =-\frac{2}{3}$ is a root of equation (i) then

$27{(-\frac{2}{3})}^3+54{(-\frac{2}{3})}^2+c(-\frac{2}{3})-10=0$

$\Rightarrow c=9$.