Question

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

Answer 1

$10x^3-n{x}^2-54x-27=0$ has roots in H.P.
Putting $x=\frac{1}{t}$
$27t^3+54t^2+nt-10=0\cdots \left(1\right)$
So the roots are in A.P.
Let the roots be $a-d,a,a+d$
Sum of roots
$a-d+a+a+d=-\frac{54}{27}\Rightarrow 3a=-2\Rightarrow a=-\frac{2}{3}$
As $a$ is one of the root of the equation $\left(1\right)$, putting $a=-\frac{2}{3}$ in the equation $\left(1\right)$, we get
$27\times (-\frac{8}{27})+54\times \left(\frac{4}{9}\right)+n\times (-\frac{2}{3})-10=0\Rightarrow -8+24-\frac{2n}{3}-10=0\Rightarrow \frac{2n}{3}=6\therefore n=9$