Question

If the roots of the equation $10x^{3}-cx^{2}-54x-27=0$ are in harmonic progression, then the value of $c$ must be equal to

Answer 1

The roots of $10x^{3}-cx^{2}-54x-27=0$ are in H.P.
Replacing $x$ by $\frac{1}{x}$ we will get an equation whose roots will be in A.P.
$\Rightarrow \frac{10}{x^{3}}-\frac{c}{x^{2}}-\frac{54}{x}-27=0\Rightarrow 27x^{3}+54x^{2}+cx-10=0$
Hence, roots of this equation are in A.P. i.e. $a-d,a,a+d$
$\left(a - d\right)+a+\left(a + d\right)=-\frac{54}{27}=-2\Rightarrow a=-\frac{2}{3}$
Since $a$ is the root of the given equation
$\Rightarrow 27\left(- \frac{2}{3}\right)^{3}+54\left(- \frac{2}{3}\right)^{2}+c\left(- \frac{2}{3}\right)-10=0$
$\Rightarrow -8+24-10=\frac{2 c}{3}\Rightarrow c=\frac{6 \times 3}{2}=9$
$c=9$