Question

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

Answer 1

Given equation is $10x^3-c{x}^2-54x-27=0$

Replacing $x$ by $\frac{1}{x}$ we get an equation whose roots will be in AP :
So the new equation obtained would be $27x^3+54x^2+cx-10=0$
Now let the roots be $a-d,a,a+d$
Sum of roots $3a=-2$
$a=\frac{-2}{3}$
Product of roots $\frac{-2}{3}(d^2-\frac{4}{9})$ $=\frac{-10}{27}$
$d=1$
Roots are $\frac{-5}{3},\frac{-2}{3},\frac{1}{3}$
$c=27(\frac{-5}{3}\frac{-2}{3}+\frac{-2}{3}\frac{1}{3}+\frac{-5}{3}\frac{1}{3})$
$\Rightarrow c=9$