Question

If roots of the equation ${6x}^3-{11x}^2+6x-1=0$ are in HP , then the sum of the first three terms of the corresponding AP is

Answer 1

Substitute $x=\frac{1}{y}$ in the given equation
then $\frac{6}{y^3}-\frac{11}{y^2}+\frac{6}{y}-1=0\Rightarrow 6-11y+6y^2-y^2=0$
$\Rightarrow y^3-6y^2+11y-6=0--------------\left(1\right)$
Now roots of (1) are in A.P.let roots be $\alpha -\beta ,\alpha ,\alpha +\beta$
Then, sum of roots $\alpha -\beta +\alpha +\alpha +\beta =6$
$\Rightarrow 3\alpha =6$
$\therefore \alpha =2$
Products of the roots ($\alpha -\beta \left)\alpha \right(\alpha +\beta )=6$
$\therefore 2(4-{\beta }^2)=6\therefore \beta =\pm 1$
Roots of (1) are $1,2,3$ or $3,2,1$
Hence, roots of the given equation are
$1$,$\frac{1}{2},\frac{1}{3}$ or $\frac{1}{3},\frac{1}{2}$ ,$1$