If the roots of $6{\mathrm{x}}^3 – 11{\mathrm{x}}^2 + 6\mathrm{x} – 1 = 0$ are in H.P then one of the roots is

$\begin{array}{l}Iftherootsof6x^3-11x^2+6x-1=0areinH.P.\\ Then6{\left(\frac{1}{x}\right)}^3-11{\left(\frac{1}{x}\right)}^2+6\left(\frac{1}{x}\right)-1=0areinA.P\\ \Rightarrow 6-11x+6x^2-x^3=0\\ \Rightarrow x^3-6x^2+11x-6=0\\ Leta-d,a,a+darerootsthen\\ a-d+a+a+d=6\\ \Rightarrow 3a=6\\ \Rightarrow a=2\\ \therefore Oneoftherootis\frac{1}{2}\end{array}$