If the roots of $6x^3-11x^2+6x-1=0$ are in H.P then one of the roots is

If roots are in HP then replace x by 1//x  then the transformed   equation roots are in AP.

The root of $x^3-6x^2+11x-6=0$are in A.P.

Sum of the roots$3\alpha =6$ ,$\alpha =2$ .

One of the root is $\frac{1}{2}$