If the roots of $3{\mathrm{x}}^3-22{\mathrm{x}}^2+48\mathrm{x}-32=0$ are in H.P, then the roots are 
 

$\begin{array}{l}Therootsof3x^3-22x^2+48x-32=0areinHP\\ Then3{\left(\frac{1}{x}\right)}^3-22{\left(\frac{1}{x}\right)}^2+48\left(\frac{1}{x}\right)-32=0areinAP\\ \Rightarrow 3-22x+48x^2-32x^3=0\\ \Rightarrow 32x^3-48x^2+22x-3=0\\ Leta-d,a,a+dareroots\\ \therefore a-d+a+a+d=\frac{48}{32}\\ \Rightarrow 3a=\frac{3}{2}\to a=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ and(a-d)a(a+d)=\frac{3}{32}\\ \Rightarrow a(a^2-d^2)=\frac{3}{32}\\ \Rightarrow \frac{1}{2}(\frac{1}{4}-d^2)=\frac{3}{32}\\ \Rightarrow \frac{1}{4}-d^2=\frac{3}{16}\\ \Rightarrow d^2=\frac{1}{4}-\frac{3}{16}\Rightarrow \frac{1}{16}\Rightarrow d=\frac{1}{4}\\ \therefore a-d=\frac{1}{2}-\frac{1}{4}\Rightarrow \frac{1}{4},a+d=\frac{1}{2}+\frac{1}{4}\Rightarrow \frac{3}{4}\\ \therefore Therootsare4,2,\frac{4}{3}\end{array}$