If the roots of ${\mathrm{x}}^3 – 13{\mathrm{x}}^2 + \mathrm{kx} – 27 = 0$ are in G.P then k =

$\begin{array}{l}Let\frac{a}{r},a,ararerootsof{x}^3-13x^2+kx-27=0\\ \therefore \frac{a}{r}.a.ar=27\\ \Rightarrow a^3=27\Rightarrow a=3\\ \because a=3isroot\Rightarrow 3^3-13{\left(3\right)}^2+k\left(3\right)-27=0\\ \Rightarrow 27-117+3k-27=0\\ \Rightarrow 3k=117\\ \Rightarrow k=\frac{117}{3}=39\end{array}$