If the roots of the equation $4{\mathrm{x}}^3–12{\mathrm{x}}^2+11\mathrm{x}+\mathrm{k}=0$ are in arithmetic progression then k =

$\begin{array}{l}Leta-d,a,a+darerootsof4x^3-12x^2+11x+k=0\\ \therefore a-d+a+a+d=\frac{12}{4}\\ \Rightarrow a-d+a+a+d=\frac{12}{4}\\ \Rightarrow 3a=3\Rightarrow a=1\\ \because a=1isroot\Rightarrow 4{\left(1\right)}^3-12{\left(1\right)}^2+11\left(1\right)+k=0\\ \Rightarrow 4-12+11+k=0\\ \Rightarrow 3+k=0\\ \Rightarrow k=-3\end{array}$