If  ${\mathrm{k}}^2+4\mathrm{k}+8,2{\mathrm{k}}^2+3\mathrm{k}+6,3{\mathrm{k}}^2+4\mathrm{k}+4$ are three consecutive terms of an A.P, then k is

We know that if a, b, c are three consecutive terms of an A.P, then $\mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b}$ i.e., $2\mathrm{b}=\mathrm{a}+\mathrm{c}$

Thus, if ${\mathrm{k}}^2+4\mathrm{k}+8,2{\mathrm{k}}^2+3\mathrm{k}+6\text{ and }3{\mathrm{k}}^2+4\mathrm{k}+4$ are three consecutive terms of an A.P., then

$\begin{array}{l}2\left(2{\mathrm{k}}^2+3\mathrm{k}+6\right)=\left({\mathrm{k}}^2+4\mathrm{k}+8\right)+\left(3{\mathrm{k}}^2+4\mathrm{k}+4\right)\\ \Rightarrow 4{\mathrm{k}}^2+6\mathrm{k}+12=4{\mathrm{k}}^2+8\mathrm{k}+12\\ \Rightarrow 2\mathrm{k}=0\Rightarrow \mathrm{k}=0\end{array}$