Let ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,......{\mathrm{a}}_{\mathrm{n}}$ be real numbers satisfying ${\mathrm{a}}_1=15$, $27-2{\mathrm{a}}_2>0$ and ${\mathrm{a}}_{\mathrm{k}}=2{\mathrm{a}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}-2}$ for $\mathrm{K}=3,4,....,11$ If $\frac{{{\mathrm{a}}_1}^2+{{\mathrm{a}}_2}^2+{{\mathrm{a}}_3}^2+....+{{\mathrm{a}}_{11}}^2}{11}=90$, then
 

 ${\mathrm{a}}_{\mathrm{k}}=2{\mathrm{a}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}-2}$
$\Rightarrow {\mathrm{a}}_1,{\mathrm{a}}_2,....,{\mathrm{a}}_{11}$ are in AP
$\therefore \frac{{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+.....+{\mathrm{a}}_{11}^2}{11}=90\Rightarrow \frac{11{\mathrm{a}}^2+35\times 11{\mathrm{d}}^2+10\mathrm{ad}}{11}=90$
$\begin{array}{l}225+35{\mathrm{d}}^2+150\mathrm{d}=90\\ 35{\mathrm{d}}^2+150\mathrm{d}+135=0\\ \mathrm{d}=-3,\frac{-9}{7}\end{array}$
But, ${\mathrm{a}}_2<\frac{27}{2}$, $\therefore \mathrm{d}=-3$
$\Rightarrow \frac{{\mathrm{a}}_1+{\mathrm{a}}_2+....+{\mathrm{a}}_{11}}{11}=\frac{11}{2}(30-10\times 3)=0$