Question

Let $a_1,a_2,a_3,...,a_{11}$ be real numbers satisfying $a_1=15$
$27-2a_2>0$ and $a_k=2a_{k-1}-a_{k-2}$ for k = 3, 4,...,11.
If $\frac{a_1^2+a_2^2+...+a_{11}^2}{11}=90,$ then the value of
$\frac{a_1+a_2+...+a_{11}}{11}$ is ......

Answer 1

Answer: 0

$a_k=2a_{k-1}-a_{k-2}\Rightarrow a_1,a_2,...,a_{11}$ are in AP.
$\therefore \frac{a_1^2+a_2^2+...+a_{11}^2}{11}=\frac{11a^2+35\times 11d^2+10ad}{11}=90$
$\Rightarrow 225+35d^2+150d=90$
$\Rightarrow 35d^2+150d+135=0\Rightarrow d=-3,-\frac{9}{7}$
Given, $a_2<\frac{27}{2}$
$\therefore d=-3$ and $d\ne -\frac{9}{7}$
$\Rightarrow \frac{a_1+a_2+...+a_{11}}{11}=\frac{11}{2}[30-10\times 3]=0$