Question

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

Answer 1

Answer: 0

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