Question

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

Answer 1

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