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 $\, \, \, \frac{a^2_1+ a^2_2 +...+ a^2_{11}}{11}=90,$ then the value of
$\frac{a_1+ a_2 +...+ a_{11}}{11}$ is ......

Answer 1

$a_k = 2a_{k-1} - a_{k-2} \Rightarrow a_1,a_2,..., a_{11}$ are in AP.
$ \therefore \frac{a^2_1+ a^2_2 +...+ a^2_{11}}{11} = \frac{11a^2+ 35 \times 11d^2+ 10ad}{11}= 90$
$\Rightarrow 225 + 35\, d^2 +150\, d = 90$
$\Rightarrow 35\, d^2 +150\, d + 135 = 0 \Rightarrow d = - 3, -\frac{9}{7}$
Given, $ a_2 < \frac{27}{2}$
$\therefore \, d =-3$ and $ d \ne -\frac{9}{7}$
$\Rightarrow \frac{a_1+ a_2 +...+ a_{11}}{11} = \frac{11}{2} [30-10 \times 3] = 0$