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

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$d = - 3$

$\frac{a_{1} + a_{2} + . .. + a_{11}}{11} = 0$

$d = \frac{- 9}{7}$

$\frac{a_{1} + a_{2} + ... + a_{11}}{11} = 9$

answer is A, B.

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Detailed Solution

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