Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15,27-2 a_20$ 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

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

Aicharya · Accepted Answer

$ 	ext { } \begin{aligned} & a_k=2 a_{k-1}-a_{k-2} \ & \Rightarrow \quad a_1, a_2, \ldots, a_{11} 	ext { are in AP } \ & 	herefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \ &=\frac{11 a^2+35 	imes 11 d^2+10 a d}{11}=90 \ & \Rightarrow \quad 225+35 d^2+150 d=90 \ & 35 d^2+150 d+135=0 \ & \Rightarrow \quad d=-3,-\frac{9}{7} \end{aligned} $ Given, $a_2 < \frac{27}{2} 	herefore d=-3$ and $d 
eq-\frac{9}{7}$ $ \Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11} $ $ =\frac{11}{2}[30-10 	imes 3]=0 $

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…

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