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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
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
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The correct answer is:
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$$
\text { } \begin{aligned}
& a_k=2 a_{k-1}-a_{k-2} \\
& \Rightarrow \quad a_1, a_2, \ldots, a_{11} \text { are in AP } \\
& \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \\
&=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \\
& \Rightarrow \quad 225+35 d^2+150 d=90 \\
& 35 d^2+150 d+135=0 \\
& \Rightarrow \quad d=-3,-\frac{9}{7}
\end{aligned}
$$
Given, $a_2 < \frac{27}{2} \therefore d=-3$ and $d \neq-\frac{9}{7}$
$$
\Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11}
$$
$$
=\frac{11}{2}[30-10 \times 3]=0
$$
\text { } \begin{aligned}
& a_k=2 a_{k-1}-a_{k-2} \\
& \Rightarrow \quad a_1, a_2, \ldots, a_{11} \text { are in AP } \\
& \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 \quad 225+35 d^2+150 d=90 \\
& 35 d^2+150 d+135=0 \\
& \Rightarrow \quad d=-3,-\frac{9}{7}
\end{aligned}
$$
Given, $a_2 < \frac{27}{2} \therefore d=-3$ and $d \neq-\frac{9}{7}$
$$
\Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11}
$$
$$
=\frac{11}{2}[30-10 \times 3]=0
$$
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