Sequences and Series 2 Question 23
24.
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……
(2010)
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Answer:
Correct Answer: 24. $(0)$
Solution:
- $a _k=2 a _{k-1}-a _{k-2}$
$\Rightarrow a _1, a _2, \ldots, a _{11}$ are in an AP.
$\therefore \quad \frac{a _1^{2}+a _2^{2}+\ldots+a _{11}^{2}}{11}=\frac{11 a^{2}+35 \times 11 d^{2}+10 a d}{11}=90$
$\Rightarrow 225+35 d^{2}+150 d=90$
$\Rightarrow 35 d^{2}+150 d+135=0 \Rightarrow d=-3,-\frac{9}{7}$
Given, $\quad a _2<\frac{27}{2}$
$\therefore \quad d=-3$ and $d \neq-\frac{9}{7}$
$\Rightarrow \quad \frac{a _1+a _2+\ldots+a _{11}}{11}=\frac{11}{2}[30-10 \times 3]=0$
