JEE Advanced · Mathematics · 5. Sequences & Series
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
- A 1
- B 2
- C 3
- D \(0\)
Answer & Solution
Correct Answer
(D) \(0\)
Step-by-step Solution
Detailed explanation
\(a_k=2 a_{k-1}-a_{k-2} \)
\( \Rightarrow a_1, a_2, \ldots, a_{11} \text { are in AP } \)
\( \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \)
\(=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \)
\( \Rightarrow 225+35 d^2+150 d=90 \)
\( 35 d^2+150 d+135=0 \)
\( \Rightarrow d=-3,-\frac{9}{7}\)
Given, \(a_2 < \frac{27}{2} \therefore d=-3\) and \(d \neq-\frac{9}{7}\)
\(
\Rightarrow \frac{a_1+a_2+\ldots+a_{11}}{11}
\)
\(
=\frac{11}{2}[30-10 \times 3]=0
\)
\( \Rightarrow a_1, a_2, \ldots, a_{11} \text { are in AP } \)
\( \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \)
\(=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \)
\( \Rightarrow 225+35 d^2+150 d=90 \)
\( 35 d^2+150 d+135=0 \)
\( \Rightarrow d=-3,-\frac{9}{7}\)
Given, \(a_2 < \frac{27}{2} \therefore d=-3\) and \(d \neq-\frac{9}{7}\)
\(
\Rightarrow \frac{a_1+a_2+\ldots+a_{11}}{11}
\)
\(
=\frac{11}{2}[30-10 \times 3]=0
\)
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