jee-advanced 2010 Q29

jee-advanced · India · paper2 Arithmetic Sequences and Series Summation of Derived Sequence from AP

Let $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots , a _ { 11 }$ be real numbers satisfying $\mathrm { a } _ { 1 } = 15 , \quad 27 - 2 \mathrm { a } _ { 2 } > 0$ and $\mathrm { a } _ { \mathrm { k } } = 2 \mathrm { a } _ { \mathrm { k } - 1 } - \mathrm { a } _ { \mathrm { k } - 2 }$ for $\mathrm { k } = 3,4 , \ldots , 11$.
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

Let $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots , a _ { 11 }$ be real numbers satisfying\\
$\mathrm { a } _ { 1 } = 15 , \quad 27 - 2 \mathrm { a } _ { 2 } > 0$ and $\mathrm { a } _ { \mathrm { k } } = 2 \mathrm { a } _ { \mathrm { k } - 1 } - \mathrm { a } _ { \mathrm { k } - 2 }$ for $\mathrm { k } = 3,4 , \ldots , 11$.

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

Paper Questions

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