JEE Problem

Algebra Level 5

{a1=15272a2>0ak=2ak1ak2for k=3,4,,11\begin{cases} a_{1} = 15 \\ 27-2a_{2} > 0 \\ a_{k} = 2a_{k-1} - a_{k-2} & \text{for } k=3, 4,\ldots, 11 \end{cases}

Let a1,a2,,a11a_{1}, a_{2}, \ldots , a_{11} be real numbers satisfying the conditions above. If a12+a22+a32++a11211=90\dfrac{a_1^2 +a_2^2+a_3^2 +\cdots + a_{11}^2}{11} = 90, then find a1+a2+a3++a1111\dfrac {a_1 +a_2+a_3 + \cdots + a_{11}}{11}.

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