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{a1=1527−2a2>0ak=2ak−1−ak−2for 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} ⎩⎪⎨⎪⎧a1=1527−2a2>0ak=2ak−1−ak−2for k=3,4,…,11
Let a1,a2,…,a11a_{1}, a_{2}, \ldots , a_{11}a1,a2,…,a11 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} = 9011a12+a22+a32+⋯+a112=90, then find a1+a2+a3+⋯+a1111\dfrac {a_1 +a_2+a_3 + \cdots + a_{11}}{11}11a1+a2+a3+⋯+a11.
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