Let a1,a2,a3,…,a11 be real numbers satisfying a1=15, 27−2a2>0 and – InterviewSolution

1.	Let a1,a2,a3,…,a11 be real numbers satisfying a1=15, 27−2a2>0 and ak=2ak−1−ak−2 for k=3,4,…,11. If (a1)2+(a2)2+⋯+(a11)211=90, then a5 is
Answer» Let $a_{1}, a_{2}, a_{3}, \dots, a_{11}$ be real numbers satisfying $a_{1} = 15,$ $27 - 2 a_{2} > 0$ and $a_{k} = 2 a_{k - 1} - a_{k - 2}$ for $k = 3, 4, \dots, 11.$ If $\frac{(a_{1})^{2} + (a_{2})^{2} + \dots + (a_{11})^{2}}{11} = 90,$ then $a_{5}$ is

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