Let a 1 , a 2 , a 3 , … , 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 , … , 11 . If a 1 2 + a 2 2 + ? + a 11 2 11 = 90 , then the value of a 1 + a 2 + ? + a 11 11 is equal to

Question

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 the value of $\frac{a_{1} + a_{2} + \dots + a_{11}}{11}$ is equal to [2010]

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$Given: a_{k} = 2 a_{k - 1} - a_{k - 2}$

$\Rightarrow \frac{a_{k - 2} + a_{k}}{2} = a_{k - 1}, 3 \leq k \leq 11$

$\Rightarrow a_{1}, a_{2}, a_{3}, \dots, a_{11} are in A.P.$

$If a is the first term and d the common difference, then$

$a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2} = 990$

$\Rightarrow 11 a^{2} + d^{2} (1^{2} + 2^{2} + \dots + 10^{2}) + 2 a d (1 + 2 + \dots + 10) = 990$

$\Rightarrow 11 a^{2} + \frac{10 \times 11 \times 21}{6} d^{2} + 2 a d \times \frac{10 \times 11}{2} = 990$

$\Rightarrow a^{2} + 35 d^{2} + 10 a d = 90$

$Since a = a_{1} = 15$

$∴$ $35 d^{2} + 150 d + 135 = 0 \Rightarrow 7 d^{2} + 30 d + 27 = 0$

$\Rightarrow (d + 3) (7 d + 9) = 0 \Rightarrow d = - 3 or - \frac{9}{7}$

$Then a_{2} = 15 - 3 = 12 or 15 - \frac{9}{7} = \frac{96}{7} > \frac{27}{2}$

$∴$ $d \neq - \frac{9}{7}$

$Hence \frac{a_{1} + a_{2} + \dots + a_{11}}{11} = \frac{\frac{11}{2} [2 \times 15 + 10 (- 3)]}{11} = 0$

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Solution

Q. 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 a12+a22+⋯+a11211=90, then the value of a1+a2+⋯+a1111 is equal to [2010]

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