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Standard XII

Mathematics

Arithmetic Progression

Let a 1, a 2,...

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 $a_{5}$ is

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Solution

$a_{k} = 2 a_{k - 1} - a_{k - 2}$
$a_{k} + a_{k - 2} = 2 a_{k - 1}$
$\Rightarrow a_{1}, a_{2}, a_{3}, \dots, a_{11}$ are in A.P.
Let the common difference be $d$
$a_{1} = 15, a_{2} = 15 + d, a_{3} = 15 + 2 d, \dots, a_{11} = 15 + 10 d$

Now, $\frac{(a_{1})^{2} + (a_{2})^{2} + \dots + (a_{11})^{2}}{11}$
$= \frac{(15)^{2} + (15 + d)^{2} + (15 + 2 d)^{2} + \dots + (15 + 10 d)^{2}}{11} = 90$
$\Rightarrow \frac{(15)^{2} \cdot 11 + 30 d (1 + 2 + \dots + 10) + d^{2} (1^{2} + 2^{2} + \dots + 10^{2})}{11} = 90$
$\Rightarrow 35 d^{2} + 150 d + 225 = 90$
$\Rightarrow 35 d^{2} + 150 d + 135 = 0$
$\Rightarrow 7 d^{2} + 30 d + 27 = 0$
$\Rightarrow 7 d^{2} + 21 d + 9 d + 27 = 0$
$\Rightarrow (d + 3) (7 d + 9) = 0$
$\Rightarrow d = - 3, \frac{- 9}{7}$
Since $a_{2} < \frac{27}{2},$ $d \neq \frac{- 9}{7}$
$∴ d = - 3$

Hence, $a_{5} = 15 + 4 (- 3) = 3$

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