Subjective Type

Let $$a_1, a_2, a_3, ..., a_{11}$$ be real numbers satisfying $$a_1 = 15,27 - 2a_2 > 0$$ and $$a_k = 2a_{k - 1} - a_{k - 2}$$ for $$k = 3, 4, ...., 11$$. If $$\dfrac{a_1^2 + a_2^2 + .... + a+{11}^2}{11} = 90$$, then the value of $$\dfrac{a_1 + a_2 + .... + a_{11}}{11}$$ is equal to.

Solution

$$a_k = 2a_{k - 1} - a_{k - 2}\Rightarrow a_1, a_2,..., a_{11}$$ are in A.P.

$$\therefore \ \ \dfrac{a_1^2 + a_2^2 + ... + a_{11}^2}{11} = \dfrac{11a^2 + 35 \times 11d^2 + 10ad}{11} = 90$$

$$\Rightarrow \ \ 225 + 35d^2 + 15d = 90$$

$$35d^2 + 150d + 135 = 0 \Rightarrow \ \ -3, \dfrac{-9}{7}$$

Given $$a_2 < \dfrac{27}{2}$$,

we get $$d = -3$$ and $$d = \dfrac{-9}{7}$$

$$\Rightarrow \ \ \dfrac{a_1 + a_2 + ... + a_{11}}{11} = \dfrac{11}{2}[30 - 10 \times 3] = 0$$

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