JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 7)
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 $$\,\,\,{{a_1^2 + a_2^2 + .... + a_{11}^2} \over {11}} = 90$$, then the value of $${{{a_1} + {a_2} + .... + {a_{11}}} \over {11}}$$ is equal to :
Answer
0
Explanation
$${a_k} = 2{a_{k - 1}} - {a_{k - 2}}$$
$$ \Rightarrow {a_1},{a_2},\,\,.....,\,\,{a_{11}}$$ are in AP
$$\therefore$$ $${{a_1^2 + a_2^2 + \,\,....\,\, + \,\,a_{11}^2} \over {11}}$$
$$ = {{11{a^2} + 35 \times 11{d^2} + 10ad} \over {11}} = 90$$
$$ \Rightarrow 225 + 35{d^2} + 150d = 90$$
$$35{d^2} + 150d + 135 = 0$$
$$ \Rightarrow d = - 3, - {9 \over 7}$$
Given, $${a_2} < {{22} \over 7}$$ $$\therefore$$ $$d = - 3$$ and $$d \ne - {9 \over 7}$$
$$ \Rightarrow {{{a_1} + {a_2} + \,\,...\,\, + \,\,{a_{11}}} \over {11}}$$
$$ = {{11} \over 2}[30 - 10 \times 3] = 0$$
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