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QJEEAdv10Ma877d: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma877d Solution: Let ......, be real numbers satisfying . if , then the value of …

Let a1, a2 , a3{a_1},\,{a_{2\,}},\,{a_3} ......, a11{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    a12+a22+....+a11211=90\,\,\,{{a_1^2 + a_2^2 + .... + a_{11}^2} \over {11}} = 90 , then the value of a1+a2+....+a1111{{{a_1} + {a_2} + .... + {a_{11}}} \over {11}} is equal to :

Step-by-Step Verified Solution

ak=2ak−1−ak−2{a_k} = 2{a_{k - 1}} - {a_{k - 2}}

⇒a1,a2,  .....,  a11\Rightarrow {a_1},{a_2},\,\,.....,\,\,{a_{11}} are in AP

∴\therefore a12+a22+  ....  +  a11211{{a_1^2 + a_2^2 + \,\,....\,\, + \,\,a_{11}^2} \over {11}}

=11a2+35×11d2+10ad11=90= {{11{a^2} + 35 \times 11{d^2} + 10ad} \over {11}} = 90

⇒225+35d2+150d=90\Rightarrow 225 + 35{d^2} + 150d = 90

35d2+150d+135=035{d^2} + 150d + 135 = 0

⇒d=−3,−97\Rightarrow d = - 3, - {9 \over 7}

Given, {a_2} < {{22} \over 7} ∴\therefore d=−3d = - 3 and d≠−97d \ne - {9 \over 7}

⇒a1+a2+  ...  +  a1111\Rightarrow {{{a_1} + {a_2} + \,\,...\,\, + \,\,{a_{11}}} \over {11}}

=112[30−10×3]=0= {{11} \over 2}[30 - 10 \times 3] = 0

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