2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    Let \[{{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{49}}\] be in A.P. such that \[\sum\limits_{k=0}^{12}{{{a}_{4k+1}}}=416\] and \[{{a}_{9}}+{{a}_{43}}=66.\] If \[a_{1}^{2}+a_{2}^{2}+.......+a_{17}^{2}=140m\], then \[m\] is equal to: [JEE Main Online 08-04-2018]

    A)  34                              

    B)  33

    C)  66                              

    D)  68

    Correct Answer: A

    Solution :

    \[{{a}_{1}}+{{a}_{5}}+.......+{{a}_{49}}=416\] \[{{a}_{1}}+{{a}_{1}}+4d+.......+{{a}_{1}}+48d=416\] \[13{{a}_{1}}+4d\left( \frac{12\times 13}{2} \right)=416\]                 \[13({{a}_{1}}+24d)=416\]                 \[\therefore {{a}_{1}}+24d=32\]??(1) Also,                 \[{{a}_{9}}+{{a}_{43}}=66\]                 \[\therefore 2{{a}_{1}}+50d=66\]                 \[{{a}_{1}}+25d=33\]                      ?..(2) From (1) & (2), we get                 \[d=1\] & \[{{a}_{1}}=8\] \[\therefore \]  \[{{a}_{1}}^{2}+{{a}_{2}}^{2}+......+{{a}_{17}}^{2}\]                 \[={{8}^{2}}+{{9}^{2}}+......+{{24}^{2}}\]                 \[=\Sigma {{24}^{2}}-\Sigma {{7}^{2}}\]                 \[=140\times 34\]


You need to login to perform this action.
You will be redirected in 3 sec spinner