JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},\]........., \[{{a}_{n}}\]are in A.P. and \[{{a}_{1}}+{{a}_{4}}+{{a}_{7}}+\]......... \[+{{a}_{16}}=114,\]then \[{{a}_{1}}+{{a}_{6}}+{{a}_{11}}+{{a}_{16}}\]is equal to : [JEE Main 10-4-2019 Morning]

    A) 38                               

    B) 98

    C) 76       

    D) 64

    Correct Answer: C

    Solution :

    \[{{a}_{1}}+{{a}_{4}}+{{a}_{7}}+{{a}_{10}}+{{a}_{13}}+{{a}_{16}}=114\] \[\Rightarrow \frac{6}{2}({{a}_{1}}+{{a}_{16}})=114\]\[\Rightarrow {{a}_{1}}+{{a}_{16}}=38\] So,\[{{a}_{1}}+{{a}_{6}}+{{a}_{11}}+{{a}_{16}}=\frac{4}{2}({{a}_{1}}+{{a}_{16}})\] \[=2\times 38=76\]                


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