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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},\]..... are in A.P. such that \[{{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40,\]then the sum of the first 15 terms of this A.P. is : [JEE Main 12-4-2019 Afternoon]

    A) 200

    B) 280

    C) 120                 

    D) 150

    Correct Answer: A

    Solution :

    \[{{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40\]           \[a+a+6d+a+15d=40\]           \[\Rightarrow 3a+21d=40\]\[\Rightarrow \]           \[{{S}_{15}}=\frac{15}{2}\left( 2a+14d \right)=15\left( a+7d \right)\]           \[{{S}_{15}}=15\times \frac{40}{3}\Rightarrow 200\,\,\,\,\,\,\,\,\]                


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