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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    If \[{{19}^{th}}\] term of a non-zero A.P. is zero, then it \[({{49}^{th}}\,\,term):({{29}^{th}}\,\,term)\]is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 1 : 3                             

    B) 2 : 1  

    C) 3 : 1                 

    D)   4 : 1

    Correct Answer: C

    Solution :

    Let a be the first term and d be the common difference. Given,\[{{a}_{19}}=a+18d=0\]\[\Rightarrow \]\[a=-18d\] \[\therefore \]\[\frac{a+48d}{a+28d}=\frac{-18d+48d}{-18d+28d}=\frac{30d}{10d}=\frac{3}{1}\]


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