JEE Main & Advanced Mathematics Sequence & Series Question Bank Mock Test - Sequences and Series

  • question_answer
    Let \[S=\frac{4}{19}+\frac{44}{{{19}^{2}}}+\frac{444}{{{19}^{3}}}+...\]up to \[\infty \]. Then S is equal to

    A) 40/9                 

    B) 38/81

    C) 36/171

    D) none of these

    Correct Answer: B

    Solution :

    [b] \[S=\frac{4}{19}+\frac{44}{{{19}^{2}}}+\frac{444}{{{19}^{3}}}+...\]         ...(1) \[\Rightarrow \frac{1}{19}S=\frac{4}{{{19}^{2}}}+\frac{44}{{{19}^{3}}}+...\]                      ...(2) Subtracting (2) from (1), we get \[\frac{18}{19}S=\frac{4}{19}+\frac{40}{{{19}^{2}}}+\frac{400}{{{19}^{3}}}+...\] \[=\frac{\frac{4}{19}}{1-\frac{10}{19}}=\frac{4}{9}\] \[\Rightarrow S=\frac{38}{81}\]


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