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

  • question_answer
    If the sum of the first ten terms of the series \[{{\left( 1\frac{3}{5} \right)}^{2}}+{{\left( 2\frac{2}{5} \right)}^{2}}+{{\left( 3\frac{1}{5} \right)}^{2}}+{{4}^{2}}+{{\left( 4\frac{4}{5} \right)}^{2}}+.....,\] is \[\frac{16}{5}m,\] then m is equal to:

    A) 100

    B) 99   

    C) 102

    D) 101

    Correct Answer: D

    Solution :

    [d] \[{{\left( \frac{8}{5} \right)}^{2}}+{{\left( \frac{12}{5} \right)}^{2}}+{{\left( \frac{16}{5} \right)}^{2}}+{{\left( \frac{20}{5} \right)}^{2}}....+{{\left( \frac{44}{5} \right)}^{2}}\] \[S=\frac{16}{25}\left( {{2}^{2}}+{{3}^{2}}+{{4}^{2}}+....+{{11}^{2}} \right)\] \[=\frac{16}{25}\left( \frac{11(11+1)(22+1)}{6}-1 \right)\] \[=\frac{16}{25}\times 505=\frac{16}{5}\times 101\] \[\Rightarrow \,\,\frac{16}{5}m=\frac{16}{5}\times 101\Rightarrow m=101.\]


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