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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    If the sum of the first ten terms of the series \[{{\left( 1\frac{3}{5} \right)}^{2}}+{{\left( 2\frac{2}{5} \right)}^{2}}+{{\left( 3\frac{1}{5} \right)}^{2}}+....,\]is\[\frac{16}{5}m,\]then m is equal to :- [JEE Main Solved Paper-2016 ]

    A)  99                                         

    B) 102

    C) 101                                        

    D) 100

    Correct Answer: C

    Solution :

                    Given series is \[S=\frac{{{8}^{2}}}{{{5}^{2}}}+\frac{{{12}^{2}}}{{{5}^{2}}}+\frac{{{16}^{2}}}{{{5}^{2}}}+...10\]terms \[=\frac{{{4}^{2}}}{{{5}^{2}}}({{2}^{2}}+{{3}^{2}}+{{4}^{2}}+....10\,\text{terms})\] \[=\frac{16}{25}\left( \frac{11.12.23}{6}-1 \right)=\frac{16}{25}\times 505\] \[\therefore \]\[m=101\]


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