SSC Sample Paper Mock Test-11 SSC CGL Tear-II Paper-1

  • question_answer
    \[\left( \frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+\frac{1}{10\cdot 13}+\frac{1}{13\cdot 16} \right)\]is equal to

    A)  \[\frac{1}{3}\]

    B)  \[\frac{5}{16}\]

    C)  \[\frac{3}{8}\]

    D)  \[\frac{41}{7280}\]

    Correct Answer: B

    Solution :

    Given expression                           
    \[=\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+\frac{1}{10\times 13}+\frac{1}{13\times 16}\]
    \[=\frac{1}{3}\left( \begin{align}   & \left( \frac{3}{1\times 4} \right)+\left( \frac{3}{4\times 7} \right)+\left( \frac{3}{7\times 10} \right)+\left( \frac{3}{10\times 13} \right)+ \\  & \left( \frac{3}{13\times 16} \right) \\ \end{align} \right)\]
    \[=\frac{1}{3}\left[ \begin{align}   & \left( 1-\frac{1}{4} \right)+\left( \frac{1}{4}-\frac{1}{7} \right)+\left( \frac{1}{7}-\frac{1}{10} \right)+\left( \frac{1}{10}-\frac{1}{13} \right) \\  & +\left( \frac{1}{13}-\frac{1}{16} \right) \\ \end{align} \right]\]
    \[=\frac{1}{3}\left( 1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16} \right)\]\[=\frac{1}{3}\left( 1-\frac{1}{16} \right)=\frac{1}{3}\left( \frac{16-1}{16} \right)=\frac{1}{3}\times \frac{15}{16}=\frac{5}{16}\]


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