J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    The sum of the series \[5-\frac{10}{3}+\frac{20}{9}-\frac{40}{27}+....\] is equal to

    A)  \[1\]                 

    B)  \[2\]

    C)  \[3\]               

    D)  None of these

    Correct Answer: C

    Solution :

    Let \[S=5-\frac{10}{3}+\frac{20}{9}-\frac{40}{27}+...\] ?.(i) \[\frac{1}{3}S=\frac{5}{3}-\frac{10}{9}+\frac{20}{27}-.....\] ?.(ii) On adding Eqs. (i) and (ii), we get \[\frac{4}{3}S=5-\frac{5}{3}+\frac{10}{9}-\frac{20}{27}\] \[=5-\frac{5}{3}\left( 1-\frac{2}{3}+\frac{4}{9}-.... \right)\] \[=5-\frac{5}{3}\left( \frac{1}{1+\frac{2}{3}} \right)\] \[=5-\frac{5}{3}\left( \frac{3}{5} \right)\] \[\Rightarrow \] \[\frac{4}{3}S=4\] \[\Rightarrow \] \[S=3\]


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