JEE Main & Advanced Mathematics Sequence & Series Question Bank Geometric Progression

  • question_answer
    The sum of infinite terms of the geometric progression \[\frac{\sqrt{2}+1}{\sqrt{2}-1},\frac{1}{2-\sqrt{2}},\frac{1}{2}.....\] is [Kerala (Engg.) 2002]

    A) \[\sqrt{2}{{(\sqrt{2}+1)}^{2}}\]

    B) \[{{(\sqrt{2}+1)}^{2}}\]

    C) \[5\sqrt{2}\]

    D) \[3\sqrt{2}+\sqrt{5}\]

    Correct Answer: A

    Solution :

    \[\frac{\sqrt{2}+1}{\sqrt{2}-1},\frac{1}{\sqrt{2}(\sqrt{2}-1)},\frac{1}{2},......\] Common ratio of the series \[=\frac{1}{\sqrt{2}(\sqrt{2}+1)}\] \[\therefore \] sum \[=\frac{a}{1-r}={\left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right)}/{\left( 1-\frac{1}{\sqrt{2}(\sqrt{2}+1)} \right)}\;\]      \[=\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}.\,\frac{\sqrt{2}\,(\sqrt{2}+1)}{(1+\sqrt{2})}\]\[=\sqrt{2}{{(\sqrt{2}+1)}^{2}}\].


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