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

  • question_answer
    For a sequence \[<{{a}_{n}}>,\ {{a}_{1}}=2\] and \[\frac{{{a}_{n+1}}}{{{a}_{n}}}=\frac{1}{3}\]. Then \[\sum\limits_{r=1}^{20}{{{a}_{r}}}\] is

    A) \[\frac{20}{2}[4+19\times 3]\]

    B) \[3\left( 1-\frac{1}{{{3}^{20}}} \right)\]

    C) \[2(1-{{3}^{20}})\]

    D) None of these

    Correct Answer: B

    Solution :

    The sequence is a G.P. with common ratio\[\frac{1}{3}\]. Now from \[\frac{a(1-{{r}^{n}})}{1-r},\,\,\,\,\frac{2\,[1-{{(1/3)}^{20}}]}{1-(1/3)}\] = \[3\,\left[ 1-\frac{1}{{{3}^{20}}} \right]\].


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