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

  • question_answer
    The value of\[{{a}^{{{\log }_{b}}x}}\], where \[a=0.2,\ b=\sqrt{5},\ x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.........\]to \[\infty \] is

    A) 1

    B) 2

    C) \[\frac{1}{2}\]

    D) 4

    Correct Answer: D

    Solution :

    \[x=\frac{1/4}{1-(1/2)}=\frac{1}{2}\] \[\therefore \]\[{{\left( \frac{1}{5} \right)}^{{{\log }_{\sqrt{5}}}\left( \frac{1}{2} \right)}}={{\left( \frac{1}{5} \right)}^{{{\log }_{5}}\left( \frac{1}{4} \right)}}={{5}^{-{{\log }_{5}}{{4}^{-1}}}}={{5}^{{{\log }_{5}}4}}=4\].


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