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KVPY Sample Paper KVPY Stream-SX Model Paper-23

  • question_answer
    Given a, b, c are positive integers forming an increasing GP, \[b-a\] is a perfect square of a natural number and \[{{\log }_{6}}a+{{\log }_{6}}b+{{\log }_{6}}c=6,\] then \[a+b+c\] is equal to

    A) 108

    B) 109      

    C) 110      

    D) 111

    Correct Answer: D

    Solution :

    We have, a, b, c are in GP.
    \[\therefore \]      \[{{b}^{2}}=ac\]
    and \[{{\log }_{6}}a+{{\log }_{6}}b+{{\log }_{6}}c=6\] \[\Rightarrow \]\[{{\log }_{6}}(abc)=6\]\[\Rightarrow \]\[abc={{6}^{6}}\] \[\Rightarrow \]\[{{b}^{3}}={{6}^{6}}\]\[\Rightarrow \]\[b=36\] \[\Rightarrow \]\[ac=36\times 36={{2}^{4}}\times {{3}^{4}}\]
    \[\Rightarrow \]\[b-a={{N}^{2}}\] \[\Rightarrow \]\[36-a={{N}^{2}}\]
    a is a factor of \[{{2}^{4}}\times {{3}^{4}}\] , \[a=27\]is possible value
    \[\therefore \]\[a=27,\]\[b=36,\]\[c=48\]
    \[a+b+c=27+36+48=111\]


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