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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    Which term of the GP\[3,3\sqrt{3},9....\]is 2187?

    A)  15                                         

    B)  14

    C)  13                         

    D)         19

    E)  20

    Correct Answer: C

    Solution :

    Let nth term of GP is 2187 \[\therefore \]  \[3{{(\sqrt{3})}^{n-1}}=2187\] \[\Rightarrow \]               \[{{3}^{(n/2-1/2+1)}}={{3}^{7}}\] \[\Rightarrow \]               \[\frac{n}{2}+\frac{1}{2}=7\] \[\Rightarrow \]               \[n=13\]


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