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

  • question_answer
    If \[x\] is added to each of numbers 3, 9, 21 so that the resulting numbers may be in G.P., then the value of \[x\] will be [MP PET 1986]

    A) 3

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

    C) 2

    D) \[\frac{1}{3}\]

    Correct Answer: A

    Solution :

    \[3+x,\ 9+x,\ 21+x\] are in G.P. \[\Rightarrow \] \[{{(9+x)}^{2}}=(3+x)(21+x)\] \[\Rightarrow \] \[81+{{x}^{2}}+18x={{x}^{2}}+24x+63\] \[\Rightarrow \] \[6x=18\] or \[x=3\]. Trick: Check for (a), \[3+3,\ 9+3,\ 21+3\] are in G.P.


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