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JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

  • question_answer
    There are four numbers of which the first three are in G.P. and the last three are in A.R, whose common difference is 6. If the first and the last numbers are equal then two other numbers are

    A) -2, 4

    B) -4, 2

    C) 2, 6

    D) none

    Correct Answer: B

    Solution :

    [b] Let the last three numbers in A.P. be a, \[a+6,\] \[a+12,\] then the first term is also \[a+12,\]. But \[a+12,\]a, \[a+6\] are in GP. \[\therefore \,\,\,\,{{a}^{2}}=(a+12)\,(a+6)\Rightarrow {{a}^{2}}={{a}^{2}}+18a+72\] \[\therefore \,\,\,\,a=-4.\] \[\therefore \] The numbers are \[8,\,\,-4,\,\,2,\,\,8.\]


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