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

  • question_answer
    If \[(1-p)(1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})\]\[=1-{{p}^{6}},\] \[p\ne 1\] then the value of \[\frac{p}{x}\] is

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

    B) 3

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

    D) 2

    Correct Answer: B

    Solution :

    [b] \[(1-p)(1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})\] \[=1-{{p}^{6}}\] \[\Rightarrow (1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=\frac{1-{{p}^{6}}}{1-p}\]\[\Rightarrow (1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})\]             \[=1+p+{{p}^{2}}+{{p}^{3}}+{{p}^{4}}+{{p}^{5}}\] Comparing we get \[p=3x\] or \[p/x=3\]


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