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

  • question_answer
    Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then \[{{P}^{2}}{{R}^{3}}:{{S}^{3}}\] is equal to

    A) 1 : 1

    B) Common ratio: 1

    C) \[{{\left( \text{first term} \right)}^{\text{2}}}\text{: }{{\left( \text{common ratio} \right)}^{\text{2}}}\]

    D) \[{{\left( common\text{ }ratio \right)}^{n}}:1\]

    Correct Answer: A

    Solution :

    [a] If the three terms of the GP. be \[\frac{a}{r},\] a and or then \[S=\frac{a}{r}+a+ar=\frac{a}{r}(1+r+{{r}^{2}})\] \[P={{a}^{3}}\] and \[R=\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}=\frac{1}{ar}({{r}^{2}}+r+1)\] Now,  \[\frac{{{P}^{2}}{{R}^{3}}}{{{S}^{3}}}=\frac{{{a}^{6}}\frac{1}{{{a}^{3}}{{r}^{3}}}{{({{r}^{2}}+r+1)}^{3}}}{\frac{{{a}^{3}}}{{{r}^{3}}}{{({{r}^{2}}+r+1)}^{3}}}\] So, the required ratio is \[1:1\].


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