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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    Consider the ten numbers\[ar,a{{r}^{2}}a{{r}^{3}},\,.......\,,\,a{{r}^{10}}\]. If their sum is 18 and the sum of their reciprocals is 6 then the product of these ten numbers, is

    A)  324                                       

    B)  343

    C)  243                                       

    D)  729

    Correct Answer: C

    Solution :

    Given \[\frac{ar({{r}^{10}}-1)}{r-1}\,=18\]             Also \[\frac{\frac{1}{ar}\,\left( 1-\frac{1}{{{r}^{10}}} \right)}{1-\frac{1}{r}}\,=6\Rightarrow \,\,\frac{1}{a{{r}^{11}}}.\,\,\frac{({{r}^{10}}-1)r}{r-1}\,=6\] \[\frac{1}{{{a}^{2}}{{r}^{11}}}.\,\frac{ar({{r}^{10}}-1)}{r-1}\,=6\]                      ?(2) Form (1) and (2) \[\frac{1}{{{a}^{2}}{{r}^{11}}}.18=6\Rightarrow \,{{a}^{2}}{{r}^{11}}=3\] Now \[P={{a}^{10}}\,{{r}^{55}}\,=({{a}^{2}}{{r}^{11}})\,={{3}^{5}}\,=243\]


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