JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    Let \[{{a}_{1}},{{a}_{2}},...,{{a}_{10}}\]be a G,P. If \[\frac{{{a}_{3}}}{{{a}_{1}}}=25,\]then \[\frac{{{a}_{9}}}{{{a}_{5}}}\]equals [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[2({{5}^{2}})\]                                  

    B)               \[4({{5}^{2}})\]  

    C)               \[{{5}^{4}}\]                                      

    D)               \[{{5}^{3}}\]

    Correct Answer: C

    Solution :

    Here, \[{{a}_{1}},{{a}_{2}},\],.......\[{{a}_{10}}\]are in G.P. Let the common ratio be r. Given,\[\frac{{{a}_{3}}}{{{a}_{1}}}=25\Rightarrow \frac{{{a}_{1}}{{r}^{2}}}{{{a}_{1}}}=25\Rightarrow {{r}^{2}}=25\Rightarrow r=\pm 5\] Now,\[\frac{{{a}_{9}}}{{{a}_{5}}}=\frac{{{a}_{1}}{{r}^{8}}}{{{a}_{1}}{{r}^{4}}}={{r}^{4}}={{5}^{4}}\]


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