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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        If\[{{a}_{1}},{{a}_{2}},.....,{{a}_{n}}\]On are in arithmetic progression, where\[{{a}_{i}}>0\]for all\[i\]. Then \[\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+...+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}\]is equal to

    A)  \[\frac{{{n}^{2}}(n+1)}{2}\]                       

    B)  \[\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}\]

    C)  \[\frac{n(n-1)}{2}\]                       

    D)  None of these

    Correct Answer: B

    Solution :

                    Since,\[{{a}_{1}},{{a}_{2}},...,{{a}_{n}}\]are in AP. Then, \[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=.....={{a}_{n}}-{{a}_{n-1}}=d\] where d is common difference Now,     \[\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{1}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{2}}}}\]                                                 \[+......+\frac{1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n-1}}}}\]                 \[=\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{d}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{d}\]                                                 \[+...+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{d}\] \[=\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}})\times \frac{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\] \[=\frac{1}{d}\left( \frac{{{a}_{n}}-{{a}_{1}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right)=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\]


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