Let a_1, a_2, a_3, ldots, a_n be n positive c | vedclass

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(A) Let the given expression be $S_n = \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$.Rationalizing each term,we get $\frac{1}{\sqrt{a_k} + \

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(A) Let the given expression be $S_n = \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$.Rationalizing each term,we get $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.Thus,$S_n = \frac{1}{d} \sum_{k=1}^{n-1} (\sqrt{a_{k+1}} - \sqrt{a_k}) = \frac{\sqrt{a_n} - \sqrt{a_1}}{d}$.The limit becomes $\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \lim_{n \rightarrow \infty} \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{nd}}$.Since $a_n = a_1 + (n-1)d$,as $n \rightarrow \infty$,$a_n \approx nd$.Therefore,the limit is $\lim_{n \rightarrow \infty} \frac{\sqrt{nd} - \sqrt{a_1}}{\sqrt{nd}} = \lim_{n \rightarrow \infty} \left( 1 - \frac{\sqrt{a_1}}{\sqrt{nd}} \right) = 1$.

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