If a_r 0, r in N and a_1, a_2, a_3, ..., a_ | vedclass

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(B) In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{2n} = a_2 + a_{2n-1} = ... = a_n + a_{n

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$\frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

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(B) In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{2n} = a_2 + a_{2n-1} = ... = a_n + a_{n+1} = k$Each term in the series is of the form $\frac{k}{\sqrt{a_i} + \sqrt{a_{i+1}}}$.Rationalizing the denominator: $\frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{a_i - a_{i+1}} = \frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{-d}$,where $d$ is the common difference.The sum becomes $\sum_{i=1}^{n} \frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{-d} = \frac{k}{-d} (\sqrt{a_1} - \sqrt{a_{n+1}})$.Since $a_{n+1} = a_1 + nd$,we have $a_1 - a_{n+1} = -nd$.Substituting $k = a_1 + a_{2n}$ and the difference,the expression simplifies to $\frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$.

If $a_r > 0, r \in N$ and $a_1, a_2, a_3, ..., a_{2n}$ are in an arithmetic progression,then $\frac{a_1 + a_{2n}}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + \frac{a_3 + a_{2n-2}}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}} = ?$

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