If a_1, a_2, a_3, ......., a_n are in A.P.,wh | vedclass

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(A) Given that $a_1, a_2, a_3, ......., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$.We know that $a_n = a_1 + (n - 1)d$,which implie

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$\frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}}$

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(A) Given that $a_1, a_2, a_3, ......., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$.We know that $a_n = a_1 + (n - 1)d$,which implies $d = \frac{a_n - a_1}{n - 1}$.Rationalizing each term of the sum:$\frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$.Summing these terms from $i = 1$ to $n-1$:$\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{1}{d} ((\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + ....... + (\sqrt{a_n} - \sqrt{a_{n-1}}))$.This is a telescoping sum,which simplifies to:$\frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.Substituting $d = \frac{a_n - a_1}{n - 1}$:$= \frac{n - 1}{a_n - a_1} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{\sqrt{a_n} + \sqrt{a_1}}$.

If $a_1, a_2, a_3, ......., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\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}} = $

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