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

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

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

Vedclass · Answer

(A) Given that $a_1, a_2, ..., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$ for all $i$.We know that $a_n = a_1 + (n-1)d$,which implies $(n-1)d = a_n - a_1$.Rationalizing each term of the sum $S = \sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}}$:$S = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$$S = \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,so $S = \frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.Multiplying the numerator and denominator by $(\sqrt{a_n} + \sqrt{a_1})$:$S = \frac{1}{d} \cdot \frac{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})}{\sqrt{a_n} + \sqrt{a_1}} = \frac{1}{d} \cdot \frac{a_n - a_1}{\sqrt{a_n} + \sqrt{a_1}}$Since $a_n - a_1 = (n-1)d$,we get:$S = \frac{1}{d} \cdot \frac{(n-1)d}{\sqrt{a_n} + \sqrt{a_1}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$.

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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