If a, b, c are in A.P.,then frac{1}{sqrt{a} + | vedclass

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(A) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Consider the terms $x = \frac{1}{\sqrt{a} + \sqrt{b}}$,$y = \frac{1}{\sqrt{a} + \sqrt{c}}

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(A) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Consider the terms $x = \frac{1}{\sqrt{a} + \sqrt{b}}$,$y = \frac{1}{\sqrt{a} + \sqrt{c}}$,and $z = \frac{1}{\sqrt{b} + \sqrt{c}}$. To check if they are in $A.P.$,we check if $2y = x + z$. Rationalizing the denominators: $x = \frac{\sqrt{b} - \sqrt{a}}{b - a}$,$y = \frac{\sqrt{c} - \sqrt{a}}{c - a}$,$z = \frac{\sqrt{c} - \sqrt{b}}{c - b}$. Since $a, b, c$ are in $A.P.$,let $b - a = d$ and $c - b = d$,so $c - a = 2d$. $x = \frac{\sqrt{b} - \sqrt{a}}{d}$,$y = \frac{\sqrt{c} - \sqrt{a}}{2d}$,$z = \frac{\sqrt{c} - \sqrt{b}}{d}$. $x + z = \frac{\sqrt{b} - \sqrt{a} + \sqrt{c} - \sqrt{b}}{d} = \frac{\sqrt{c} - \sqrt{a}}{d}$. $2y = 2 	imes \frac{\sqrt{c} - \sqrt{a}}{2d} = \frac{\sqrt{c} - \sqrt{a}}{d}$. Since $x + z = 2y$,the terms are in $A.P.$

If $a, b, c$ are in $A.P.$,then $\frac{1}{\sqrt{a} + \sqrt{b}}, \frac{1}{\sqrt{a} + \sqrt{c}}, \frac{1}{\sqrt{b} + \sqrt{c}}$ are in

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