If a, b, c are in Arithmetic Progression,then | vedclass

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

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Arithmetic Progression

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(A) Given that $a, b, c$ are in Arithmetic Progression,we have $2b = a + c$. Consider the terms $x = \frac{1}{\sqrt{b} + \sqrt{c}}$,$y = \frac{1}{\sqrt{c} + \sqrt{a}}$,and $z = \frac{1}{\sqrt{a} + \sqrt{b}}$. Rationalizing the denominators: $x = \frac{\sqrt{c} - \sqrt{b}}{c - b}$,$y = \frac{\sqrt{a} - \sqrt{c}}{a - c}$,$z = \frac{\sqrt{b} - \sqrt{a}}{b - a}$. Since $a, b, c$ are in $A$.$P$.,let $c - b = b - a = d$. Then $a - c = -2d$. $x = \frac{\sqrt{c} - \sqrt{b}}{d}$,$y = \frac{\sqrt{a} - \sqrt{c}}{-2d}$,$z = \frac{\sqrt{b} - \sqrt{a}}{d}$. For these to be in $A$.$P$.,$2y = x + z$ must hold. $x + z = \frac{\sqrt{c} - \sqrt{b} + \sqrt{b} - \sqrt{a}}{d} = \frac{\sqrt{c} - \sqrt{a}}{d}$. $2y = 2 	imes \frac{\sqrt{a} - \sqrt{c}}{-2d} = \frac{\sqrt{c} - \sqrt{a}}{d}$. Since $2y = x + z$,the terms are in Arithmetic Progression.

If $a, b, c$ are in Arithmetic Progression,then $\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}$ are in:

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