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  4. If x, y, z are in A.P., then (1/√y + √z), (1/√z +√x), (1/√x+√y) are in

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Q. If $x, y, z$ are in $A.P$., then $\frac{1}{\sqrt{y} + \sqrt{z}}, \frac{1}{\sqrt{z} +\sqrt{x}}, \frac{1}{\sqrt{x}+\sqrt{y}}$ are in

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Solution:

$\frac{1}{\sqrt{z} + \sqrt{x}}- \frac{1}{\sqrt{y} +\sqrt{z}}=\frac{ \sqrt{y } -\sqrt{x}}{\left(\sqrt{z} + \sqrt{x}\right)\left(\sqrt{y} +\sqrt{z}\right)}$

$ = \frac{\left(\sqrt{y } +\sqrt{x}\right)\left(\sqrt{y} -\sqrt{x}\right)}{\left(\sqrt{x} +\sqrt{y}\right)\left(\sqrt{z} +\sqrt{x}\right)\left(\sqrt{y} +\sqrt{z}\right)}$

$ =\frac{ y-x}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}$
Similarly,

$ \frac{1}{\sqrt{x}+\sqrt{y}} - \frac{1}{\sqrt{z}+\sqrt{x}} =\frac{ z-y}{\left(\sqrt{x} +\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)} $

Since, $x, y, z$ are in $A. P$.

$ \Rightarrow y- x = z-y $

$\Rightarrow \frac{1}{\sqrt{y}+\sqrt{z}}, \frac{1}{\sqrt{z} +\sqrt{x}}, \frac{1}{\sqrt{x} +\sqrt{y}}$ are in $A. P$.