Question

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

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is A A.P

Given that $x,y,z$ are in $A.P$

$\Rightarrow y-x=z-y$

$\Rightarrow x-y=y-z$

$\Rightarrow \frac{1}{x-y}=\frac{1}{y-z}$

This can be written as

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

$\Rightarrow \frac{\sqrt{y}-\sqrt{z}}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{y}+\sqrt{z}}$

Add and subtract $\sqrt{x}$ and $\sqrt{z}$ on $L.H.S$ and $R.H.S$ respectively

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

Divide on both sides with $(\sqrt{z}+\sqrt{x})$

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

$\Rightarrow \frac{1}{\sqrt{z}+\sqrt{x}}-\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{y}-\sqrt{z}}-\frac{1}{\sqrt{z}+\sqrt{x}}$

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