Question

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

• A. $A.P.$
• B. $G.P.$
• C. $A.G.P$.
• D. no definite sequence

Answer 1

Answer: (A)

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

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

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