Question

If the numbers $x,y,z$ are in H.P then $\frac{\sqrt{yz}}{\sqrt{y}+\sqrt{z}},\frac{\sqrt{xz}}{\sqrt{x}+\sqrt{z}},\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}$ are in

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

Answer 1

The correct option is A A.P

$x,y,z$ are in HP
$\Rightarrow \frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in AP

Now, $\frac{\sqrt{yz}}{\sqrt{y}+\sqrt{z}}+\frac{\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{1}{\sqrt{\frac{1}{z}}+\sqrt{\frac{1}{y}}}+\frac{1}{\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}}$

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

$=\frac{\sqrt{\frac{1}{z}}-\sqrt{\frac{1}{x}}}{d}$ where $d=\frac{1}{z}-\frac{1}{y}=\frac{1}{y}-\frac{1}{x}$

$=\frac{\sqrt{\frac{1}{z}}-\sqrt{\frac{1}{x}}}{\frac{1}{2}(\frac{1}{z}-\frac{1}{x})}$

$=2\left(\frac{1}{\sqrt{\frac{1}{z}}+\sqrt{\frac{1}{x}}}\right)$

$=\frac{2\sqrt{xz}}{\sqrt{x}+\sqrt{z}}$
$\therefore \frac{\sqrt{yz}}{\sqrt{y}+\sqrt{z}}+\frac{\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{2\sqrt{xz}}{\sqrt{x}+\sqrt{z}}$

$\therefore$ terms are in AP.