Question

For some positive and distinct real numbers $x ,y$, and $z$ , if $\frac{1}{\sqrt{ y}+ \sqrt{z}}$ is the arithmetic mean of $\frac{1}{\sqrt{x}+ \sqrt{z}}$ and $\frac{1}{\sqrt{x} +\sqrt{y}}$ , then the relationship which will always hold true, is

• A. $y ,x$, and $z$ are in arithmetic progression
• B. $\sqrt{x}, \sqrt{y}$ , and $\sqrt{z}$ are in arithmetic progression
• C. $x ,y$, and $z$ are in arithmetic progression
• D. $\sqrt x, \sqrt z$ , and $\sqrt y$ are in arithmetic progression

Answer 1

Answer: (A)

$y ,x$, and $z$ are in arithmetic progression

Answer 2

Given
$\frac{1}{\sqrt{y}+\sqrt{z}}\ \text{is}$ the arithmetic mean of \frac{1}{\sqrt{x}+\sqrt{z}}\ \text{and}$$\frac{1}{\sqrt{x}+\sqrt{y}}
$\frac{2}{\sqrt{y}+\sqrt{z}}=\frac{1}{\sqrt{x}+\sqrt{z}}+\frac{1}{\sqrt{x}+\sqrt{y}}$

⇒$$2(\sqrt{x} + \sqrt{z})(\sqrt{x} + \sqrt{y}) = (\sqrt{y} + \sqrt{z})(\sqrt{x} + \sqrt{y} + \sqrt{x} + \sqrt{z})

⇒$$2(x + \sqrt{xy} + \sqrt{xz} + \sqrt{yz}) = 2\sqrt{xy} + y + \sqrt{yz} + 2\sqrt{xz} + \sqrt{yz} + z

$⇒\ 2x=y+z$
Therefore, x is the arithmetic mean of y and z, y, x, and z are in A.P
The correct option is (A): $y ,x$ and $z$ are in arithmetic progression.