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

The Correct Option is A

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.