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

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

Given that $\frac{1}{\sqrt{y}+\sqrt{z}}$ is the arithmetic mean of $\frac{1}{\sqrt{x}+\sqrt{z}}$ and $\frac{1}{\sqrt{x}+\sqrt{y}}$, we have: $\frac{2}{\sqrt{y}+\sqrt{z}}=\frac{1}{\sqrt{x}+\sqrt{z}}+\frac{1}{\sqrt{x}+\sqrt{y}}$

$⇒$$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$

$⇒\ 2x=y+z$
This implies that x is the arithmetic mean of y and z, meaning y, x, and z are in an arithmetic progression. The correct option is (A): $y ,x$ and $z$ are in arithmetic progression.

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

The Correct Option is A

Solution and Explanation

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