If a and x are positive integers such that x<a and a-x, x, a+x are in A.P., then least possible value of a is

If a and x are positive integers such that x

A
5
B
7
C
11
D
none of these

Solution:

We have $2 \sqrt{x} = \sqrt{a + x} + \sqrt{a - x}$
$\begin{array}{l} \Rightarrow 2 \sqrt{x} = \frac{2 x}{\sqrt{a + x} - \sqrt{a - x}} \Rightarrow \sqrt{a + x} - \sqrt{a - x} = \sqrt{x} \\ a + x + a - x - 2 \sqrt{a^{2} - x^{2}} = x \Rightarrow 2 a - x = 2 \sqrt{a^{2} - x^{2}} \\ \Rightarrow 4 a^{2} + x^{2} - 4 a x = 4 a^{2} - 4 x^{2} \Rightarrow 5 x^{2} = 4 a x \\ ∴ a = \frac{5}{4} x \end{array}$
As $a$ and $x$ are positive integers, least possible value of $x$ is $4$ and that of $a$ is $5 .$

If $a$ and $x$ are positive integers such that $x < a$ and $\sqrt{a - x}, \sqrt{x}, \sqrt{a + x}$ are in A.P., then least possible value of $a$ is

Solution:

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