Question

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

• A. $5$
• B. $7$
• C. $11$
• D. None of these

Answer 1

Answer: (A)

$5$

$\sqrt{a-x},\sqrt{x}$ and $\sqrt{x+a}$ are in A.P
$\Rightarrow 2\sqrt{x}=\sqrt{x+a}+\sqrt{a-x}$.....(1)

$\Rightarrow 2\sqrt{x}=\frac{(\sqrt{x+a}+\sqrt{a-x})(\sqrt{x+a}-\sqrt{a-x})}{(\sqrt{x+a}-\sqrt{a-x})}$

$\Rightarrow 2\sqrt{x}=\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}$
$⟹\sqrt{a+x}-\sqrt{a-x}=\sqrt{x}$.....(2)

Adding equation (1) and (2) we get
$⟹3\sqrt{x}=2\sqrt{a+x}$
$⟹a=\frac{5x}{4}$
Now using the condition that both 'a' and 'x' are positive integers, the least possible value of $a=5$.