Q. Let $a$ be a positive number such that the arithmetic mean of $a$ and 2 exceeds their geometric mean by 1. Then, the value of $a$ is
A) 3
B) 5
C) 9
D) 8
Explanation
(d)
$\frac{a + 2}{2} = \sqrt{2a} + 1$
$\Rightarrow \ \frac{a}{2} = \sqrt{2a}$
$\Rightarrow \ \frac{a^{2}}{4} = 2a$
$\Rightarrow \ a = 8$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.