Invariance of Geometric Mean under AM-HM Iteration

Let the two initial positive numbers be $a$ and $b$ ($a \neq b$).
Then $A_1 = \frac{a+b}{2}$, $G_1 = \sqrt{ab}$, and $H_1 = \frac{2ab}{a+b}$.
Notice that $A_1 H_1 = \left(\frac{a+b}{2}\right)\left(\frac{2ab}{a+b}\right) = ab = G_1^2$.

For $n \ge 2$, $G_n$ is the geometric mean of $A_{n-1}$ and $H_{n-1}$:
$G_n = \sqrt{A_{n-1} H_{n-1}}$

Also, by definition of arithmetic and harmonic means of two quantities $x, y$, their product is always equal to the square of their geometric mean ($xy = G^2$). Thus:
$A_{n-1} H_{n-1} = G_{n-1}^2 \implies G_n = \sqrt{G_{n-1}^2} = G_{n-1}$

By mathematical induction:
$G_1 = G_2 = G_3 = \dots = \sqrt{ab}$

Correct Answer: C