1. **Stating the problem:** We start with a point $S = (a,b)$ in the plane where $0 < b < a$. We generate a sequence of points $(x_n, y_n)$ defined by: $$x_0 = a, \quad y_0 = b,$$ $$x_{n+1} = \frac{x_n + y_n}{2}, \quad y_{n+1} = \frac{2 x_n y_n}{x_n + y_n}.$$ 2. **Invariant identification:** We are told that the product $x_n y_n$ remains constant for all $n$, i.e., $$x_n y_n = ab.$$ This is the invariant. 3. **Monotonicity and bounds:** Initially, $y_0 < x_0$. We want to show this inequality remains true for all $n$. - Since $x_{n+1}$ is the arithmetic mean of $x_n$ and $y_n$, and $y_{n+1}$ is the harmonic mean, we know the harmonic mean is always less than the arithmetic mean when $x_n \neq y_n$. - Therefore, $y_{n+1} < x_{n+1}$ for all $n$. 4. **Difference behavior:** Consider the difference: $$x_{n+1} - y_{n+1} = \frac{x_n + y_n}{2} - \frac{2 x_n y_n}{x_n + y_n} = \frac{(x_n - y_n)^2}{2(x_n + y_n)}.$$ Since $x_n, y_n > 0$, this difference is positive and decreases with $n$. 5. **Limit of the sequences:** Because the difference $x_n - y_n$ is positive and decreasing, both sequences converge to the same limit $x$. Using the invariant $x_n y_n = ab$, at the limit we have: $$x^2 = ab,$$ so $$x = \sqrt{ab}.$$ **Final answer:** The sequences $(x_n)$ and $(y_n)$ converge to $\sqrt{ab}$.