Three. (20 points) Given the parabola $C: y=\frac{1}{2} x^{2}$, $A_{1}\left(x_{1}, 0\right)$ and $A_{2}\left(x_{2}, 0\right)$ are two points on the $x$-axis $\left(x_{1}+x_{2} \neq 0, x_{1} x_{2} \neq 0\right)$. Perpendicular lines to the $x$-axis are drawn through points $A_{1}$ and $A_{2}$, intersecting the parabola $C$ at points $A_{1}^{\prime}$ and $A_{2}^{\prime}$, respectively. The line $A_{1}^{\prime} A_{2}^{\prime}$ intersects the $x$-axis at point $A_{3}\left(x_{3}, 0\right)$. In this way, $x_{1}$ and $x_{2}$ determine $x_{3}$. Similarly, $x_{2}$ and $x_{3}$ can determine $x_{4}, \cdots$. Given $x_{1}=6, x_{2}=2$, find the value of $x_{6}$.
Three, from the given conditions, we have
$$
A_{1}^{\prime}\left(x_{1}, \frac{1}{2} x_{1}^{2}\right), A_{2}^{\prime}\left(x_{2}, \frac{1}{2} x_{2}^{2}\right) \text {. }
$$

Then \( k_{x / \dot{k} / 2}=\frac{\frac{1}{2}\left(x_{2}^{2}-x_{1}^{2}\right)}{x_{2}-x_{1}}=\frac{1}{2}\left(x_{2}+x_{1}\right) \). Therefore, \( l_{x /\{:\{}: y-\frac{1}{2} x_{2}^{2}=\frac{1}{2}\left(x_{2}+x_{1}\right)\left(x-x_{2}\right) \). Let \( y=0 \), we get \( \frac{1}{x}=\frac{1}{x_{1}}+\frac{1}{x_{2}} \), which means \( \frac{1}{x_{3}}=\frac{1}{x_{1}}+\frac{1}{x_{2}} \).
Similarly, \( \frac{1}{x_{4}}=\frac{1}{x_{2}}+\frac{1}{x_{3}}, \frac{1}{x_{5}}=\frac{1}{x_{3}}+\frac{1}{x_{4}}, \cdots \cdots \).
Thus, \( \frac{1}{x_{6}}=\frac{1}{x_{4}}+\frac{1}{x_{5}}=\frac{2}{x_{4}}+\frac{1}{x_{3}} \) \( =\frac{3}{x_{3}}+\frac{2}{x_{2}}=\frac{5}{x_{2}}+\frac{3}{x_{1}} \).
Therefore, with \( x_{1}=6, x_{2}=2 \), we get \( x_{6}=\frac{1}{3} \).