Three. (20 points) Given point $A(\sqrt{5}, 0)$ and the curve $y=$ $\sqrt{\frac{x^{2}}{4}-1}(2 \leqslant x \leqslant 2 \sqrt{5})$ with points $P_{1} 、 P_{2}, \cdots$ 、 $P_{n}$. If $\left|P_{1} A\right| 、\left|P_{2} A\right| 、 \cdots 、\left|P_{n} A\right|$ form an arithmetic sequence with common difference $d \in\left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$, find the maximum value of $n$.
The given curve is a segment of the following hyperbola, i.e.,
$$
\frac{1}{4} x^{2}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0) \text {. }
$$
$A(\sqrt{5}, 0)$ is its right focus, as shown in Figure 5 (where the line $l$ is the right directrix $x=\frac{4}{\sqrt{5}}$, and the point $P(2 \sqrt{5}, 2)$, with the eccentricity $\left.e=\frac{\sqrt{5}}{2}\right)$.
It is easy to see that $\left|P_{n} A\right|_{\text {min }}=\sqrt{5}-2$,
$$
\left|P_{n} A\right|_{\text {max }}=e|P H|=\frac{\sqrt{5}}{2}\left(2 \sqrt{5}-\frac{4}{\sqrt{5}}\right)=3 \text {. }
$$

According to the problem, we can set the first term of the arithmetic sequence $a_{1}=\sqrt{5}-2$, and the $n$-th term $a_{n}=3$. Then $3=(\sqrt{5}-2)+(n-1) d$.
This gives $d=\frac{5-\sqrt{5}}{n-1}(n>1)$.
By the problem, $\frac{1}{5}<d<\frac{1}{\sqrt{5}}$, i.e., $\frac{1}{5}<\frac{5-\sqrt{5}}{n-1}<\frac{1}{\sqrt{5}}$.

This yields $5 \sqrt{5}-4<n<26-5 \sqrt{5}$.
And $7=5 \times 2.2-4<5 \sqrt{5}-4$,
and $26-5 \sqrt{5}<26-5 \times 2.2=15$,
thus $7<n<15$.
Therefore, $n$ can be at most 14.