14. Given $A(\sqrt{5}, 0)$ and points $P_{1}, P_{2}, \cdots, P_{n}$ on the curve $\frac{x^{2}}{4}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0)$. Does there exist an $n$ such that $\left|P_{1} A\right|,\left|P_{2} A\right|, \cdots,\left|P_{n} A\right|$ form an arithmetic sequence with a common difference $d \in \left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$? If it exists, find all possible values of $n$; if not, explain the reason. (Take $\sqrt{5}=2.24$)
14. Since $d>0$, the sequence is increasing, so $\left|P_{\mid} A\right|$ is the smallest, and $\left|P_{n} A\right|$ is the largest.
As shown in Figure 5, the curve $\frac{x^{2}}{4}$
$$
-y^{2}=1(2 \leqslant x \leqslant
$$
$2 \sqrt{5}, y \geqslant 0)$ is part of a hyperbola,
with $A(\sqrt{5}, 0)$ being
its right focus, and the equation of the right directrix $l$ is $x=$
$$
\frac{4}{\sqrt{5}}, e=\frac{\sqrt{5}}{2} \text {. }
$$

According to the problem, the first term of the arithmetic sequence is $\left|P_{1} A\right|=\sqrt{5}-2$, and the $n$-th term is $\left|P_{n}^{*} A\right|=3$. Therefore, we have
$$
3=(\sqrt{5}-2)+(n-1) d \text {. }
$$

Solving for $d$ gives $d=\frac{5-\sqrt{5}}{n-1}(n>1)$.
Since $d \in\left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$, we have $\frac{1}{5}<\frac{5-\sqrt{5}}{n-1}<\frac{1}{\sqrt{5}}$.
Solving for $n$ gives $n \in(5 \sqrt{5}-4,26-5 \sqrt{5})$.
Since $n$ is a positive integer, the maximum value of $n$ is 14.
Thus, when $n \leqslant 14$, any $n$ consecutive points among these 14 points can form an arithmetic sequence.
Since it is an arithmetic sequence, we have $n \geqslant 3$.
Therefore, $n=3,4,5, \cdots, 14$.