6. Given point $A(\sqrt{5}, 0)$ and the curve $y=\sqrt{\frac{x^{2}}{4}-1} \quad(2 \leqslant x \leqslant 2 \sqrt{5})$ 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 $d \in\left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$, then the maximum value of $n$ is ( ).
A. 13
B. 14
C. 15
D. 16
6. B

It is easy to find that the directrix of the known hyperbola is $x=\frac{4}{\sqrt{5}}$.
Assume the distances from $P_{n}$ to the directrix are $d_{n}$, then $\left|P_{n} A\right|=\frac{\sqrt{5}}{2} x_{n}-2$ where $P_{n}\left(x_{n}, y_{n}\right)$, and $A(\sqrt{5}, 0)$ is the focus, so $x_{1}, x_{2}, \cdots, x_{n}$ or $A P$, and the smaller $d$ is, the larger $n$ is. Let $d=\frac{1}{5} \div e=\frac{2}{5 \sqrt{5}}$.
Given $x_{1}=2$, $d=\frac{2}{5 \sqrt{5}}$, $x_{n}=x_{1}+(n-1) d=2+\frac{2}{5 \sqrt{5}}(n-1)$,

From $x_{n} \leqslant 2 \sqrt{5}$, we get: $n \leqslant 5 \sqrt{5}(\sqrt{5}-1)+1 \approx 14.82$, so we take $n=14$.