6. Given a line $l$ in the plane and a point $P$ at a distance of 1 from $l$, 100 points $A_{1}, A_{2}, \cdots, A_{100}$ on $l$ satisfy
$$
\begin{array}{l}
\overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{i+1}}=-\frac{2}{i}(i=1,2, \cdots, 99) . \\
\text { Then } \overrightarrow{P A_{100}} \cdot \overrightarrow{P A_{1}}=
\end{array}
$$
6. $-\frac{51}{50}$.

Draw a perpendicular from point $P$ to line $l$, with the foot of the perpendicular being $O$. Establish a Cartesian coordinate system with $l$ as the $x$-axis and $OP$ as the $y$-axis.
Then, $P(0,1)$.

Let $A_{i}\left(x_{i}, 0\right)(i=1,2, \cdots, 100)$.
Thus, $\overrightarrow{P A_{i}}=\left(x_{i},-1\right)$.
$$
\begin{array}{l}
\text { By } \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{i+1}}=-\frac{2}{i} \\
\Rightarrow x_{i} x_{i+1}+1=-\frac{2}{i} \\
\Rightarrow x_{i} x_{i+1}=-\frac{i+2}{i}(i=1,2, \cdots, 999) . \\
\text { Therefore, } x_{1} x_{100}=\frac{\prod_{i=1}^{40} x_{2 i-1} x_{2 i}}{\prod_{i=1}^{49} x_{2 i} x_{2 i+1}}=\frac{\prod_{i=1}^{50}\left(-\frac{2 i+1}{2 i-1}\right)}{\prod_{i=1}^{49}\left(-\frac{2 i+2}{2 i}\right)} \\
=-\frac{101}{50} \\
\Rightarrow \overrightarrow{P A_{100}} \cdot \overrightarrow{P A_{1}}=x_{100} x_{1}+1=-\frac{51}{50} .
\end{array}
$$