6. Given $A(a, b), B(c, d)$, and
$$
(a-c)^{2}+(b-d)^{2} \neq 0,
$$

point $P_{n}\left(n \in \mathbf{N}_{+}\right)$ satisfies $A P_{1}=\frac{1}{2} A B, B P_{2}=$ $\frac{1}{2} B P_{1}, P_{n} P_{n+2}=\frac{1}{2} P_{n} P_{n+1}$. Then $\lim _{n \rightarrow \infty} A P_{n}$ is ( ).
(A) $\left(\frac{c-a}{3}, \frac{d-b}{3}\right)$
(B) $\left(\frac{c-a}{2}, \frac{d-b}{2}\right)$
(C) $\left(\frac{2(c-a)}{3}, \frac{2(d-b)}{3}\right)$
(D) $\left(\frac{3(c-a)}{4}, \frac{3(d-b)}{4}\right)$
6.C.

It is easy to see that the limit position of $P_{n}$ is at the point dividing segment $A B$ in the ratio $2: 1$.