6. The smallest element of the set of positive integers $A_{k}$ is 1, and the largest element is 2007, and the elements can be arranged in an arithmetic sequence with a common difference of $k$. Then the number of elements in the union $A_{17} \cup A_{59}$ is ( ).
(A) 119
(B) 120
(C) 151
(D) 154
6. C.

Let $\left|A_{k}\right|$ denote the number of elements in the set $A_{k}$, and suppose
$$
\left|A_{k}\right|=n+1 \text {. }
$$

From $2007=1+n k$, we get $n=\frac{2006}{k}$. Therefore,
$$
\begin{array}{l}
\left|A_{17}\right|=\frac{2006}{17}+1=119, \\
\left|A_{59}\right|=\frac{2006}{59}+1=35, \\
\left|A_{17} \cap A_{59}\right|=\left|A_{1003}\right|=\frac{2006}{17 \times 59}+1=3 . \\
\text { Hence }\left|A_{17} \cup A_{99}\right|=\left|A_{17}\right|+\left|A_{59}\right|-\left|A_{1003}\right| \\
=119+35-3=151 .
\end{array}
$$