If the difference $r$ divides $a_1$, then $a_1 = dr$, $d \in \mathbb{N}$ and $a_n = (d + n - 1)r$, and a multiple of a positive integer $k$ is obtained when $d + n - 1$ is a multiple of $k$.

For the converse, observe first that if $r = 0$, the sequence is not complete. Because $r \neq 0$ and by the assumption there is a multiple of $r$ of the form $a_1 + (n-1)r$, with $n \in \mathbb{N}^*$, we conclude $r \mid a_1$.