First notice that for even $d$ the sequence is of the form $a_n = 1 + nd$, i.e. all the terms of the sequence are odd and the sequence is monotonically increasing. Therefore $a_n \neq 1$ for $n > 0$.

Let $d$ be an arbitrary odd number. We can easily show by induction that $a_n < d$ if $a_n$ is odd, and $a_n < 2d$ if $a_n$ is even. Thus the sequence is limited, so it is periodic.

Let $r$ be the smallest index such that $a_r = a_s$ for some $s \neq r$. Assume $r > 0$.
If $a_r \le d$, that means that $a_r$ (and then also $a_s$) is derived from the previous term by dividing by 2, i.e. $a_r = a_{r-1}/2$, $a_s = a_{s-1}/2$, so it follows that $a_{r-1} = a_{s-1}$, and this is in contradiction with minimality of $r$.
If $a_r > d$, from $a_n \le 2d$ we conclude that $a_r$ and $a_s$ are derived from previous terms of the sequence by adding $d$, so again it follows that $a_{r-1} = a_{s-1}$, and again we have a contradiction with the minimality of $r$.

Hence, $r = 0$ and $a_s = a_0 = 1$ for some $s > 0$ for every odd $d$.