Show this congruence holds

Question

I need to show that if $d|p-1$, $d<p-1$ and $g$ is a primitive root modulo $p$ then, $$\sum_{l=1}^{(p-1)/d} g^{dl}\equiv 0\pmod{p}.$$ I have a gut feeling it has something to do with the fact that $g^{(p-1)/2}=-1$ and that this occurs in the sum for all values of $d$, but I can't figure out how to formalize this thought. Though this may be off the mark, if it is could I be pushed in the right direction?

Answer 1

Multiply the sum by $g^d$ and notice that we have the same summands. We conclude $xg^d\equiv x$, so $x\equiv 0$