I'm working on the following additive combinatorics problem and I'm not sure if I understand the question correctly and how to approach it:
Let $A\subset \mathbb{Z}_p$ be a $k$-term arithmetic progression if there are residue classes $a,d \in \mathbb{Z}_p$ such that $A = \{a + ld : 0 \leq l < k\}$. Prove that if $A \subset \mathbb{Z}_p$ is an arithmetic progression then so is $\mathbb{Z}_p \setminus A$.
So are we supposed to consider the terms in the sequence to "wrap" around $\mathbb{Z}_p$? And when we take out $A$ from $\mathbb{Z}_p$, are we taking out all the elements modulo $p$?
Let $A$ be a $k$-term arithmetic progression in $Z_p$, where $1 \le k < p$, and let $A'=Z_p\setminus A$.
If $d=0$, then $A=\{a\}$ is a $1$-term arithmetic progression.
In this case, we can represent $A'$ as the $(p-1)$-term arithmetic progression defined by
$$A' = \{a' + ld' \mid 0 \le l < p-1\}$$
where $a'=a+1$, and $d'=1$.
In other words, just start from $a+1$ and progress $1$ unit at a time, wrapping around when necessary, stopping just before cycling back to $a$.
If $d \ne 0$, then $A = \{a + ld \mid 0 \leq l < k\}$ is a $k$-term arithmetic progression, for some $k$ with $1 \le k < p$.
In this case, we can represent $A'$ as the $(p-k)$-term arithmetic progression defined by
$$A' = \{a' + ld' \mid 0 \le l < p-k\}$$
where $a'=a+kd$, and $d'=d$.
In other words, $A'$ just continues the arithmetic progression $A$ from where it left off, wrapping around when necessary. Since $d \ne 0$, it follows that $\gcd(d,p) = 1$ (in $\mathbb{Z}$), hence the multiples of $d$ in $Z_p$ yield all elements of $Z_p$, and so the same is true if we shift the multiples of $d$ by $a$. Thus, the elements of $\{a + ld \mid 0 \le l < p\}$ are distinct, hence, $A'$ doesn't have to worry about either cycling back to itself or running into $A$.