Solution Verfication on a problem regarding progressions

Question

Let $(a_k)_{k \geq 1}$ be an AP of common difference $r\neq 0$ composed of pair wise distinct natural numbers. Show that it has infinite number of composite naturals in it.
Proof: Let us take $p=a_k$ be a term in the sequence where $k$ is any natural number such that $p \neq 1$. $$a_k=a_1+(k-1)r\Rightarrow a_k+rp=p+rp=a_1+(k+p-1)r=a_{k+p}=p(1+r)$$
$r$ must be positive since if $r<0$,we take $l=\frac{a_1}{-r}$ then we can find $n\in \mathbb N $ such that $$l<n \Rightarrow a_1<(-r)n \Rightarrow a_1+rn<0 \Rightarrow a_1+r(q-1)<0(q=n+1) \Rightarrow a_q<0$$ but all the terms are natural numbers. We can see clearly now that $a_{k+p}$ is a composite natural as $p>1$ and $r+1>1$. Thus, for every $k$ that we take, we can find a natural number $s>k$ such that $a_s$ is composite.Thus,infinite number of composite naturals are in this sequence.
I am getting a feel that it is incomplete. Can someone verify this proof for me and add necessary corrections?

Answer 1

Looks good.

For readability wise, I personally prefer to have the argument of $r$ must be positive comes first and then we illustrate how to find a composite number.

Note that we can generalize the result to AP of which there are infinitely many positive numbers with $r \ne 0$.