problem solving - Calculate the probability with a finite arithmetic progression

We have a finite arithmetic progression $ a_n $, where $ n \geq 3 $ and its $r\neq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.

My proposition :

$ \Omega={n!\over (n-3)!}$

$ \mathbf{A}= {n \choose 3} \cdot 2$

But I think that my way of thinking about way of counting $ \mathbf{A}$ is incorrect.

Any suggestions how can I count it? Thanks in advice, Kuba!

Answer

$\Omega = n(n-1)(n-2)$

We're interested in triples whose elements differ by $r$, $2r$ ... up to $\lceil n/3\rceil r$
Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.
My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2\lceil n/3\rceil)$ triples whose elements differ by $\lceil n/3\rceil r$. Then:
$P(A) = \frac{2(n-2+n-4...+n-2 \lceil n/3\rceil)}{n(n-1)(n-2)}$