The problem is equivalent to the following one: Determine all integers $(a,k)$ such that for all sufficiently large prime numbers $p > N$, then $a + kP$ is also a prime. (Note : $a + k > 0$ and obviously $k \ge 0$. ) Firstly, if $k = 0$, taking $a = P$, any prime number, clearly satisfies the problem. We'll now suppose that $k > 0$. If $k > 1$, this ensures that $a \not= 0$. Take a large prime $Q$ such that $\gcd(ak,Q) = 1$. Therefore, by Dirichlet Theorem, there exists infinitely many prime $P$ such that \[ P \equiv -\frac{a}{k} \ (\text{mod} \ Q) \]This forces $a + kP$ being a prime. Hence, $a + kP = Q$ for all such primes $P$, which is clearly a contradiction. Therefore, we must have $k = 1$. Hence, we have $P + a$ being a prime for all arbitrarily large primes $P > N$. This forces $a = 0$. Therefore, all arithmetic progressions that satisfy are $a_i = P, \ \forall i \in \mathbb{N}$ and $a_i = i, i \in \mathbb{N}$.