1999 AIME Problem 1

If the common difference is not a multiple of $3,$ one of the first three terms is divisible by $3$

The difference must be a multiple of $6,$ so try five-term progressions with difference $6$ starting from small primes

Let the terms be $p,$ $p + d,$ $\ldots,$ $p + 4d.$ The even prime cannot be the first term: an even difference makes the second term even, while an odd difference makes the third term even. If $d$ were odd, consecutive terms would have opposite parity, so some term other than the first would be even and greater than $2$ — impossible. If $d$ were not a multiple of $3,$ then $p,$ $p + d,$ $p + 2d$ would cover all residues mod $3,$ so some term would be divisible by $3;$ that term would have to be $3$ itself, forcing $p = 3,$ but then $p + 3d = 3(1 + d)$ is composite. Hence $d$ is divisible by $6.$

With $d \ge 6$ the fifth term is at least $p + 24.$ Trying $p = 5$ and $d = 6$ gives $5, 11, 17, 23, 29,$ all prime, and no smaller fifth term is possible since $p \ge 5$ (the starts $p = 2$ and $p = 3$ fail as above). The answer is $29.$