By Terence Tao's result, there exist increasing arithmetic progressions made of prime numbers, as long as wanted. Its terms being prime, are then also pairwise relatively prime. Just joking (although how can you disagree with such a proof?) Take $a = n! + 1$ and $r=n!$. Consider the arithmetic progression $\{a + kr \ ; \ k=1,2,\ldots,n\}$. Assume some prime $p$ divides both $a+ir$ and $a+jr$ for some $1\leq i < j \leq n$. Then $p \mid (a+jr) - (a+ir) = (j-i)r$. If $p \mid j-i < n$, then also $p \mid n! = r$, so in all cases $p \mid r$. But then also $p \mid (a+ir) - ir = a$, impossible since $\gcd(a,r)=1$. Therefore the terms of the progression are pairwise coprime.