there is already a good thread which discusses some corollaries of the Green-Tao Theorem, here:
Constructing arithmetic progressions
The question I was wondering about is of a similar flavor but isn't mentioned, namely, does there exist a $c \in \mathbb{R}$ such that given any $k \in \mathbb{N}$, can you always find a progression of primes $p$, $p+d$, $p+2d$, $\dots p+(k-1)d$ satisfying $d < cp$?
A (much) weaker algebraic variant, which is really what I'm ultimately thinking about, is the following. It is much more likely to be known and/or easy, because I think there is a fairly well-developed theory for growth in numerical semigroups.
Let $S\langle n_1, \dots, n_k\rangle$ denote the (numerical) sub-semigroup of $\mathbb{N}$ generated by $n_1, \dots, n_k$.
Is it true that for any $k \in \mathbb{N}$, there exist primes $p_1 < p_2 < \cdots < p_k$ such that $p_i \notin S\langle p_1, \dots, p_{i-1}\rangle$ for all $2 \leq i \leq k$?
This is almost certainly true, although I don't think I can prove it rigorously. The idea is to take the $p_i$ to be large but close together. By the PNT the asymptotic density of primes about as large as $N$ is about $\frac{1}{\log N}$. Hence we should be able to find $k$ primes $p_1 < \dots < p_k$ of size around $N$ and within $O(k \log N)$ of each other, so taking $N$ to be suitably large compared to $k$ gives the result.
Edit: Okay, here's a proof, although maybe it's overkill. Thanks to the recent work on prime gaps, we now know that there always exist numbers $H_m$ such that there are infinitely many intervals of length $H_m$ containing at least $m + 1$ primes. Pick a sufficiently large such interval.