It is known that in the sequence of primes there exists arithmetic progressions of primes of arbitrary length. This was proved by Ben Green and Terence Tao in 2006. However the proof given is a nonconstructive one.
I know the following theorem from Burton gives some criteria on how large the common difference must be.
Let $n > 2$. If all the terms of the arithmetic progression
$$
p, p+d, \ldots, p+(n-1)d
$$
are prime numbers then the common difference $d$ is divisible by every prime $q <n$.
So for instance if you want a sequence of primes in arithmetic progression of length $5$ ie
$$
p, p+d, \ldots, p+4d
$$
you need that $d \geq 6$. Using this we can get that the prime $p=5$ and $d = 6$ will result in a sequence primes in arithmetic progression of length $5$.
So my question is what are the known techniques for constructing a sequence of primes of length $k$? How would one find the "first" prime in the sequence or even the "largest prime" that would satisfy the sequence (assuming there is one)? Also, while the theorem gives a lower bound for $d$, is it known if it is the sharpest lowest bound there is?
NOTE: This is not my area of research so this question is mostly out of curiosity.
Best Answer
There is no largest prime that would satisfy the sequence, since there are infinitely many progressions of length $k$. It is reasonable to ask for an upper bound for the smallest $k$-term progression in the primes. Green and Tao indeed obtained such a bound. To say it's astronomically large would be a gross understatement; it's mathematically large. :) $$N_k < c2^{2^{2^{2^{2^{2^{2^{100k}}}}}}}$$ There is a nice overview in the dissertation http://www.maths.bris.ac.uk/~matfb/dissertation.pdf , and reference [6] therein answers your question.