Suppose you want to construct an arithmetic sequence of six primes and the difference is not a multiple of $5$. You start with, let us say, $7$ and proceed with a difference of $12$ to generate $7, 19, 31, 43, 55, 67.$ Uh-oh, you got a multiple of $5$.
That's because when the difference between successive terms is not a multiple of $5$, the sequence modulo $5$ will inevitably cycle through all residues modulo $5$ and thus you inevitably hit zero modulo $5.$
You could avoid this by using $5$ itself as the multiple of $5$, but then $5$ has to begin the sequence and you find the sixth term is a multiple of $5$ again, this time composite. With a difference of $12$ like before, you get $5, 17, 29, 41, 53, 65.$ You started with $5$ and then made five increments with a common difference of $12$, so could not avoid another multiple of $5.$
You get similar problems with differences that are not multiples of $2$ or $3.$ To get a sequence of primes as long as six terms, the difference must be divisible by all of $2, 3, 5$ and thus divisible by $5$# $=30.$ The sequence $7, 37, 67, 97, 127, 157$ makes it with a difference of exactly $30.$
Now try this logic with a sequence that's $24$ primes long and infer the difference between successively terms has to have all primes factors up to and including $23.$
Best Answer
As quasi points out in a comment, when $\delta=a$ the element is $a(b+1)$ which is composite unless possibly $a=1$.
More generally just take $\delta = kb+k+a$ for any $k$ whatever. Then the element is $$kb^2 +(k+a)b + a = (kb+a)(b+1)$$ which is composite.