Arithmetic sequence where every term is prime?

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.

I answered similar questions here and here.

"An Introduction To The Theory Of Numbers" by Hardy, Theorem 21, page 18.

THEOREM 21. No polynomial $f(n)$ with integral coefficients, not a constant, can be prime for all $n$, or for all sufficiently large $n$.

In this case $f(\delta)=a+\delta b$. It's a more powerful result, but the proof is very simple, as you will see.