The first term of an A.P. is 2 and the last term is 50.

Given:

$a=2, l=50, S_{n}=442$

Now, $S_{n}=442$

$\Rightarrow \frac{n}{2}[a+l]=442$

$\Rightarrow \frac{n}{2}[2+50]=442$

$\Rightarrow n=17$

$\therefore a_{17}=50$

$\Rightarrow a+(17-1) \times d=50$

$\Rightarrow 2+16 d=50$

$\Rightarrow d=3$