The sum of the first \( n \) terms of an arithmetic progression is given by:
\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]
Here, \( a = 2 \), \( d = 4 - 2 = 2 \), and \( n = 50 \):
\[
S_{50} = \frac{50}{2} (2(2) + (50-1) \cdot 2)
\]
\[
= 25 (4 + 98) = 25 \times 102 = 2550
\]