Find the sum of the first 20 terms of the arithmetic progression 16, 9, 2, (-5), ...

The given arithmetic progression is $16, 9, 2, -5, \dots$
The first term is $a = 16$.
The number of terms is $n = 20$.
The common difference $d$ is found by subtracting any term from its succeeding term:

$$d = 9 - 16 = -7$$

We can verify this with the next terms: $2 - 9 = -7$ and $-5 - 2 = -7$.

The formula for the sum $S_n$ of the first $n$ terms of an arithmetic progression is:

$$S_n = \frac{n}{2} [2a + (n-1)d]$$

Substitute $a=16$, $n=20$, and $d=-7$ into the formula:

$$S_{20} = \frac{20}{2} [2(16) + (20-1)(-7)]$$ $$S_{20} = 10 [32 + (19)(-7)]$$ $$S_{20} = 10 [32 - 133]$$ $$S_{20} = 10 [-101]$$ $$S_{20} = -1010$$

The sum of the first 20 terms is $\boxed{-1010}$.