The sum of the first 16 terms of the A.P.: 10, 6, 2, ... is

Given the arithmetic progression (A.P.): $10,6,2,\dots$
* First term ($a$) = $10$
* Common difference ($d$) = $6-10=-4$
* Number of terms ($n$) = $16$

The formula for the sum of the first $n$ terms ($S_n$) of an A.P. is:

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

Substituting the given values into the formula:

$$S_{16}=\frac{16}{2}[2(10)+(16-1)(-4)]$$

$$S_{16}=8[20+15(-4)]$$

$$S_{16}=8[20-60]$$

$$S_{16}=8[-40]=-320$$

Therefore, the sum of the first 16 terms is $-320$.