The sum of first 16 terms of the AP 10, 6, 2, ..., is

(b) - 320

Here, a = 10, d = (6 - 10) = -4 and n = 16

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get:

$S_{16}=\frac{16}{2}[2 \times 10+(16-1) \times(-4)] \quad[\because a=10, d=-4$ and $n=16]$

$=8 \times[20-60]=8 \times(-40)=-320$

Hence, the sum of the first 16 terms of the given AP is -320.