Question

The sum of first 16 terms of the AP 10, 6, 2, ..., is
(a) 320
(b) −320
(c) −352
(d) −400

Answer 1

(b) - 320

Here, a = 10, d = (6 - 10) = -4 and n = 16
Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get:
$$\begin{gathered} S_{16}=\frac{16}{2}\left[2\times 10+\left(16-1\right)\times (-4)\right][\because a=10,d=-4\mathrm{and}n=16] \\ =8\times \left[20-60\right]=8\times \left(-40\right)=-320 \end{gathered}$$
Hence, the sum of the first 16 terms of the given AP is -320.