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

Question:

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

Solution:

(b) - 320

Here, a = 10, = (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.

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