Question:
The sum of first 16 terms of the AP 10, 6, 2, … is
(a)-320
(b) 320
(c)-352
(d)-400
Solution:
(a) Given, AP is 10, 6, 2,…
$\therefore$ $S_{16}=\frac{16}{2}[2 a+(16-1) d]$ $\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$
$=8[2 \times 10+15(-4)]$
$=8(20-60)=8(-40)=-320$