Question:
If the first term of an AP is – 5 and the common difference is 2, then the
sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Solution:
$(a)$ Given $\quad a=-5$ and $d=2$
$\therefore$ $S_{B}=\frac{6}{2}[2 a+(6-1) d] \quad\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$
$=3[2(-5)+5(2)]$
$=3(-10+10)=0$