Sum of n Terms of an AP - Practice Questions & MCQ

The sum, $\mathrm{S}_{\mathrm{n}}$ of n terms of an AP with the first term ' a ' and common difference ' d ' is given by

$
S_n=\frac{n}{2}[2 a+(n-1) d]
$
OR

$
S_n=\frac{n}{2}[a+l]
$

$a \rightarrow$ first term
$d \rightarrow$ common difference
$n \rightarrow$ number of terms

$
S_n=a+(a+d)+(a+2 d)+\ldots . .+(a+(n-2) d)+(a+(n-1) d) \ldots \ldots(i)
$
Rewrite

$
\begin{aligned}
& S_n=(a+(n-1) d)+(a+(n-2) d)+\ldots \ldots \ldots+(a-2 d)+(a-d)+a \ldots \ldots \\
& (i)+(i i) \\
& 2 S_n=(2 a+(n-1) d)+(2 a+(n-1) d+\ldots \ldots \ldots \ldots \ldots \text { upto } n \text { term } \\
& S_n=\frac{n}{2}[(2 a+(n-1) d)]
\end{aligned}
$
Also, $a+(n-1) d=l=$ last term
we can also write

$
S_n=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}[a+l]
$