Question

What is the formula for the sum of an arithmetic sequence?

Answer 1

Finding the formula for an Arithmetic sequence:

Consider an $AP$ sequence with $n$ terms in which first term $=a$ common difference $=d$ and last term $=l$

Then, $T_n=a+(n-1)d$ $...\left[1\right]$

$l=a+(n-1)d$ [$\because l=n^{th}$ term]

Also, we may write the considered $AP$ sequence as

$a,(a+d),(a+2d)---(l-2d),(l-d),l$.

Let $S_n$ be the sum of the $n$terms of this $AP$ sequence.

$\therefore$ $S_n=a+(a+d)+(a+2d)+---(l-2d)+(l-d)+l$ $...\left[2\right]$

Writing $\left[2\right]$ in reverse order, we get:

$S_n=l+(l-d)+(l-2d)+---(a+2d)+(a+d)+a$. $...\left[3\right]$

Adding the corresponding terms of $\left[2\right]$ and $\left[3\right]$, we get:

$2S_n=(a+l)+(a+l)+(a+l)+---(a+l)+(a+l)+(a+l)$

$\Rightarrow$ $2S_n=(a+l)+(a+l)+---n$ terms

$\Rightarrow$ $2S_n=n(a+l)$

$\Rightarrow$ $S_n=\frac{n}{2}(a+l)$

$\Rightarrow$ $S_n=\frac{n}{2}\left[a+a(n-1)d\right]$ [From $\left(1\right)$]

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

Hence, the sum of a $AP$ sequence is given by $S_n=\frac{n}{2}(a+l)$ and $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.