# How to Derive a Formula of Arithmetic Series | Arithmetic Progression

An $\textit{arithmetic series}$ is the sum of the terms of an arithmetic sequence.
For example:

• $4, 7, 10, 13, \cdots,61$ is a finite arithmetic sequence.
• $4+7+10+13+ \cdots +61$ is the corresponding arithmetic series.

If the first term is $u_{1}$ and the common difference is $d$, the terms are:
$$u_{1},u_{1}+d,u_{1}+2d,u_{1}+3d,\cdots$$
\begin{align} \displaystyle \require{AMSsymbols} \require{color} u_{1} &= u_{1} \\ u_{2} &= u_{1} + d \\ u_{3} &= u_{1} + 2d \\ &\vdots \\ u_{n-2} &= u_{n}-2d \\ u_{n-1} &= u_{n}-d \\ u_{n} &= u_{n} \\ S_{n} &= u_{1} + (u_{1}+d) + (u_{1}+2d) + \cdots + (u_{n}-2d) + (u_{n}-d) + u_{n} \\ S_{n} &= u_{n} + (u_{n}-d) + (u_{n}-2d) + \cdots + (u_{1}+2d) + (u_{1}+d) + u_{1} \color{red}\text{ reversing} \\ 2S_{n} &= \overbrace{(u_{1}+u_{n})+(u_{1}+u_{n})+(u_{1}+u_{n})+ \cdots + (u_{1}+u_{n})+(u_{1}+u_{n})+(u_{1}+u_{n})}^{n} \color{red}\text{ adding these}\\ 2S_{n} &= n(u_{1}+u_{n}) \\ \therefore S_{n} &= \dfrac{n}{2}(u_{1}+u_{n}) \\ S_{n} &= \dfrac{n}{2}\Big[u_{1}+u_{1}+(n-1)d\Big] \ \color{red}u_{n}=u_{1}+(n-1)d\\ \therefore S_{n} &= \dfrac{n}{2}\Big[2u_{1}+(n-1)d\Big] \\ \end{align}

If we know the first term $u_{1}$, the common difference $d$ and the number of terms $n$ that we wish to add together, we can calculate the sum directly without adding up all the individual terms.

It is worthwhile also to note that $S_{n+1} = S_{n} + u_{n+1}$. This tells us that the next term in the series $S_{n+1}$ is the present sum, $S_{n}$, plus the next term in the sequence, $u_{n+1}$. This result is useful in spreadsheets where one column gives the sequence, and an adjacent column is used to give the series.

## Example 1

Find the sum of $3+5+7+ \cdots$ to $60$ terms.

\begin{align} \displaystyle u_{1} &= 3 \\ d &= 2 \\ d &= 60 \\ S_{60} &= \dfrac{60}{2}\left[2 \times 3 + (60-1) \times 2\right] \\ &= 3720 \end{align}

## Example 2

Find the sum of $5+8+11+ \cdots +599$.

\begin{align} \displaystyle u_{1} &= 5 \\ d &= 3 \\ u_{n} &= 599 \\ u_{1}+(n-1)d &= 599 \\ 5 + (n-1) \times 3 &= 599 \\ (n-1) \times 3 &= 594 \\ n-1 &= 198 \\ n &= 199 \\ S_{n} &= \dfrac{n}{2}{(u_{1}+u_{n})} \\ &= \dfrac{199}{2}(5+599) \\ &= 60\ 098 \end{align}