An $\textit{arithmetic series}$ is the sum of the terms of an arithmetic sequence.
For example:
Arithmetic Series
Sum of a Finite Arithmetic Series
If the first term is $u_{1}$ and the common difference is $d$, the terms are:
$$u_{1},u_{1}+d,u_{1}+2d,u_{1}+3d,\cdots$$
\( \begin{align} \displaystyle \require{color}
u_{1} &= u_{1} \\
u_{2} &= u_{1} + d \\
u_{3} &= u_{1} + 2d \\
&\vdots \\
u_{n-2} &= u_{n} - 2d \\
u_{n-1} &= u_{n} - d \\
u_{n} &= u_{n} \\
S_{n} &= u_{1} + (u_{1}+d) + (u_{1}+2d) + \cdots + (u_{n}-2d) + (u_{n}-d) + u_{n} \\
S_{n} &= u_{n} + (u_{n}-d) + (u_{n}-2d) + \cdots + (u_{1}+2d) + (u_{1}+d) + u_{1} \color{red}\text{ reversing} \\
2S_{n} &= \overbrace{(u_{1}+u_{n})+(u_{1}+u_{n})+(u_{1}+u_{n})+ \cdots + (u_{1}+u_{n})+(u_{1}+u_{n})+(u_{1}+u_{n})}^{n} \color{red}\text{ adding these}\\
2S_{n} &= n(u_{1}+u_{n}) \\
\therefore S_{n} &= \dfrac{n}{2}(u_{1}+u_{n}) \\
S_{n} &= \dfrac{n}{2}[u_{1}+u_{1}+(n-1)d] \ \color{red}u_{n}=u_{1}+(n-1)d\\
\therefore S_{n} &= \dfrac{n}{2}[2u_{1}+(n-1)d] \\
\end{align} \)
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 having to add 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.
Example 2
Find the sum of $5+8+11+ \cdots +599$.
internet intelligent tutors