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 $5+8+11+ \cdots +599$.
\( \begin{align} \displaystyle
u_{1} &= 5 \\
d &= 3 \\
u_{n} &= 599 \\
u_{1}+(n-1)d &= 599 \\
5 + (n-1) \times 3 &= 599 \\
(n-1) \times 3 &= 594 \\
n-1 &= 198 \\
n &= 199 \\
S_{n} &= \dfrac{n}{2}{(u_{1}+u_{n})} \\
&= \dfrac{199}{2}(5+599) \\
&= 60\ 098
\end{align} \)
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