This page covers arithmetic series. The infinite series page is a general introduction to series, and arithmetic series are related to the other kinds of infinite series linked here.

Arithmetic and geometric series are some of the simplest to understand. You should study them first. Power series and Taylor series are best understood after studying differential calculus. p-series are related to geometric series, and Fourier series are infinite series based on trigonometric functions.

Consider the infinite series below, written as a partial sum of terms and abbreviated in summation notation.

$$s = \frac{1}{4} + \frac{2}{4} + \frac{3}{4} + \frac{4}{4} + \frac{5}{4} + \dots = \sum_{n = 0}^{\infty} \; \frac{n}{4}$$

The series is an arithmetic **constant difference** between successive terms. The constant difference is the hallmark of the arithmetic series. Note that this series could start just as well with n = 1, so there is some flexibility in this kind of notation. The notation isn't the series, it just captures what the series is in a kind of shorthand.

The series **s**, above, **diverges**. It doesn't reach a some finite limit as we add more terms; the value of **s** grows and grows. In fact **all arithmetic series diverge**, except the trivial series with all terms equal to zero. You might think we could construct an arithmetic series with a negative constant difference to get it to converge, but that would just lead to a sum that grew infinitely in the negative direction. Nevertheless, arithmetic series, partiuclarly their **partial sums** are useful in certain circumstances.

There is a story about Carl Freidrich Gauss, a very important mathematician/physicist, that goes like this: Gauss' teacher gave the class a "busy-work" assignment - to sum up the numbers between 1 and 100, inclusive. The teacher thought that would take long enough for him to get a bit of rest, but Gauss thought about it for a few seconds and gave the answer: "5050." When the teacher asked how he'd done it, Gauss explained:

- Take the first and last numbers, 1 and 100. They sum to 101.
- Now take the second and second-to-last, 2 + 49. They also sum to 101.
- And we have 3 + 48 = 101, 4 + 47 = 101, and so on, right down to 50 + 51 = 101.

Gauss simply multiplied 101 by 50 to get the sum, 5050. You can try it with any range, as long as there's a **constant difference** between terms.

*[Note: as with all stories of this kind, there is some doubt about whether it really happened ... still.. it's a good story. ]*

It turns out that this is mathematically the same as finding the **average** of the first and last terms of the series, (1 + 100)/2 = 50.5, and multiplying this by the number of terms to be summed: 50.5 x 100 = 5050.

Carl Friedrich Gauss, Image: Wikipedia commons

Sometimes it's useful to find the **sum** of the first * m* terms of an arithmetic series. To find that sum, we just find the average of the terms across the range, then multiply by the number of terms. That leads us to a nice formula:

$$\sum_{n = 1}^m \; x_n = \frac{m}{2} (x_1 + x_m)$$

m is the number of terms.

$$\text{Sum of first m terms} = \sum_{n=1}^m x_n = \frac{m}{2}(x_1 + x_m)$$

$m$ is the number of terms.

$\frac{x_1 + x_m}{2}$ is the average of the first and last terms.

Before we go any further with series, let's review some **algebra** that might make some series operations easier. The relationships in the box below follow from the distributive property of multiplication and commutative property of addtion, respectively, and they'll help you to simplify the series with which you work.

By the way, the plural of series is series. Just one of those things.

$$\sum_{n=1}^m \color{magenta}{a} x_n = \color{magenta}{a} x_1 + \color{magenta}{a} x_2 + \color{magenta}{a} x_3 + \dots = \color{magenta}{a} \cdot (x_1 + x_2 + x_3 + \dots) = \color{magenta}{a} \sum_{n=1}^m x_n$$

$$ \begin{align} \sum_{n=1}^m x_n + \color{magenta}{y_n} &= x_1 + \color{magenta}{y_1} + x_2 + \color{magenta}{y_2} + x_3 + \color{magenta}{y_n} + \dots \\[5pt] &= x_1 + x_2 + y_2 + \dots + \color{magenta}{y_1} + \color{magenta}{y_2} + \color{magenta}{y_4} + \dots \\[5pt] \sum_{n=1}^m x_n + \color{magenta}{y_n} &= \sum_{n=1}^m x_n + \color{magenta}{\sum_{n=1}^m y_n} \end{align}$$

Determine the constant difference of these arithmetic series and calculate the partial sum:

1. |
$$\sum_{n = 1}^{25} \; 2n$$ ## Solution$$ \begin{align} \sum_{n = 1}^{25} \; 2n &= 2(1) + 2(2) + \dots + 2(25) \\[5pt] &= 2 + 4 + 6 + 4 + \dots + 50 \\[5pt] &= \frac{25}{2} \left( 2 + 50 \right) \\[5pt] &= \frac{25}{2} (52) = 25(26) = \bf 650 \end{align}$$ The constant difference is 2. |

2. |
$$\sum_{n = 1}^{10} \; \frac{n}{4} - n$$ ## Solution$$ \begin{align} \sum_{n = 1}^{10} \; \frac{n}{4} - n &= \left( \frac{1}{4} - 1 \right) + \left( \frac{2}{4} - 2 \right) + \left( \frac{3}{4} - 3 \right) \\[5pt] &+ \dots + \left( \frac{10}{4} - 10 \right) \\[5pt] &= -\frac{3}{4} - \frac{6}{4} - \frac{9}{4} - \frac{12}{4} - \dots - \frac{30}{4} \\[5pt] &= \frac{10}{2} \left( -\frac{3}{4} - \frac{30}{4} \right) \\[5pt] &= 5 \left( \frac{-33}{4} \right) = \bf \frac{-165}{4} \end{align}$$ The constant difference is $\frac{-3}{4}$. |

3. |
$$\sum_{n = 1}^{21} \; n + 100$$ ## Solution$$ \begin{align} \sum_{n = 1}^{21} \; n + 100 &= 101 + 102 + 103 + \dots + 121 \\[5pt] &= \frac{21}{2} - \left( 101 + 121 \right) = \frac{21}{2}(222) \\[5pt] &= 21(111) = \bf 2,331 \end{align}$$ The constant difference is 1. |

4. |
$$\sum_{n = 1}^{50} \; \frac{n}{6}$$ ## Solution$$ \begin{align} \sum_{n = 1}^{50} \; \frac{n}{6} &= \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \dots + \frac{50}{6} \\[5pt] &= \frac{50}{2} - \left( \frac{1}{6} + \frac{50}{6} \right) \\[5pt] &= \frac{50(51)}{6} = \bf 425 \end{align}$$ The constant difference is $\frac{1}{6}.$ |

5. |
$$\sum_{n = 1}^{100} \; \frac{n + 1}{2}$$ ## Solution$$ \begin{align} \sum_{n = 1}^{50} \; \frac{n + 1}{26} &= \frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \dots + \frac{101}{2} \\[5pt] &= \frac{100}{2} - \left( 1 + \frac{101}{2} \right) \\[5pt] &= \frac{100(103)}{2} \\[5pt] &= 50(103) = \bf 47 \end{align}$$ The constant difference is $\frac{1}{2}.$ |

6. |
$$\sum_{n = 0}^{50} \; \frac{2 - n}{2}$$ ## Solution$$ \begin{align} \sum_{n = 0}^{50} \; \frac{2 - n}{2} &= \frac{2-0}{2} + \frac{2-1}{2} + \frac{2-2}{2} + \frac{2-3}{2} \\[5pt] &+ \dots + \frac{2 - 50}{2} \\[5pt] &= 1 + \frac{1}{2} + 0 - \frac{1}{2} - \frac{3}{2} - \dots - \frac{48}{2} \\[5pt] &= \frac{51}{2} \left( 1 - \frac{48}{2} \right) \\[5pt] &= \frac{51}{2} \left( \frac{2 - 48}{2} \right) \\[5pt] &= \frac{51(-46)}{4} = \bf 586.5 \end{align}$$ The constant difference is $\frac{1}{2}.$ |

7. |
A falling object falls 4.9 m in the first second of freefall, another 14.7 m in the second, another 24.5 m in the third second and 34.3 in the fourth second. If this trend continues, use what you know about arithmetic series to calculate how far the object will fall (a) in total after 10 s, and (b) during the 10 ## Solution
The constant difference is $9.8$. $$ \begin{align} d_{10} &= d_1 + (n-1) d \tag{1} \\[5pt] &= 4.9 + 9(9.8) \\[5pt] &= 93.1, \end{align}$$ where (1) is the formula for the $n^{th}$ term of an arithmetic series. Then we have: $$s_{10} = \frac{10}{2} (4.9 + 93.1) = \bf 490$$ |
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8. |
An arithmetic series begins with -10 and has a common difference of 1.5. Find the 50 ## Solution$$a_n = a_1 + (n-1) d$$ $$ \begin{align} a_{50} &= -10 + (50-1) 1.5 \\[5pt] &= -10 + 73.5 \\[5pt] &= \bf 63.5 \end{align}$$ $\longleftarrow$ This is the formula for the $n^{th}$ term of an arithmetic series. d is the common difference between terms. |

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