Finite Arithmetic Series

An arithmetic series is the sum of all the specified terms in an arithmetic sequence. These series may be finite or infinite. All arithmetic series may be defined in the form $\sum\limits_{n=1}^{k}i$ (sigma notation), where $k$ is a specified term in the arithmetic sequence $a_1+a_2+a_3+...+a_k$ (or $k$ can be $\infty$, in which case the sum is $\infty$ as well) and $i$ represents the linear equation used to derive the arithmetic sequence (ex. $a_n=3n+5$).

Now, to calculate an arithmetic series, we use the formula $\sum\limits_{n=1}^{k}i=\frac{k}{2}(a_1+a_k)$.

---

ex. $\sum\limits_{n=1}^{5}3n-1$

The "long way":

$a_1=3(1)-1=2$

$a_2=3(2)-1=5$

$a_3=3(3)-1=8$

$a_4=3(4)-1=11$

$a_5=3(5)-1=14$

$\sum\limits_{n=1}^{5}3n-1=2+5+8+11+14=40$

Using the formula:

$\sum\limits_{n=1}^{5}3n-1=\frac{5}{2}(2+14)=40$

Proof:

Arithmetic series $\sum\limits_{n=1}^{k}i$ can be expressed as $$a_1+a_2+a_3+...+a_k\\=a_1+(a_1+d)+(a_1+2d)+...+(a_1+(k-1)d)$$ where $d$ is the common difference.

Case 1: Even number of terms, or $k|2$.

Grouping the first and last, second and second to last, etc. we get

$$(a_1+(a_1+(k-1)d))+((a_1+d)+(a_1+(k-2)d))+...\\...+((a_1+(\frac{k}{2}-1)d)+(a_1+(\frac{k}{2})d))\\ =(2a_1+kd-d)+(2a_1+kd-d)+...+(2a_1+kd-d)$$

regrouping the terms, we get

$$(a_1+(a_1+(k-1)d))+(a_1+(a_1+(k-1)d))+...\\...+(a_1+(a_1+(k-1)d))\\ =\frac{k}{2} (a_1+(a_1+(k-1)d))\\ =\sum_{n=1}^{k}i=\frac{k}{2}(a_1+a_k)$$

Case 2: Odd number of terms, or $(k+1)|2$.

Grouping the first and last, second and second to last, etc. we get the same as with an even number of terms, but with the middle term alone:

$$(a_1+(a_1+(k-1)d))+((a_1+d)+(a_1+(k-2)d))+...\\...+(a_1+(\frac{k+1}{2})d)\\ =(2a_1+kd-d)+(2a_1+kd-d)+...+(a_1+(\frac{k+1}{2})d)\\ =(2a_1+kd-d)+(2a_1+kd-d)+...+\frac{1}{2}(2a_1+kd-d)$$

Regrouping the terms, we get

$$(a_1+(a_1+(k-1)d))+(a_1+(a_1+(k-1)d))+...\\...+\frac{1}{2}(a_1+(a_1+(k-1)d))\\ =\frac{k}{2}(a_1+(a_1+(k-1)d))\\ =\sum_{n=1}^{k}i=\frac{k}{2}(a_1+a_k)$$

Take your favorite arithmetic sequence, such as $10, 15, 20, 25, 30, 35, 40$.

$$10+15+20+25+30+35+40$$

Consider what we can do faster: find its sum or add up all the numbers.

(Such a sum is called an arithmetic series.)

Rather than go for the sum directly, let's ask a different question first: find the average of the arithmetic sequence $10, 15, 20, 25, 30, 35, 40$

When there's an odd number of terms, the average is just the middle term of the arithmetic sequence (this happens because the other terms are evenly spaced around it). When there's an even number of terms, there's no middle term. So we go halfway between the two middle terms.

(You may have seen this called the median of a collection of numbers. In general, the median and average are different; for an arithmetic sequence, they are the same.)

The formula $\frac{a_1+a_n}{2}$, where $a_1$ and $a_n$ are the first and last terms, always gives us either the middle term or the middle point between the two. Think about why this works.

In general, the average of a sequence $a_1, a_2, \dots, a_n$ is given by $$\frac{a_1 + a_2 + \dots + a_n}{n}.$$ We know how to compute the average of an arithmetic series, so how can we use that to find the sum?

Since the average is the sum divided by $n$ (the number of terms), we can get the sum by multiplying the average by $n$. We can write this down as a formula:

If $a_1, a_2, \dots, a_n$ is an arithmetic sequence, then $$a_1 + a_2 + \dots + a_n = n \cdot \frac{a_1 + a_n}{2}.$$ As a special case, we have the formula for the sum of the first $n$ positive integers (which form an arithmetic sequence): $$1 + 2 + \dots + n = \frac{n(n+1)}{2}.$$

Arithmetic series are defined to be the sums of the terms of arithmetic sequences. For example, in the sequence $2,5,8,11$, the sum of the terms is 26.

Let's take a closer look at the sequence $2,5,8,11,...$.

The sum of the first $1$ terms is $2$.

The sum of the first $2$ terms is $7$.

The sum of the first $3$ terms is $15$.

The sum of the first $4$ terms is $26$.

The sum of the first $5$ terms is $40$.

There's something special about the numbers $2,7,15,26,40$.

Note that $2= (2+2)*\frac{1}{2}$.

Note that $7= (2+5)*\frac{2}{2}$.

Note that $15= (2+8)*\frac{3}{2}$.

Note that $26= (2+11)*\frac{4}{2}$.

Let's think about why this pattern keeps on going. We'll need to take a look at arithmetic sequences. They are in the form $a, a+d, a+2d, ... a+(n-2)d, a+(n-1)d$.

Note that there are $n$ terms. The first and last terms add up to $2a+nd$. The second and second-to-last terms do too! In general, the kth term and kth-from-the-back terms always add up to $2a+nd$.

Then we can evaluate the sum of the sequence $a+d, a+2d, a+3d, ... a+(n-1)d$.

Adding in pairs, we see that it equals $(2a+nd) \frac{n}{2}$.

This formula works for any arithmetic sequence with first term $a$, common difference $d$, and $n$ terms.

Finite Arithmetic Series

As we learned earlier, a series is the sum of all terms in an arithmetic sequence. We've learned that with geometric series. We can easily calculate the series of a finite arithmetic sequence by individually adding up all of the terms, but there is also a straightforward formula we can use.

Arithmetic sequences can be written using recursive sequences. Each term is found by adding the same constant to the previous term. You'll see this topic further when you learn about triangular numbers.

Using this formula, we can find the sum of an arithmetic sequence. In this formula:

• $n$ is the number of terms.

• $a_1$ is the initial term.

• $a_n$ is the $n^{th}$ term.

We can build on this topic with word problems!

(Video) Finding the Sum or an Arithmetic Series

by Brian McLogan

In this video by Brian McLogan you will use summation notation to find the arithmetic series.