# 7.8.1: Sums of Finite Arithmetic Series

- Page ID
- 14806

## Finding the Sum of a Finite Arithmetic Series

A theater's seating is arranged such that each row has two more seats than the one in front of it. The first row has five seats and there are 30 rows of seats in the theater. How many total seats are in the theater?

## Sum of a Finite Arithmetic Series

The method of using the calculator to evaluate the sum of a series can be used to find the sum of an arithmetic series as well. However, in this concept we will explore an algebraic method unique to arithmetic series. As we discussed earlier in the unit a series is simply the sum of a sequence so an arithmetic series is a sum of an arithmetic sequence. Let’s look at a problem to illustrate this and develop a formula to find the sum of a finite arithmetic series.

Let's find the sum of the arithmetic series: 1+3+5+7+9+11+…+35+37+39.

Now, while we could just add up all of the terms to get the sum, if we had to sum a large number of terms that would be very time consuming. A famous German mathematician, Johann Carl Friedrich Gauss, used the method described here to determine the sum of the first 100 integers in grade school. First, we can write out all the numbers twice, in ascending and descending order, and observe that the sum of each pair of numbers is the same:

\(\ \begin{array}{cccccccc}

1 & 3 & 5 & 7 & 9 & 11 & \ldots & 35 & 37 & 39 \\

39 & 37 & 35 & 33 & 31 & 29 & \ldots & 5 & 3 & 1 \\

&&&&& {\vdots} \\

40 & 40 & 40 & 40 & 40 & 40 & \ldots & 40 & 40 & 40

\end{array}\)

Notice that the sum of the corresponding terms in reverse order is always equal to 40, which is the sum of the first and last terms in the sequence.

What Gauss realized was that this sum can be multiplied by the number of terms and then divided by two (since we are actually summing the series twice here) to get the sum of the terms in the original sequence. For the problem he was given in school, finding the sum of the first 100 integers, he was able to just use the first term, \(\ a_{1}=1\), the last term, \(\ a_{n}=100\), and the total number of terms, \(\ n=100\), in the following formula:

\(\ \frac{n\left(a_{1}+a_{n}\right)}{2}=\frac{100(1+100)}{2}=5050\)

In our problem we know the first and last terms, but how many terms are there? We need to find \(\ n\) to use the formula to find the sum of the series. We can use the first and last terms and the \(\ n^{t h}\) term to do this.

\(\ \begin{array}{l}

a_{n}=a_{1}+d(n-1) \\

39=1+2(n-1) \\

38=2(n-1) \\

19=n-1 \\

20=n

\end{array}\)

Now the sum is \(\ \frac{20(1+39)}{2}=400\)

## Proof of the Arithmetic Sum Formula

The rule for finding the \(\ n^{t h}\) term of an arithmetic sequence and properties of summations can be used to prove the formula algebraically. First, we will start with the \(\ n^{t h}\) term rule \(\ a_{n}=a_{1}+(n-1) d\). We need to find the sum of numerous \(\ n^{t h}\) terms (\(\ n\) of them to be exact) so we will use the index, \(\ i\), in a summation as shown below:

\(\ \sum_{i=1}^{n}\left[a_{1}+(i-1) d\right]\) Keep in mind that \(\ a_{1}\) and \(\ d\) are constants in this expression.

We can separate this into two separate summations as shown: \(\ \sum_{i=1}^{n} a_{1}+\sum_{i=1}^{n}(i-1) d\)

Expanding the first summation, \(\ \sum_{i=1}^{n} a_{1}=a_{1}+a_{1}+a_{1}+\ldots+a_{1}\) such that \(\ a_{1}\) is added to itself \(\ n\) times. We can simplify this expression to \(\ a_{1} n\).

In the second summation, \(\ d\) can be brought out in front of the summation and the difference inside can be split up as we did with the addition to get: \(\ d\left[\sum_{i=1}^{n} i-\sum_{i=1}^{n} 1\right]\). Using rules you have seen before, \(\ \sum_{i=1}^{n} i=\frac{1}{2} n(n+1)\) and \(\ \sum_{i=1}^{n} 1=n\). Putting it all together, we can write an expression without any summation symbols and simplify.

\(\ \begin{array}{left}a_{1} n+d\left[\frac{1}{2} n(n+1)-n\right]\\

=a_{1} n+\frac{1}{2} d n(n+1)-d n \quad &\text{Distribute } d\\

=\frac{1}{2} n\left[2 a_{1}+d(n+1)-2 d\right] &\text{Factor out } \frac{1}{2} n\\

=\frac{1}{2} n\left[2 a_{1}+d n+d-2 d\right]\\

=\frac{1}{2} n\left[2 a_{1}+d n-d\right]\\

=\frac{1}{2} n\left[2 a_{1}+d(n-1)\right] \quad &\leftarrow \text{This version of the equationis very useful if you don't know the } n^{t h}\text{ term.}\\

=\frac{1}{2} n\left[a_{1}+\left(a_{1}+d(n-1)\right)\right]\\

=\frac{1}{2} n\left(a_{1}+a_{n}\right)

\end{array}\)

Now, let's find the sum of the first 40 terms in the arithmetic series \(\ 35+31+27+23+\) …

For this particular series we know the first term and the common difference, so let’s use the rule that doesn't require the \(\ n^{t h}\) term: \(\ \frac{1}{2} n\left[2 a_{1}+d(n-1)\right]\), where \(\ n=40\), \(\ d=-4\) and \(\ a_{1}=35\).

\(\ \frac{1}{2}(40)[2(35)+(-4)(40-1)]=20[70-156]=-1720\)

We could also find the \(\ n^{t h}\) term and use the rule \(\ \frac{1}{2} n\left(a_{1}+a_{n}\right)\), where \(\ a_{n}=a_{1}+d(n-1)\).

\(\ a_{40}=35+(-4)(40-1)=35-156=-121\), so the sum is

\(\ \frac{1}{2}(40)(35-121)=20(-86)=-1720\)

Next, given that in an arithmetic series \(\ a_{21}=165\) and \(\ a_{35}=277\), let's find the sum of terms 21 to 35.

This time we have the “first” and “last” terms of the series, but not the number of terms or the common difference. Since our series starts with the \(\ 21^{s t}\) term and ends with the \(\ 35^{t h}\) term, there are 15 terms in this series. Now we can use the rule to find the sum as shown.

\(\ \frac{1}{2}(15)(165+277)=3315\)

Finally, let's find the sum of the arithmetic series \(\ \sum_{i=1}^{8}(12-3 i)\).

From the summation notation, we know that we need to sum 8 terms. We can use the expression \(\ 12-3 i\) to find the first and last terms as and the use the rule to find the sum.

First term: \(\ 12-3(1)=9\)

Last term: \(\ 12-3(8)=-12\)

\(\ \sum_{i=1}^{8}(12-3 i)=\frac{1}{2}(8)(9-12)=4(-3)=-12\)

We could use the calculator in this problem as well: *sum*(*seq*(12−3x,x,1,8))=−12

## Examples

Earlier, you were asked to find the total number of seats in the theater.

**Solution**

For this particular series we know the first term and the common difference, so let’s use the rule that doesn't require the \(\ n^{t h}\) term: \(\ \frac{1}{2} n\left[2 a_{1}+d(n-1)\right]\), where \(\ n=30\), \(\ d=2\) and \(\ a_{1}=5\).

\(\ \frac{1}{2}(30)[2(5)+(2)(30-1)]=15[10+58]=1020\)

Therefore, there are a total of 1020 seats in the theater.

Find the sum of the series \(\ 87+79+71+63+\ldots+-105\).

**Solution**

\(\ d=8\), so

\(\ \begin{aligned}

-105 &=87+(-8)(n-1) \\

-192 &=-8 n+8 \\

-200 &=-8 n \\

n &=25

\end{aligned}\)

and then use the rule to find the sum is \(\ \frac{1}{2}(25)(87-105)=-225\)

Find \(\ \sum_{i=10}^{50}(3 i-90)\).

**Solution**

\(\ 10^{t h}\) term is \(\ 3(10)-90=-60\), \(\ 50^{t h}\) term is \(\ 3(50)-90=60\) and \(\ n=50-10+1=41\)

(add 1 to * include* the \(\ 10^{t h}\) term). The sum of the series is \(\ \frac{1}{2}(41)(-60+60)=0\). Note that the calculator is a great option for this problem:

*sum*(

*seq*(3x−90,x,10,50))=0

Find the sum of the first 30 terms in the series \(\ 1+6+11+16+\ldots\)

**Solution**

\(\ d=5\), use the sum formula, \(\ \frac{1}{2} n\left(2 a_{1}+d(n-1)\right)\), to get

\(\ \frac{1}{2}(30)[2(1)+5(30-1)]=15[2+145]=2205\)

## Review

Find the sums of the following arithmetic series.

- \(\ -6+-1+4+\ldots+119\)
- \(\ 72+60+48+\ldots+-84\)
- \(\ 3+5+7+\ldots+99\)
- \(\ 25+21+17+\ldots+-23\)
- Find the sum of the first 25 terms of the series \(\ 215+200+185+\ldots\)
- Find the sum of the first 14 terms in the series \(\ 3+12+21+\ldots\)
- Find the sum of the first 32 terms in the series \(\ -70+-65+-60+\ldots\)
- Find the sum of the first 200 terms in \(\ -50+-49+-48+\ldots\)

Evaluate the following summations.

- \(\ \sum_{i=4}^{10}(5 i-22)\)
- \(\ \sum_{i=2}^{25}(-3 i+37)\)
- \(\ \sum_{i=11}^{48}(i-20)\)
- \(\ \sum_{i=5}^{40}(50-2 i)\)

Find the sum of the series bounded by the terms given. Include these terms in the sum.

- \(\ a_{7}=39\) and \(\ a_{23}=103\)
- \(\ a_{8}=1\) and \(\ a_{30}=-43\)
- \(\ a_{4}=-15\) and \(\ a_{17}=24\)
- How many cans are needed to make a triangular arrangement of cans if the bottom row has 35 cans and successive row has one less can than the row below it?
- Thomas gets a weekly allowance. The first week it is one dollar, the second week it is two dollars, the third week it is three dollars and so on. If Thomas puts all of his allowance in the bank, how much will he have at the end of one year?

## Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.7.