Alternative Format — Lesson 5: Arithmetic Series

Let's Start Thinking

Arithmetic and Geometric Sequences

Previously we were introduced to arithmetic and geometric sequences. Here's a brief review of each.

Arithmetic sequences

A sequence in which the differences\(\) of consecutive terms is constant.

​​E.g., The sequence \(11,~8,~5,~2,\ldots\)  is arithmetic since it has \(d=-3\), where \(d\) is the common difference.

Recall that the general term of an arithmetic sequence is

 \(t_n=a+(n-1)d\)

where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

Now let's look at geometric sequences.

Geometric sequences

A sequence in which the ratios of consecutive terms is constant.

E.g., The sequence \(3,-6,~12,-24,\ldots\) is geometric since it has \(r=-2\), where \(r\) is the common ratio.

Recall that the general term of a geometric sequence is

 \(t_n=ar^{n-1}\)

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

In this lesson, we will look at ways to determine the sum of the terms of an arithmetic sequence.

Try This

The sixth term of an arithmetic sequence is \(2\). The sum of the first ten terms of the sequence is \(24\).

Determine \(\)\(t_{21}\), the twenty-first term of the sequence.

Deriving the Arithmetic Series Formulas

What Is a Series?

The mathematical idea of a series is closely related to that of a sequence. Recall what is meant by a sequence:

Some examples of sequences are:

• \(1,~1,~2,~3,~5,~8,~13,\ldots\)
• spring, summer, fall, winter, spring, summer, \(\ldots\)
• \(8,~18,~28,~38,~48,\ldots\)

For a numerical sequence, if we replace the idea of a list of numbers with a sum of numbers, then we have a series:

 Example 1

Solution

1. The first six terms of the sequence are \(1,4,9,16,25,36\).

   Therefore, \(S_6=1+4+9+16+25+36\).

2. \(S_8\) is the sum of the first \(8\) terms of the series.

   Thus, \(S_8=1+4+9+16+25+36+49+64\) so that \(S_8=204\).

3. We verify the formula by evaluating the right side and comparing the result to part b). For \(n=8\), the right side is: 

   \(\begin{align*} \dfrac{n(n+1)(2n+1)}{6}&=\dfrac{8(8+1)[2(8)+1]}{6}\\ &= \dfrac{8(9)(17)}{6}\\ &=204 \end{align*}\)

   Thus, according to the formula, \(1^2+2^2+3^2+\cdots+8^2=204\).

   Therefore, we have verified the formula is correct for \(n=8\) since the formula yields the same result as we calculated in part b).

In the remainder of this lesson, we will derive and apply formulas for calculating the sum of an arithmetic series. Some of this work dates back to the late 1700s, around the time of a mathematician named Carl Friedrich Gauss.

Carl Friedrich Gauss

There is a famous story about arithmetic series dating back to the late 1700s. The story is about a boy named Carl Friedrich Gauss and something he did when he was in primary school.

Source: Gauss - johan10/iStock/Getty Images

The name Gauss may be familiar to you, as he grew to become a famous mathematician, so famous that centuries later, tens of thousands of Grades 7 and 8 students write a math contest every year called the Gauss Contest.

Source: The Centre for Education in Mathematics and Computing (2017). Gauss Contest Grade 8. Retrieved from: https://cemc.uwaterloo.ca/contests/past_contests/2017/2017Gauss8Contest.pdf

So what happened to Gauss in primary school? Well, as the story goes, when Gauss was around Grade 2 or 3, his teacher wanted to keep his students busy for a while. So the teacher told the class to add all the integers from \(1\) to \(100\). But according to the story, almost immediately, Gauss had the answer, \(5050\).

Gauss and \(1+2+3+\cdots+100\)

So how did Gauss quickly add all the integers from \(1\) to \(100\)?

\(1+2+3+\cdots+98+99+100=~?\)

Well, he rearranged the terms of the sum into pairs. He decided to pair together \(1\) and \(100\), \(2\) and \(99\), \(3\) and \(98\), and so on until the two middle terms, \(50\) and \(51\), were paired together.

\( 1+2+3+\cdots+98+99+100 = (1+100)+(2+99)+(3+98)+\cdots+(50+51) \)

Notice that every number from \(1\) to \(100\) has been paired together, and this happens because there are an even number of terms in the sum that Gauss had to calculate. If there were an odd number of terms, the middle term would not have a pair, and Gauss would have needed to account for that.

Now by pairing this way, what Gauss recognized was that each pair has a sum of \(101\). Since there are \(50\) pairs, we get

\(\begin{align*} 1+2+3+\cdots+98+99+100 &= (1+100)+(2+99)+(3+98)+\cdots+(50+51) \\ &= 101+101+101+\cdots+101 \\ & = 50(101) \\ &= 5050 \end{align*}\)

Now, Gauss went on to become one of the most important mathematicians of all time, making contributions to the world of mathematics, physics, astronomy, and other mathematical sciences.

We can tweak Gauss's method slightly to avoid the complication introduced when there are an odd number of terms in the sum. This next variation works for any number of terms.

A slightly different approach:

First, we write the sum that is to be calculated and we'll call the sum \(S\).

\( S\; =\;\;\;\;1+\;\;\;2+\;\;\;3+\cdots+\;\;98+\;\;99+100 \tag{1} \)

We again use the idea of pairing numbers together so that each pair has the same sum. But this time, let's do it by writing \(S\) again, but with the terms in reverse order.

\( S\; = 100+\;\;99+\;98+\cdots+\;\;\;\;3+\;\;\;2+\;\;\;\;1 \tag{2} \)

Next, we can add these two equations together. We get Equation \((1)\) \(+\) Equation  \((2)\):

\(\begin{align*} S\; & =\;\;\;\;1+\;\;\;2+\;\;\;3+\cdots+\;\;98+\;\;99+100 \tag{1} \\ S\;& = 100+\;\;99+\;98+\cdots+\;\;\;\;3+\;\;\;2+\;\;\;\;1 \tag{2} \\ \hline \;2S\;&=101+101+101+\cdots+101+101+101 \tag{1)+(2} \end{align*}\)

On the left side, we get \(2S\). And on the right side, each pair of terms has sum \(101\). Since there are \(100\) terms, we get that \(2S=100(101) = 10~100\).

But as the left side says, this is double the sum that we want. That's because we wrote the initial sum twice. So we divide both sides by \(2\). Putting this all together, we get that

\(\begin{align*} S\; & =\;\;\;\;1+\;\;\;2+\;\;\;3+\cdots+\;\;98+\;\;99+100 \tag{1} \\ S\;& = 100+\;\;99+\;98+\cdots+\;\;\;\;3+\;\;\;2+\;\;\;\;1 \tag{2} \\ \hline \;2S\;&=101+101+101+\cdots+101+101+101 \tag{1)+(2}\\ 2S\;&=100(101)\\ 2S\; &= 10~100\\ S\; & = 5050 \end{align*}\)

Gauss Extended to \(1+2+3+\cdots+n\)

We can generalize this to find the sum of not just the first \(100\) positive integers, but the first \(n\) positive integers, that is \(1+2+3+\cdots+n\).

As before we write our sum, and notice that since the last term is \(n\), the second last term is \(n-1\), and the third last term is \(n-2\). Then we write the sum again, this time in reverse order.

\(\begin{align*} S\;&=\;\;\;\;\;\;\;\;\;\;1+\;\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;\;\;\;3+\cdots+(n-2)+(n-1)+\;\;\;\;\;\;\; n \tag{1}\\ S\;&\;= \;\;\;\;\;\;\;\;\;n+(n-1)+(n-2)+\cdots+\;\;\;\;\;\;\;\;\;\;3+\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;1 \tag{2} \end{align*}\)

We add these two equations. We get Equation \((1)\) \(+\) Equation  \((2)\):

\(\begin{align*} S\;&=\;\;\;\;\;\;\;\;\;\;1+\;\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;\;\;\;3+\cdots+(n-2)+(n-1)+\;\;\;\;\;\;\; n \tag{1}\\ S\;&\;= \;\;\;\;\;\;\;\;\;n+(n-1)+(n-2)+\cdots+\;\;\;\;\;\;\;\;\;\;3+\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;1 \tag{2} \\ \hline \;2S\;&=\;\;(n+1)+(n+1)+(n+1)+\cdots+(n+1)+(n+1)+(n+1) \tag{1)+(2} \end{align*}\)

On the left side we get \(2S\). On the right side, we get that the first pair has sum \(n+1\), and the second pair also has sum \(n+1\), since \(2+(n-1)=n+1\). Likewise, the third pair is \(n+1\). In fact, every pair has sum \(n+1\).

Now the right side has \(n\) terms, where each term is \(n+1\). So the right side is \(n(n+1)\). We need to divide both sides by \(2\). Putting this all together, we get that

\(\begin{align*} S\;&=\;\;\;\;\;\;\;\;\;\;1+\;\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;\;\;\;3+\cdots+(n-2)+(n-1)+\;\;\;\;\;\;\; n \tag{1}\\ S\;&\;= \;\;\;\;\;\;\;\;\;n+(n-1)+(n-2)+\cdots+\;\;\;\;\;\;\;\;\;\;3+\;\;\;\;\;\;\;\;\;2+\;\;\;\;\;\;1 \tag{2} \\ \hline \;2S\;&=\;\;(n+1)+(n+1)+(n+1)+\cdots+(n+1)+(n+1)+(n+1) \tag{1)+(2}\\ 2S\;&\;=n(n+1)\\ S\; & \;=\dfrac{n(n+1)}2 \end{align*}\)

Thus, we have get the following:

Gauss Extended to Any Arithmetic Series

We can further generalize Gauss's approach to any arithmetic series.

So a general arithmetic sequence is the ordered list

\(a,~a+d,~a+2d,\ldots\)

Since a series is just the sum of the terms of a sequence, this sequence corresponds to this general arithmetic series, where the commas have been replaced by addition signs:

\(a+(a+d) + (a+2d) + \cdots\)

We can also write that the sum of the first \(n\) terms, \(S_n\), is 

\(S_n=a+(a+d) + (a+2d) + \cdots+[a+(n-1)d]\)

Notice that the sum that Gauss needed to calculate is

The sum \(1+2+3+\cdots+100\) is the sum of the arithmetic series with \(a=1\) since the first term is \(1\), \(d=1\) since the common difference is \(1\), and \(n=100\) since there are \(100\) terms in the series.

So let's apply Gauss's modified method to the general arithmetic series to determine a formula for the sum of the first \(n\) terms of an arithmetic series.

We first write the sum of the first \(n\) terms starting at the first term, \(a\). Next, we write the terms in reverse order. 

\(\begin{align*} S_n& \;=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a+\;\;\;\;\;\;\;\;\;\;\;\;(a+d)+\cdots+[a+(n-1)d] \tag{1}\\ S_n& \;=\;\;\;[a+(n-1)d]+\;\;[a+(n-2)d]+\cdots+\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a \tag{2} \end{align*}\)

Take a look for a moment at the second term in the reverse order sequence, \([a+(n-2)d]\). This term has one less \(d\) than the previous term and \((n-2)d\)'s instead of \((n-1)d\).

Then, as before, we add these two equations together. We get Equation \((1)\) \(+\) Equation  \((2)\):

\(\begin{align*} S_n& \;=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a+\;\;\;\;\;\;\;\;\;\;\;\;(a+d)+\cdots+[a+(n-1)d] \tag{1}\\ S_n& \;=\;\;\;[a+(n-1)d]+\;\;[a+(n-2)d]+\cdots+\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a \tag{2} \\ \hline 2S_n&=\;[2a+(n-1)d]+[2a+(n-1)d]+\cdots+[2a+(n-1)d] \tag{1)+(2} \end{align*}\)

The left side is \(2S\), and on the right side, the first term has \(2a\)'s and \((n-1)d\)'s. The second term is going to be the sum of \(a+d\) and \(a + (n-2)d\). So it also has has \(2a\)'s and \((n-1)d\)'s, since \(d+(n-2)d= (n-1)d\).

This ends up being the case for all of the terms. There are \(n\) of these terms. Since each is the same, we have that \(2S_n=n[2a+(n-1)d]\). We can divide both sides by \(2\). Putting this all together, we get

\(\begin{align*} S_n& \;=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a+\;\;\;\;\;\;\;\;\;\;\;\;(a+d)+\cdots+[a+(n-1)d] \tag{1}\\ S_n& \;=\;\;\;[a+(n-1)d]+\;\;[a+(n-2)d]+\cdots+\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a \tag{2} \\ \hline 2S_n&=\;[2a+(n-1)d]+[2a+(n-1)d]+\cdots+[2a+(n-1)d] \tag{1)+(2}\\ 2S_n&\;=n[2a+(n-1)d]\\ S_n &\;= \dfrac{n[2a+(n-1)d]}2 \\ S_n &\;= \dfrac{n}2[2a+(n-1)d] \end{align*}\)

Thus, we have get the following:

Arithmetic Series Summary

An arithmetic series is the sum of the terms of an arithmetic sequence:

\(S=a+(a+d)+(a+2d)+\cdots\)

where \(a\) is the first term and \(d\) is the common difference.

The notation \(S_n\), for \(n\in\mathbb{N}\), can be used to represent the sum of the first \(n\) terms of an arithmetic series: 

\(\begin{align*} S_n&=t_1+t_2+t_3+\cdots+t_n\\ &= a+(a+d)+(a+2d)+\cdots+[a+(n-1)d] \end{align*}\)

For an arithmetic series, \(S_n\) is given by:

\(S_n=\dfrac{n}2[2a+(n-1)d]\)

Example 2

Solution

The series is arithmetic with \(a=50\) and \(d=-9\).

Applying the formula:

\(\begin{align*} S_n &= \dfrac{n}{2}[2a+(n-1)d]\\[5px] S_{20} &= \dfrac{20}{2}[2(50)+(20-1)(-9)]\\[5px] &= 10(100-171)\\[5px] &= -710\\ \end{align*}\)

Therefore, the sum of the first \(20\) terms of the series is \(-710\).

Check Your Understanding 1

Question

Determine the sum of the first \(22\) terms of the arithmetic series \(7+0-7- \ldots\).

Answer

The sum of the first \(22\) terms is \(-1463\).

Feedback

The series is arithmetic with \(a=7\) and \(d=-7\). Therefore, the sum of the first \(22\) terms of the series is:

\(\begin{align*} S_n &= \dfrac{n}{2} [2a+(n-1)d] \\ S_{22} &= \dfrac{22}{2}[2(7) +(22-1)(-7)] \\ &= -1463\end{align*}\)

A Second Formula

An alternative formula for the sum of the first \(n\) terms of an arithmetic series can be found using the fact that the \(n\)^th term is \(t_n=a+(n-1)d\).

Then:

\(\begin{align*} S_n&=\dfrac{n}{2}[2a+(n-1)d]\\ &= \dfrac{n}{2}[a+a+(n-1)d]\\ &= \dfrac{n}{2}(a+[a+(n-1)d])\\ &= \dfrac{n}{2}(t_1+t_n)\\ \end{align*}\)

Alternatively, this can be written as:

\(\begin{align*} S_n &= \dfrac{n}{2}(t_1+t_n)\\ &= \dfrac{n(t_1+t_n)}2\\ &= n\left(\dfrac{t_1+t_n}{2} \right) \end{align*}\)

Notice that the expression \(\dfrac{t_1+t_n}{2}\) is the average of the first and last terms of the arithmetic series.

Applying the Arithmetic Series Formulas

Arithmetic Series

In this part of the lesson, we will solve several types of questions that make use of the formulas for determining the sum of an arithmetic series.

Example 3

Solution — Part A

Since this series is arithmetic, the difference between consecutive terms is constant and denoted by \(d\). 

Thus,

\(\begin{align*} d &= t_2-t_1\\[5px] &= (3m+7)-(m+4)\\[5px] &= 3m+7-m-4\\[5px] &= 2m+3 \end{align*}\)

Or, 

\(\begin{align*} d &= t_3-t_2\\[5px] &= (5m+10)-(3m+7)\\[5px] &= 5m+10-3m-7\\[5px] &= 2m+3 \end{align*}\)

Therefore, we calculate the required sum using \(a=m+4\), \( d=2m+3\) and \(n=8\):

\(\begin{align*} S_n &= \dfrac{n}2[2a+(n-1)d]\\[10px] S_8 &= \dfrac82[2(m+4)+(8-1)(2m+3)]\\[5px] &= 4(2m+8+14m+21)\\[5px] &= 4(16m+29)\\[5px] &= 64m+116 \end{align*}\)

Therefore, the sum of the first eight terms, \(S_8\), of the series \((m+4)+(3m+7)+(5m+10)+\cdots\) is \(64m+116\).

Solution — Part B

Since this series is arithmetic, the difference between consecutive terms is constant and denoted by \(d\). 

Thus,

\(\begin{align*} d &= t_2-t_1\\[5px] &= (8p-q)-(7p-3q)\\[5px] &= 8p-q-7p+3q\\[5px] &= p+2q \end{align*}\)

Or, 

\(\begin{align*} d &= t_3-t_2\\[5px] &= (9p+q)-(8p-q)\\[5px] &= 9p+q-8p+q\\[5px] &= p+2q \end{align*}\)

Therefore, we calculate the required sum using \(a=7p-3q\), \( d=p+2q\) and \(n=8\):

\(\begin{align*} S_n &= \dfrac{n}2[2a+(n-1)d]\\[10px] S_8 &= \dfrac82[2(7p-3q)+(8-1)(p+2q)]\\[5px] &= 4(14p-6q+7p+14q)\\[5px] &= 4(21p+8q)\\[5px] &= 84p+32q \end{align*}\)

Therefore, the sum of the first eight terms, \(S_8\), of the series \((7p-3q)+(8p-q)+(9p+q)+\cdots\) is \(84p+32q\).

Check Your Understanding 2

Question

Determine the sum of the first \(11\) terms of the following arithmetic series:

\((-9q-p) + (-11q-2p) + (-13q-3p)+ \ldots\)

Answer

The sum of the first \(11\) terms is \(-209q-66p\).

Feedback

Since this series is arithmetic, the difference between any two consecutive terms is constant. If \(d\) represents the constant difference, then one way to calculate \(d\) is

\(\begin{align*} d&=t_2-t_1 \\ &= (-11q-2p)-(-9q-p) \\ &= -11q-2p+9q+p \\ &= -2q-p\end{align*}\)

Therefore, this is an arithmetic series with \(a=-9q-p\), \(d=-2q-p\), and \(n=11\), so that the sum is given by

\(\begin{align*} S_n&=\dfrac{n}{2} [2a+(n-1)d] \\ S_{11} &= \dfrac{11}{2} [2(-9q-p) +(11-1)(-2q-p)] \\ &= \dfrac{11}{2} (-18q-2p-20a-10p) \\ &= \dfrac{11}{2} (-38q-12p)\\ &= -209q -66p \end{align*}\)

Therefore, the sum of the first \(11\) terms of this series is \(S_{11}=-209q-66p\).

Let's try another example of using the arithmetic series formulas.

Example 4

Solution

We first note that the given series is arithmetic with \(a=13\), \(d=6\), and \(t_n=307\).

To use either arithmetic series formula, we need to know the term number \(n\) of the last term.

We proceed using two steps:

1. Determine the number of terms, \(n\).
2. Determine \(S_n\).

Step 1: Determine the number of terms, \(n\)

We require the term number of the last term, \(n\), such that \(t_n=307\).

So we can get an equation from this. We start with writing the general term. Then we can substitute \(a=13\) and \(d=6\) into this expression. But we're actually only interested at this point in the case \(t_n=307\). So we substitute \(307\) for \(t_n\). We get

\(\begin{align*} t_n & = a+(n-1)d \\ t_n & = 13+6(n-1) \end{align*}\)

At the same time, the \(6\) has been expanded into the bracket. We now have a single variable linear equation that we solve by isolating. So we find that

\(\begin{align*} t_n & = a+(n-1)d \\ t_n & = 13+6(n-1) \\ 307 & = 13+6n-6 \\ 307 &= 6n+7 \\ 300 & = 6n \\ n &= 50\end{align*}\)

Now we can go onto the second step.

Step 2: Determine \(S_{50}\)

We have two formulas for the sum of an arithmetic series. We can apply either arithmetic series formula. We will show the solution for both formulas.

First, using \(S_n=\dfrac{n}2[2a+(n-1)d]\) with \(a=13\), \( d=6\), and \(n=50\), we write out the formula, substitute our given values, and after some arithmetic we find that

\(\begin{align*} S_n & = \dfrac{n}{2}[2a+(n-1)d] \\ S_{50} & = \dfrac{50}{2}[2(13)+(50-1)(6)] \\ &= (25)(26+294) \\ &=8000\end{align*} \)

Alternatively, using \(S_n=n\left(\dfrac{t_1+t_n}{2}\right)\) with \(t_1=13\),  \(t_n=307\), and \(n=50\), we again write out the formula, substitute the known values, and evalue. We find that 

\(\begin{align*} S_n & = n\left(\dfrac{t_1+t_n}{2}\right) \\ S_{50} & = 50\left(\dfrac{13+307}{2}\right) \\ &= 50(160) \\ &=8000 \end{align*} \)

As expected, we get the same result. Now this time because we knew the last term, the second formula was more efficient. Another advantage of this second formula, \(S_n=n\left(\dfrac{t_1+t_n}{2}\right)\), is that it is easier to remember. For an arithmetic series, you average the first and last terms and then multiply by the number of terms.

Therefore, the sum \(13+19+25+\cdots+307\) is equal to \(8000\).

Example 5

Solution

This question requires that we determine the sum \(15+14+13+\cdots+4\), an arithmetic series with first term \(a=15\) and common difference \(d=-1\).

As in a previous example, we proceed using two steps:

1. Determine \(n\).
2. Determine \(S_n\).

Step 1: Determine \(n\)

We require the term number of the last term.

In other words, we require \(n\) such that \(t_n=4\):

\(\begin{align*} t_n &= a+(n-1)d\\ 4 &= 15-1(n-1)\\ 4 &= 15 - n +1\\ 4 &= 16- n\\ n &= 12 \end{align*}\)

Step 2: Determine \(S_{12}\)

We can apply either arithmetic series formula.

Using \(S_n=\dfrac{n}2[2a+(n-1)d]\) with \(a=15\), \(d=-1\) and \(n=12\):

\(\begin{align*} S_n &= \dfrac{n}{2}[2a+(n-1)d]\\[5px] S_{12}&= \dfrac{12}{2}[2(15)+(12-1)(-1)]\\[5px] &= (6)(30-11)\\[5px] &=114 \end{align*}\)

Using \(S_n=n\left(\dfrac{t_1+t_n}{2}\right)\) with \(t_1=15\), \(t_n=4\) and \(n=12\):

\(\begin{align*} S_n &= n\left(\dfrac{t_1+t_n}{2}\right)\\[5px] S_{50} &= 12\left(\dfrac{15+4}{2}\right)\\[5px] &=114 \end{align*}\)

Therefore, there are \(114\) cans stacked against the wall.

Check Your Understanding 3

Question

Determine the sum of the arithmetic series \(-17-11-5+\ldots+91\).

Answer

The sum of the series is \(703\).

Feedback

Before reading the solution, ensure you followed these two steps:

1. Determine \(n\) using \(t_n=91\).
2. Determine \(S_n\).

Step 1: Determine \(n\)

Apply \(t_n = a + (n-1)d\) with \(t_1=-17\), \(t_n=91\), and \(d=6\): 

\(\begin{align*} t_n &= a+(n-1)d \\ t_n &= -17+6(n-1) \\ 91 &= -17 + 6n - 6 \\ 91 & = -23 + 6n \\ 114 &= 6n \\ n &= 19\end{align*} \)

Step 2: Determine \(S_{19}\)

Complete this question by doing one of the following two things. 

• Apply \( S_n = \dfrac{n}{2} [2a+(n-1)d] \) with \(a=-17\), \(d=6\), and \(n=19\). \[\begin{align*} S_n &= \dfrac{n}{2} [2a+(n-1)d] \\ S_{19} &= \dfrac{19}{2}[2(-17) +(19-1)(6)] \\ &= \dfrac{19}{2}(-34+108)\\ &= 703 \end{align*}\]

OR

• Apply \(S_n = n \left( \dfrac{t_1+t_2}{2} \right)\) with \(t_1=-17\), \(t_n=91\), and \(n=19\).\[\begin{align*} S_n &= n \left( \dfrac{t_1+t_2}{2} \right) \\ &= 19 \left( \dfrac{-17+19}{2} \right) \\ &= 19(37) \\ &=703\end{align*}\]

Using either method, we find that the sum of the series \(-17-11-5+ \ldots + 91\) is equal to \(703\).

Try This Revisited

Let's return to the Try This problem presented at the beginning of this lesson.

If you attempted this question when you began the lesson, you may have had to resort to trial and error. Let's now see an algebraic solution.

Solution

Since the sixth term of the arithmetic sequence is \(2\):

\(\begin{align*} t_n &= a+(n-1)d\\[5px] t_6 &= a+(6-1)d\\[5px] 2 &= a+5d \tag{1} \end{align*}\)

Since the sum of the first ten terms is \(24\):

\(\begin{align*} S_n &= \dfrac{n}{2}[2a+(n-1)d]\\[5px] S_{10} &= \dfrac{10}{2}[2a+(10-1)d]\\[5px] 24&= 5(2a+9d)\\[5px] 24&= 10a+45d \tag{2}\\ \end{align*}\)

Equation \((1)\) \(\times\) \(10\):

\(\begin{align*} 10a+50d &= 20 \tag{3} \end{align*}\)Equation \((3)\) \(-\) Equation \((2)\):

\(\begin{align*} 10a+50d &= 20 \tag{3}\\ 10a+45d &= 24 \tag{2}\\ \hline 5d &= -4 \tag{3) - (2}\\ d &= -\dfrac45 \end{align*}\)

Substituting \(d=-\dfrac{4}{5}\) into \((1)\):

\(\begin{align*} a+5d &= 2\\[5px] a+ 5\left(-\dfrac45 \right) &= 2\\[5px] a-4 &= 2\\[5px] a &= 6 \end{align*}\)

Thus, the sequence described in the Try This is:

\(6,~\dfrac{26}{5},~\dfrac{22}{5},~\dfrac{18}{5},~\dfrac{14}{5},~2, \ldots\)

The \(21\)^st term of this sequence is:

\(\begin{align*} t_n &= a+ (n-1)d\\ t_{21} &= 6 + (21-1)\left(-\dfrac45 \right)\\ &= 6-16\\ &= -10 \end{align*}\)

Therefore, \(t_{21}=-10\).

Check Your Understanding 4

Question

The \(9\)^th term of an arithmetic sequence is \(10\). The sum of the first \(9\) terms of the sequence is \(54\). Determine the first term, \(a\), and the common difference \(d\).

Answer

The value of \(a\) is \(2\) and the value of \(d\) is \(1\).

Feedback

Since \(t_9=10\): 

\(\begin{align*} t_n&=a+(n-1)d \\ t_9 &= a+(9-1)d \\ 10 &= a +8d \tag{1} \end{align*}\)

Since \(S_9 = 54\):

\(\begin{align*} S_n&=\dfrac{n}{2} [2a+(n-1)d] \\ S_{9} &= \dfrac{9}{2} [2a +(9-1)d] \\ 54 &= \dfrac{9}{2}(2a+8d) \\ 54 &= 9a+36d \tag{2} \end{align*}\)

Solve \((1)\) and \((2)\) using elimination: 

• equation \((1) \times 9 \rightarrow\) equation \((3)\)
• equation \((3) \; - \) equation \((2)\)
  \(\begin{align*} 9a + 72d & = 90 \tag{3}\\ 9a + 36d &= 54 \tag{2} \\ \hline\;\;\;\;\;\;\;\;\;\;36d &= 36 \tag{3)-(2} \\ d&= 1 \end{align*}\)
• Substitute \(d=1\) into \((1)\):\[\begin{align*} a+ 8(1) &= 10 \\ a+8 &= 10 \\ a &=2 \end{align*}\]

Therefore, the value of the first term, \(a\), is \(2\) and the value of the common difference, \(d\), is \(1\).

Example 6

Solution 

Let \(c\) be the number of chairs in the auditorium.

Then,

\(c=50+51+52+\cdots+150\)

Here are the two methods we will use to determine the sum of this series:

1. Use the arithmetic series formula \(S_n=n \left(\dfrac{t_1+t_n}{2}\right)\)
2. Use the sum of the first \(n\) positive integers formula, \(1+2+3+\cdots=\dfrac{n(n+1)}2\)

Method 1: Arithmetic Series

We begin by determining \(n\) such that \(t_n=150\).

Since the series \(c=50+51+52+\cdots+150\) is arithmetic with \(a=50\) and \(d=1\):

\(\begin{align*} t_n &= a+(n-1)d\\[5px] t_n &= 50 + (n-1)(1)\\[5px] 150 &= 50 +n-1\\[5px] 150 &= n+49\\[5px] n &= 101 \end{align*}\)

This should not surprise us: since the terms in the series are consecutive integers, there are \(150-50+1=101\) terms.

Therefore, the sum is:

\(\begin{align*} S_n&=n\left(\dfrac{t_1+t_n}{2}\right)\\[5px] c&=101\left( \dfrac{50+150}{2}\right)\\[5px] &=101(100)\\[5px] &=10~100\\ \end{align*}\)

Therefore, there are \( 10~100\) chairs in the auditorium.

Method 2: Sum of First \(n\) positive integers

The series \(c=50+51+52+\ldots+150\) can be rewritten as a series that starts at \(1\) and goes up to \(150\), subtract the series that starts at \(1\) and goes up to \(49\):

\(\begin{align*} c&=50+51+52+\ldots+150\\ &=(1+2+3+\cdots+49+50+51+52+\cdots+150)-(1+2+3+\cdots+49) \end{align*}\)

Then, since \(1+2+3+\cdots+n = \dfrac{n(n+1)}{2}\), the series becomes:

\(\begin{align*} c&=(1+2+3+\cdots+49+50+51+52+\cdots+150)-(1+2+3+\cdots+49)\\[5px] &=(1+2+3+\cdots+150)-(1+2+3+\cdots+49)\\[5px] &= \dfrac{(150)(151)}{2}-\dfrac{(49)(50)}{2}\\[5px] &= 10~100 \end{align*}\)

Therefore, there are \( 10~100\) chairs in the auditorium.

Example 7

Solution

Since \(S_6=16\):

\(\begin{align*} S_n &= \dfrac{n}{2}[2a+(n-1)d]\\[5px] S_6 &= \dfrac62[2a+(6-1)d]\\[5px] 16&= 3(2a+5d)\\[5px] 16 &= 6a+15d \tag{1} \end{align*}\)

Since \(S_{10}=40\):

\(\begin{align*} S_n &= \dfrac{n}{2}[2a+(n-1)d]\\[5px] S_{10} &= \dfrac{10}2[2a+(10-1)d]\\[5px] 40&= 5(2a+9d)\tag{2}\\[5px] \dfrac{40}{5}&= \dfrac{5(2a+9d)}{5} \\[5px] 8 &= 2a+9d\tag{3} \end{align*}\)

Equation \((3)\) \(\times\) \(3\):
\(6a +27d=24 \tag{4}\)
Equation \((4)\) \(-\) Equation \((1)\) yields:

\(\begin{align*} 6a+27d &= 24 \tag{4}\\ 6a+15d &= 16 \tag{1}\\ \hline 12d &= 8 \tag{4) - (1}\\ d &= \dfrac{8}{12}\\ d &= \dfrac23 \end{align*}\)

Substituting \(d=\dfrac23\) into \((3)\):

\(\begin{align*} 2a+9d&=8\\[5px] 2a+9\left(\dfrac23 \right) &= 8\\[5px] 2a+6 &= 8\\[5px] 2a &= 2\\[5px] a &= 1 \end{align*}\)

Therefore, this is an arithmetic series with first term \(a=1\) and common difference \(d=\dfrac23\).

Wrap-Up

Take It With You

The Try This problem at the beginning of this lesson presented us with an arithmetic series with \(t_6=2\) and \(S_{10}=24\). For this arithmetic series, we were able to determine the first term, \(a\), and common difference, \(d\).

Consider the very slight modification of an arithmetic series with \(t_6=2\) and \(S_{11}=24\).

1. What happens when you try to determine \(a \) and \(d\) in this situation? Why?
2. What happens if you use \(t_6=2\) and \(S_{11}=22\) instead?

Extension: Summation Notation

Summation notation is a shorthand way of writing a series whose general term is known.

For example, consider the series \(1+4+9+16+25+36\).

Notice that the terms of this series are of the form \(t_n=n^2\).

Notation to represent a sum like this one has been developed. Here is the summation notation for \(1+4+9+16+25+36\):

\(\begin{align*} \sum_{i=1}^{6}i^2 \end{align*}\)

Here is the interpretation:

• The symbol \(\Sigma\) means, in this context, "find the sum of". 
• The expression \(i^2\) indicates the general term of the series.
• The \(i=1\) written under the \(\Sigma\) is the lower bound of the sum. It indicates that the first term in the series corresponds to \(i=1\); in this case, the first term is \((1)^2=1\).
• The \(6\) written above the \(\Sigma\) is the upper bound of the sum. It indicates that the last term in the series corresponds to \(i=6\); in this case, the last term is \((6)^2=36\).
• The \(i\) is called the index of the summation.
• The complete series is the sum of the terms that correspond to \(i=1, 2, 3, \ldots, 6\) (i.e., all integer values of the index from the lower bound to the upper bound, inclusive).

\(\Sigma\) is the uppercase Greek letter sigma; as a result, summation notation can also be referred to as "sigma notation".

Note that the index \(i\) can be any letter as long as the same letter is used below the \(\Sigma\) and in the general term.

Example 8

Solution — Part A

To write \(\begin{align*} \sum_{k=4}^{7}2^k \end{align*}\) without summation notation, we write the series of terms of the form \(2^k\) from \(k=4\) to \(k=7\):

\(\begin{align*} \sum_{k=4}^{7}2^k =2^4+2^5+2^6+2^7 \end{align*}\)

Solution — Part B

To write \(\begin{align*} \sum_{n=3}^{5}(4n^2+n) \end{align*}\)without summation notation, we write the series of terms of the form \(4n^2+n\) from \(n=3\) to \(n=5\):

\(\begin{align*} \sum_{n=3}^{5}(4n^2+n) = [4(3)^2+3]+[4(4)^2+4]+[4(5)^2+5] \end{align*}\)

Check Your Understanding 5

Question

Write \(\displaystyle \sum_{k=5}^7 (7k^3 - k)\) without using summation notion. Do not simplify.

Answer

\([7(5)^3 -5]+[7(6)^3-6] + [7(7)^3 - 7]\)

Feedback

To write \(\displaystyle \sum_{k=5}^7 (7k^3 - k)\) without summation notion, write the series of terms that are found by 

• substituting \(k=5\) into \((7k^3 -k)\) gives \([7(5)^3 -5]\),
• substituting \(k=6\) into \((7k^3 -k)\) gives \([7(6)^3-6]\), and 
• substituting \(k=7\) into \((7k^3 -k)\) gives \([7(7)^3 - 7]\).

Thus, \(\displaystyle \sum_{k=5}^7 (7k^3 - k) = [7(5)^3 -5]+[7(6)^3-6] + [7(7)^3 - 7]\).

Example 9

Solution

Writing the series without summation notation:

\(\begin{align*} \sum_{k=12}^{57}(4k+3) &= [4(12)+3]+[4(13)+3]+[4(14)+3]+\cdots+[4(57)+3]\\ &= 51+55+59+\cdots+231 \end{align*}\)

Notice that the series is arithmetic since the common difference is \(4\). This is not a surprise, since the general \(\)term, \(4k + 3\), (given in the summation notation) is linear with slope \(4\).

Notice further that since this series starts at term \(k=12\) and ends at term \(k=57\), there are \(57-12+1=46\) terms in the series. Therefore:

\(\begin{align*} S_n &= n\left(\dfrac{t_1+t_n}{2}\right)\\ S_{46} &= 46 \left(\dfrac{51+231}{2}\right)\\ &= 46(141)\\ &= 6486 \end{align*}\)

Properties of Summation Notation

Here are a few of the properties of summation notation. 

For all sequences of real numbers, \(t_n\) and \(s_n\), and for all real numbers \(c\) and any positive integers \(n\):

Property 1

\(\begin{align*} \sum_{k=1}^{n}c=cn \end{align*}\)

The series is the sum of \(n\) constants that are all equal to \(c\).

There are \(n\) \(c\)s:

\(\begin{align*} \sum_{k=1}^{n}c&=c+c+c+\cdots+c\\ &= nc \end{align*}\)

Property 2

\(\begin{align*} \sum_{k=1}^{n}ct_k=c\sum_{k=1}^{n}t_k \end{align*}\)

\(c\) can be common factored from the sum:

\(\begin{align*} \sum_{k=1}^{n}ct_k &= ct_1+ct_2+ct_3+ \cdots+ct_n\\ &= c(t_1+t_2+t_3+\cdots+t_n)\\ &=c\sum_{k=1}^{n}t_k \end{align*}\)

Property 3

\(\begin{align*} \sum_{k=1}^{n}(t_k+s_k)=\sum_{k=1}^{n}t_k + \sum_{k=1}^{n}s_k \end{align*}\)

For a sum of finite length, the terms of an addition can be rearranged into any order:

\(\begin{align*} \sum_{k=1}^{n}(t_k+s_k) &=(t_1+s_1) + (t_2+s_2) + (t_3+s_3) +\cdots +(t_n+s_n)\\ &= (t_1+t_2+t_3+\cdots+t_n)+(s_1+s_2+s_3+\cdots+s_n)\\ &= \sum_{k=1}^{n}t_k + \sum_{k=1}^{n}s_k \end{align*}\)

Property 4

\(\begin{align*} \sum_{k=1}^{n}(t_k-s_k) = \sum_{k=1}^{n}t_k - \sum_{k=1}^{n}s_k \end{align*}\)

This is a consequence of Properties 2 and 3:

\(\begin{align*} \sum_{k=1}^{n}(t_k-s_k) &= \sum_{k=1}^{n}(t_k+(-s_k))\\ &= \sum_{k=1}^{n}t_k+\sum_{k=1}^{n}(-s_k) ~~~~~~~~~~~~~~\text{By Property 3}\\ &= \sum_{k=1}^{n}t_k-\sum_{k=1}^{n}s_k ~~~~~~~~~~~~~~~~~~~~\text{By Property 2}\\ \end{align*}\)