23.2: The arithmetic sequence

We have already encountered examples of arithmetic sequences in the previous section. An arithmetic sequence is a sequence for which we add a constant number to get from one term to the next, for example:

$$8,\underset{+3}{\hookrightarrow } 11,\underset{+3}{\hookrightarrow } 14, \underset{+3}{\hookrightarrow } 17, \underset{+3}{\hookrightarrow } 20, \underset{+3}{\hookrightarrow } 23, \dots \nonumber$$

Definition: Arithmetic Sequence

A sequence $\{a_n\}$ is called an arithmetic sequence if any two consecutive terms have a common difference $d$. The arithmetic sequence is determined by $d$ and the first value $a_1$. This can be written recursively as:

$$a_n=a_{n-1}+d \quad \quad \text{for }n\geq 2 \nonumber$$

Alternatively, we have the general formula for the $n$th term of the arithmetic sequence:

$$\label{EQU:arithmetic-sequence-general-term} \boxed{a_n=a_1+d\cdot (n-1)}$$

Example $\PageIndex{1}$

Determine if the sequence is an arithmetic sequence. If so, then find the general formula for $a_n$ in the form of equation $\ref{EQU:arithmetic-sequence-general-term}$.

1. $7,13, 19, 25, 31, \dots$
2. $13, 9, 5, 1, -3, -7, \dots$
3. $10, 13, 16, 20, 23, \dots$
4. $a_n=8\cdot n +3$

Solution

1. Calculating the difference between two consecutive terms always gives the same answer $13-7=6$, $19-13=6$, $25-19=6$, etc. Therefore the common difference $d=6$, which shows that this is an arithmetic sequence. Furthermore, the first term is $a_1=7$, so that the general formula for the $n$th term is $a_n=7+6\cdot (n-1)$.
2. One checks that the common difference is $9-13=-4$, $5-9=-4$, etc., so that this is an arithmetic sequence with $d=-4$. Since $a_1=13$, the general term is $a_n=13-4\cdot(n-1)$.
3. We have $13-10=3$, but $20-16=4$, so that this is not an arithmetic sequence.
4. If we write out the first couple of terms of $a_n=8n+3$, we get the sequence:

$$11, 19, 27, 35, 43, 51, \dots \nonumber$$

From this it seems reasonable to suspect that this is an arithmetic sequence with common difference $d=8$ and first term $a_1=11$. The general term of such an arithmetic sequence is

$$a_1+d(n-1)=11+8(n-1)=11+8n-8=8n+3=a_n \nonumber$$

This shows that $a_n=8n+3=11+8(n-1)$ is an arithmetic sequence.

Example $\PageIndex{2}$

Find the general formula of an arithmetic sequence with the given property.

1. $d=12$, and $a_6=68$
2. $a_1=-5$, and $a_{9}=27$
3. $a_{5}=38$, and $a_{16}=115$

Solution

1. According to equation $\ref{EQU:arithmetic-sequence-general-term}$ the general term is $a_n=a_1+d(n-1)$. We know that $d=12$, so that we only need to find $a_1$. Plugging $a_6=68$ into $a_n=a_1+d(n-1)$, we may solve for $a_1$:

$$68=a_6=a_1+12\cdot (6-1)=a_1+12\cdot 5=a_1+60 \,\, \stackrel{(-60)}{\implies} \,\, a_1=68-60=8 \nonumber$$

The $n$th term is therefore, $a_n=8+12\cdot (n-1)$.

2. In this case, we are given $a_1=-5$, but we still need to find the common difference $d$. Plugging $a_9=27$ into $a_n=a_1+d(n-1)$, we obtain

$$\begin{aligned} 27=a_9&=-5+d\cdot(9-1)\\&= -5+8d \\ \stackrel{(+5)}{\implies} 32&=8d \\ \stackrel{(\div 8)}{\implies} 4&=d \end{aligned} \nonumber$$

The $n$th term is therefore, $a_n=-5+4\cdot (n-1)$.

3. In this case we neither have $a_1$ nor $d$. However, the two conditions $a_{5}=38$ and $a_{16}=115$ give two equations in the two unknowns $a_1$ and $d$.

$$\left\{\begin{array} { c } { 3 8 = a _ { 5 } = a _ { 1 } + d ( 5 - 1 ) } \\ { 1 1 5 = a _ { 1 6 } = a _ { 1 } + d ( 1 6 - 1 ) } \end{array} \quad \Longrightarrow \left\{\begin{array}{c} 38=a_{1}+4 \cdot d \\ 115=a_{1}+15 \cdot d \end{array}\right.\right. \nonumber$$

To solve this system of equations, we need to recall the methods for doing so. One convenient method is the addition/subtraction method. For this, we subtract the top from the bottom equation:

$$\begin{matrix} & 115 & = & a_1 & +15\cdot d & \\ - (& 38 & = & a_1& +4\cdot d & ) \\ \hline & 77 & = & & +11\cdot d & \\ \end{matrix} \begin{matrix} \, \\ \, \\ \quad\quad \stackrel{(\div 11)}{\implies} \quad\quad 7=d \end{matrix} \nonumber$$

Plugging $d=7$ into either of the two equations gives $a_1$. We plug it into the first equation $38=a_1+4d$:

$$38=a_1+4\cdot 7 \quad\implies \quad 38=a_1+28 \quad \stackrel{(-28)}{\implies} \quad 10=a_1 \nonumber$$

Therefore, the $n$th term is given by $a_n=10+7\cdot (n-1)$.

We can sum the first $k$ terms of an arithmetic sequence using a trick, which, according to lore, was found by the German mathematician Carl Friedrich Gauss when he was a child at school.

Example $\PageIndex{3}$

Find the sum of the first $100$ integers, starting from $1$.

Solution

In other words, we want to find the sum of $1+2+3+\dots + 99+100$. First, note that he sequence $1, 2, 3, \dots$ is an arithmetic sequence. The main idea for solving this problem is a trick, which will indeed work for any arithmetic sequence:

Let $S=1+2+3+\cdots+98+99+100$ be what we want to find. Note that

$$2S=\begin{array}{ccccccccccccc}&1&+&2&+&3&\cdots&+&98&+&99&+&100\\ +&100&+&99&+&98&\cdots&+&3&+&2&+&1 \end{array} \nonumber$$

Note that the second line is also $S$ but is added in the reverse order. Adding vertically we see then that

$$2S=101+101+101+\cdots+101+101+101, \nonumber$$

where there are $100$ terms on the right hand side. So

$$2S=100\cdot101{\text{ and therefore }}S=\frac{100\cdot101}{2}=5050 \nonumber$$

The previous example generalizes to the more general setting starting with an arbitrary arithmetic sequence.

Observation: Arithmetic Sequence

Let $\{a_n\}$ be an arithmetic sequence, whose $n$th term is given by the formula $a_n=a_1+d(n-1)$. Then, the sum $a_1+a_2+\dots+a_{k-1}+a_k$ is given by adding $(a_1+a_k)$ precisely $\dfrac k 2$ times:

$$\label{EQU:arithmetic-series} \boxed{\sum_{i=1}^k a_i =\dfrac{k}{2}\cdot (a_1+a_k)}$$

In order to remember the formula above, it may be convenient to think of the right hand side as $k\cdot\dfrac{a_1+a_k}{2}$ (this is, $k$ times the average of the first and the last term).

Proof

For the proof of equation $\ref{EQU:arithmetic-series}$, we write $S= a_1+a_2+\dots +a_{k-1}+a_{k}$. We then add it to itself but in reverse order:

$$2S=\begin{array}{ccccccccccccc}&a_1&+&a_2&+&a_3&\cdots&+&a_{k-2}&+&a_{k-1}&+&a_k\\ +&a_{k}&+&a_{k-1}&+&a_{k-2}&\cdots&+&a_3&+&a_2&+&a_1 \end{array} \nonumber$$

Now note that in general $a_l+a_m=2a_1+d(l+m-2)$. We see that adding vertically gives

$$2S=k(2a_1+d(k-1))=k(a_1+(a_1+d(k-1))=k(a_1+a_k) \nonumber$$

Dividing by $2$ gives the desired result.

Example $\PageIndex{4}$

Find the value of the arithmetic series.

1. Find the sum $a_1+\dots +a_{60}$ for the arithmetic sequence $a_n=2+13(n-1)$.
2. Determine the value of the sum: $\sum\limits_{j=1}^{1001} (5-6j)$
3. Find the sum of the first $35$ terms of the sequence $$4, 3.5, 3, 2.5, 2, 1.5, \dots \nonumber$$

Solution

1. The sum is given by the formula $\ref{EQU:arithmetic-series}$: $\sum_{i=1}^k a_i=\dfrac k 2 \cdot (a_1+a_k)$. Here, $k=60$, and $a_1=2$ and $a_{60}=2+13\cdot(60-1)=2+13\cdot 59=2+767=769$. We obtain a sum of

$$a_1+\dots +a_{60}=\sum_{i=1}^{60} a_i=\dfrac{60}{2}\cdot (2+769)=30\cdot 771=23130 \nonumber$$

We may confirm this with the calculator as described in example 23.1.5 in the previous section.

$$\text{Enter: } \operatorname{sum}(\operatorname{seq}((2+13 \cdot(n-1), n, 1,60)) \nonumber$$

2. Again, we use the above formula $\sum_{j=1}^k a_j=\dfrac k 2 \cdot (a_1+a_k)$, where the arithmetic sequence is given by $a_j=5-6j$ and $k=1001$. Using the values $a_1=5-6\cdot 1=5-6=-1$ and $a_{1001}=5-6\cdot 1001=5-6006=-6001$, we obtain:

$$\begin{aligned} \sum_{j=1}^{1001} (5-6j)& = \dfrac{1001}{2}(a_1+a_{1001})\\&=\dfrac{1001}{2}((-1)+(-6001))\\ &= \dfrac{1001}{2}\cdot (-6002)\\& = 1001\cdot (-3001)\\& =-3004001\end{aligned} \nonumber$$

3. First note that the given sequence $4, 3.5, 3, 2.5, 2, 1.5, \dots$ is an arithmetic sequence. It is determined by the first term $a_1=4$ and common difference $d=-0.5$. The $n$th term is given by $a_n=4-0.5\cdot (n-1)$, and summing the first $k=35$ terms gives:

$$\sum_{i=1}^{35} a_i=\dfrac{35}{2}\cdot (a_1+a_{35}) \nonumber$$

We see that we need to find $a_{35}$ in the above formula:

$$a_{35}=4-0.5\cdot (35-1)=4-0.5\cdot 34=4-17=-13 \nonumber$$

This gives a total sum of

$$\sum_{i=1}^{35} a_i=\dfrac{35}{2}\cdot (4+(-13))=\dfrac{35}{2}\cdot (-9)=\dfrac{-315}{2} \nonumber$$

The answer may be written as a fraction or also as a decimal, that is: $\sum_{i=1}^{35}a_i=\dfrac{-315}2=-157.5$.