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23.2: The arithmetic sequence

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    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, \(a_n=8+12\cdot (n-1)\).

    1. 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, \(a_n=-5+4\cdot (n-1)\).

    1. 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.

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

    clipboard_e533312df3c127d0a2733a76c1c69cc3b.png

    1. 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:

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

    1. 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: \(\sum_{i=1}^{35}a_i=\dfrac{-315}2=-157.5\).


    This page titled 23.2: The arithmetic sequence is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.