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Intermediate Algebra
Sequences, Series and Binomial Theorem
Arithmetic Sequences

Arithmetic Sequences

By the end of this section, you will be able to: determine if a sequence is arithmetic, find the general term (nnth term) of an arithmetic sequence, and find the sum of the first nn terms of an arithmetic sequence.

Determine if a Sequence is Arithmetic

The last section introduced sequences and now we will look at two specific types of sequences that each have special properties. In this section we will look at arithmetic sequences and in the next section, geometric sequences.

An arithmetic sequence is a sequence where the difference between consecutive terms is constant. The difference between consecutive terms, an−an−1a_n-a_{n-1}, is dd, the common difference, for nn greater than or equal to two.

Arithmetic Sequence

An arithmetic sequence is a sequence where the difference between consecutive terms is always the same.

The difference between consecutive terms, an−an−1a_n-a_{n-1}, is dd, the common difference, for nn greater than or equal to two.

For the sequence

7, 10, 13, 16, 19, 22,…,7,\ 10,\ 13,\ 16,\ 19,\ 22,\ldots,

the consecutive differences are

10−7=3,13−10=3,16−13=3,19−16=3,22−19=3.10-7=3,\quad 13-10=3,\quad 16-13=3,\quad 19-16=3,\quad 22-19=3.

For the sequence

8, 4, 0, −4, −8, −12,…,8,\ 4,\ 0,\ -4,\ -8,\ -12,\ldots,

the consecutive differences are

4−8=−4,0−4=−4,−4−0=−4,−8−(−4)=−4,−12−(−8)=−4.4-8=-4,\quad 0-4=-4,\quad -4-0=-4,\quad -8-(-4)=-4,\quad -12-(-8)=-4.

In each of these sequences, the difference between consecutive terms is constant, and so the sequence is arithmetic.

Example 12.13. Determine if each sequence is arithmetic. If so, indicate the common difference.

(a) 5,9,13,17,21,25,…5,9,13,17,21,25,\ldots

(b) 4,9,12,17,20,25,…4,9,12,17,20,25,\ldots

(c) 10,3,−4,−11,−18,−25,…10,3,-4,-11,-18,-25,\ldots

Solution. To determine if the sequence is arithmetic, we find the difference of the consecutive terms shown.

(a)

9−5=413−9=417−13=421−17=425−21=4 \begin{array}{rcl} 9-5&=&4\\[4pt] 13-9&=&4\\[4pt] 17-13&=&4\\[4pt] 21-17&=&4\\[4pt] 25-21&=&4 \end{array}

The sequence is arithmetic. The common difference is d=4d=4.

(b)

9−4=512−9=317−12=520−17=325−20=5 \begin{array}{rcl} 9-4&=&5\\[4pt] 12-9&=&3\\[4pt] 17-12&=&5\\[4pt] 20-17&=&3\\[4pt] 25-20&=&5 \end{array}

The sequence is not arithmetic as all the differences between the consecutive terms are not the same. There is no common difference.

(c)

3−10=−7−4−3=−7−11−(−4)=−7−18−(−11)=−7−25−(−18)=−7 \begin{array}{rcl} 3-10&=&-7\\[4pt] -4-3&=&-7\\[4pt] -11-(-4)&=&-7\\[4pt] -18-(-11)&=&-7\\[4pt] -25-(-18)&=&-7 \end{array}

The sequence is arithmetic. The common difference is d=−7d=-7.

Determine whether each sequence is arithmetic. If so, indicate the common difference.

For 9,20,31,42,53,64,…9,20,31,42,53,64,\ldots, which statement is correct?

For 12,6,0,−6,−12,−18,…12,6,0,-6,-12,-18,\ldots, which statement is correct?

For 7,1,10,4,13,7,…7,1,10,4,13,7,\ldots, which statement is correct?

Apply the same test to each sequence in the next set.

For −4,4,2,10,8,16,…-4,4,2,10,8,16,\ldots, which statement is correct?

For −3,−1,1,3,5,7,…-3,-1,1,3,5,7,\ldots, which statement is correct?

For 7,2,−3,−8,−13,−18,…7,2,-3,-8,-13,-18,\ldots, which statement is correct?

If we know the first term, a1a_1, and the common difference, dd, we can list a finite number of terms of the sequence.

Example 12.14. Write the first five terms of the sequence where the first term is 5 and the common difference is d=−6d=-6.

Solution. We start with the first term and add the common difference. Then we add the common difference to that result to get the next term, and so on.

a1=5a2=5+(−6)=−1a3=−1+(−6)=−7a4=−7+(−6)=−13a5=−13+(−6)=−19 \begin{array}{rcl} a_1&=&5\\[4pt] a_2&=&5+(-6)=-1\\[4pt] a_3&=&-1+(-6)=-7\\[4pt] a_4&=&-7+(-6)=-13\\[4pt] a_5&=&-13+(-6)=-19 \end{array}

The sequence is 5,−1,−7,−13,−19,…5,-1,-7,-13,-19,\ldots.

Write the first five terms of the sequence where the first term is 77 and the common difference is d=−4d=-4. Enter the terms separated by commas.

Write the first five terms of the sequence where the first term is 1111 and the common difference is d=−8d=-8. Enter the terms separated by commas.

Find the General Term (nnth Term) of an Arithmetic Sequence

Just as we found a formula for the general term of a sequence, we can also find a formula for the general term of an arithmetic sequence.

Let’s write the first few terms of a sequence where the first term is a1a_1 and the common difference is dd. We will then look for a pattern.

As we look for a pattern we see that each term starts with a1a_1:

a1=a1a2=a1+da3=a1+2da4=a1+3da5=a1+4d. \begin{array}{rcl} a_1&=&a_1\\[4pt] a_2&=&a_1+d\\[4pt] a_3&=&a_1+2d\\[4pt] a_4&=&a_1+3d\\[4pt] a_5&=&a_1+4d. \end{array}

The first term adds 0d0d to the a1a_1, the second term adds 1d1d, the third term adds 2d2d, the fourth term adds 3d3d, and the fifth term adds 4d4d. The number of dd’s that were added to a1a_1 is one less than the number of the term. This leads us to the following.

an=a1+(n−1)da_n=a_1+(n-1)d

General Term (nnth term) of an Arithmetic Sequence

The general term of an arithmetic sequence with first term a1a_1 and the common difference dd is

an=a1+(n−1)d.a_n=a_1+(n-1)d.

We will use this formula in the next example to find the fifteenth term of a sequence.

Example 12.15. Find the fifteenth term of a sequence where the first term is 3 and the common difference is 6.

Solution. To find the fifteenth term, a15a_{15}, use the formula with a1=3a_1=3 and d=6d=6.

Use the formula.an=a1+(n−1)dSubstitute in the values.a15=3+(15−1)6Simplify.a15=3+(14)6=87 \begin{array}{lrcl} \text{Use the formula.} & a_n&=&a_1+(n-1)d\\[4pt] \text{Substitute in the values.} & a_{15}&=&3+(15-1)6\\[4pt] \text{Simplify.} & a_{15}&=&3+(14)6\\[4pt] &&=&87 \end{array}

Find the twenty-seventh term of a sequence where the first term is 77 and the common difference is 99.

Find the eighteenth term of a sequence where the first term is 1313 and the common difference is −7-7.

Sometimes we do not know the first term and we must use other given information to find it before we find the requested term.

Example 12.16. Find the twelfth term of a sequence where the seventh term is 10 and the common difference is −2-2. Give the formula for the general term.

Solution. To first find the first term, a1a_1, use the formula with a7=10a_7=10, n=7n=7, and d=−2d=-2.

Use the formula.an=a1+(n−1)dSubstitute in the values.10=a1+(7−1)(−2)Simplify.10=a1+(6)(−2)10=a1−12a1=22 \begin{array}{lrcl} \text{Use the formula.} & a_n&=&a_1+(n-1)d\\[4pt] \text{Substitute in the values.} & 10&=&a_1+(7-1)(-2)\\[4pt] \text{Simplify.} & 10&=&a_1+(6)(-2)\\[4pt] &10&=&a_1-12\\[4pt] &a_1&=&22 \end{array}

Find the twelfth term, a12a_{12}, using the formula with a1=22a_1=22, n=12n=12, and d=−2d=-2.

Use the formula.an=a1+(n−1)dSubstitute in the values.a12=22+(12−1)(−2)Simplify.a12=22+(11)(−2)a12=0 \begin{array}{lrcl} \text{Use the formula.} & a_n&=&a_1+(n-1)d\\[4pt] \text{Substitute in the values.} & a_{12}&=&22+(12-1)(-2)\\[4pt] \text{Simplify.} & a_{12}&=&22+(11)(-2)\\[4pt] &a_{12}&=&0 \end{array}

The twelfth term of the sequence is 0, a12=0a_{12}=0.

To find the general term, substitute the values into the formula.

an=a1+(n−1)dan=22+(n−1)(−2)an=22−2n+2an=−2n+24. \begin{array}{rcl} a_n&=&a_1+(n-1)d\\[4pt] a_n&=&22+(n-1)(-2)\\[4pt] a_n&=&22-2n+2\\[4pt] a_n&=&-2n+24. \end{array}

The general term is an=−2n+24a_n=-2n+24.

Find the eleventh term of a sequence where the ninth term is 88 and the common difference is −3-3.

For the sequence where the ninth term is 88 and the common difference is −3-3, give the formula for the general term.

Now use the same process with a different known term and common difference.

Find the nineteenth term of a sequence where the fifth term is 11 and the common difference is −4-4.

For the sequence where the fifth term is 11 and the common difference is −4-4, give the formula for the general term.

Sometimes the information given leads us to two equations in two unknowns. We then use our methods for solving systems of equations to find the values needed.

Example 12.17. Find the first term and common difference of a sequence where the fifth term is 19 and the eleventh term is 37. Give the formula for the general term.

Solution. Since we know two terms, we can make a system of equations using the formula for the general term.

an=a1+(n−1)da_n=a_1+(n-1)d

We know the value of a5a_5 and a11a_{11}, so we will use n=5n=5 and n=11n=11:

{a5=a1+(5−1)da11=a1+(11−1)d. \left\{ \begin{array}{rcl} a_5&=&a_1+(5-1)d\\[4pt] a_{11}&=&a_1+(11-1)d. \end{array} \right.

Substitute in the values, a5=19a_5=19 and a11=37a_{11}=37, and simplify.

{19=a1+4d37=a1+10d. \left\{ \begin{array}{rcl} 19&=&a_1+4d\\[4pt] 37&=&a_1+10d. \end{array} \right.

Prepare to eliminate the a1a_1 term by multiplying the top equation by −1-1. Add the equations.

−19=−a1−4d37=a1+10d18=6d3=d \begin{array}{rcl} -19&=&-a_1-4d\\[4pt] 37&=&a_1+10d\\ \hline 18&=&6d\\[4pt] 3&=&d \end{array}

Substituting d=3d=3 back into the first equation,

19=a1+4⋅319=a1+127=a1. \begin{array}{rcl} 19&=&a_1+4\cdot3\\[4pt] 19&=&a_1+12\\[4pt] 7&=&a_1. \end{array}

Use the formula with a1=7a_1=7 and d=3d=3.

Use the formula.an=a1+(n−1)dSubstitute in the values.an=7+(n−1)3Simplify.an=7+3n−3an=3n+4 \begin{array}{lrcl} \text{Use the formula.} & a_n&=&a_1+(n-1)d\\[4pt] \text{Substitute in the values.} & a_n&=&7+(n-1)3\\[4pt] \text{Simplify.} & a_n&=&7+3n-3\\[4pt] &a_n&=&3n+4 \end{array}

The first term is a1=7a_1=7. The common difference is d=3d=3. The general term of the sequence is an=3n+4a_n=3n+4.

For a sequence where the fourth term is 1717 and the thirteenth term is 5353, find the first term.

For a sequence where the fourth term is 1717 and the thirteenth term is 5353, find the common difference.

For a sequence where the fourth term is 1717 and the thirteenth term is 5353, give the formula for the general term.

Repeat the process for another pair of known terms.

For a sequence where the third term is 22 and the twelfth term is −25-25, find the first term.

For a sequence where the third term is 22 and the twelfth term is −25-25, find the common difference.

For a sequence where the third term is 22 and the twelfth term is −25-25, give the formula for the general term.

Find the Sum of the First nn Terms of an Arithmetic Sequence

As with the general sequences, it is often useful to find the sum of an arithmetic sequence. The sum, SnS_n, of the first nn terms of any arithmetic sequence is written as

Sn=a1+a2+a3+⋯+an.S_n=a_1+a_2+a_3+\cdots+a_n.

To find the sum by merely adding all the terms can be tedious. So we can also develop a formula to find the sum of a sequence using the first and last term of the sequence.

We can develop this new formula by first writing the sum by starting with the first term, a1a_1, and keep adding a dd to get the next term as

Sn=a1+(a1+d)+(a1+2d)+⋯+an.S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+a_n.

We can also reverse the order of the terms and write the sum by starting with ana_n and keep subtracting dd to get the next term as

Sn=an+(an−d)+(an−2d)+⋯+a1.S_n=a_n+(a_n-d)+(a_n-2d)+\cdots+a_1.

If we add these two expressions for the sum of the first nn terms of an arithmetic sequence, we can derive a formula for the sum of the first nn terms of any arithmetic series.

Sn=a1+(a1+d)+(a1+2d)+⋯+an+Sn=an+(an−d)+(an−2d)+⋯+a12Sn=(a1+an)+(a1+an)+(a1+an)+⋯+(a1+an) \begin{array}{rrrrrrrrr} S_n&=&a_1&+&(a_1+d)&+&(a_1+2d)&+\cdots+a_n\\[4pt] +S_n&=&a_n&+&(a_n-d)&+&(a_n-2d)&+\cdots+a_1\\ \hline 2S_n&=&(a_1+a_n)&+&(a_1+a_n)&+&(a_1+a_n)&+\cdots+(a_1+a_n) \end{array}

Because there are nn sums of (a1+an)(a_1+a_n) on the right side of the equation, we rewrite the right side as n(a1+an)n(a_1+a_n).

2Sn=n(a1+an)2S_n=n(a_1+a_n)

We divide by two to solve for SnS_n.

Sn=n2(a1+an)S_n=\tfrac n2(a_1+a_n)

This gives us a general formula for the sum of the first nn terms of an arithmetic sequence.

Sum of the First nn Terms of an Arithmetic Sequence

The sum, SnS_n, of the first nn terms of an arithmetic sequence is

Sn=n2(a1+an),S_n=\tfrac n2(a_1+a_n),

where a1a_1 is the first term and ana_n is the nnth term.

We apply this formula in the next example where the first few terms of the sequence are given.

Example 12.18. Find the sum of the first 30 terms of the arithmetic sequence: 8,13,18,23,28,…8,13,18,23,28,\ldots

Solution. To find the sum, we will use the formula Sn=n2(a1+an)S_n=\tfrac n2(a_1+a_n). We know a1=8a_1=8, d=5d=5, and n=30n=30, but we need to find ana_n in order to use the sum formula.

Use the general-term formula.an=a1+(n−1)dFind an where a1=8,d=5,n=30.a30=8+(30−1)5Simplify.a30=8+(29)5a30=153Use the sum formula.Sn=n2(a1+an)Substitute in the values.S30=302(8+153)Simplify.S30=15(161)S30=2,415 \begin{array}{lrcl} \text{Use the general-term formula.} & a_n&=&a_1+(n-1)d\\[4pt] \text{Find }a_n\text{ where }a_1=8,d=5,n=30. & a_{30}&=&8+(30-1)5\\[4pt] \text{Simplify.} & a_{30}&=&8+(29)5\\[4pt] &a_{30}&=&153\\[4pt] \text{Use the sum formula.} & S_n&=&\tfrac n2(a_1+a_n)\\[4pt] \text{Substitute in the values.} & S_{30}&=&\tfrac{30}{2}(8+153)\\[4pt] \text{Simplify.} & S_{30}&=&15(161)\\[4pt] &S_{30}&=&2{,}415 \end{array}

Find the sum of the first 3030 terms of the arithmetic sequence 5,9,13,17,21,…5,9,13,17,21,\ldots.

Find the sum of the first 3030 terms of the arithmetic sequence 7,10,13,16,19,…7,10,13,16,19,\ldots.

In the next example, we are given the general term for the sequence and are asked to find the sum of the first 50 terms.

Example 12.19. Find the sum of the first 50 terms of the arithmetic sequence whose general term is an=3n−4a_n=3n-4.

Solution. To find the sum, we will use the formula Sn=n2(a1+an)S_n=\tfrac n2(a_1+a_n). We know n=50n=50, but we need to find a1a_1 and ana_n in order to use the sum formula.

an=3n−4Find a1 by substituting n=1.a1=3⋅1−4a1=−1Find an by substituting n=50.a50=3⋅50−4Simplify.a50=146Use the sum formula.Sn=n2(a1+an)Substitute in the values.S50=502(−1+146)Simplify.S50=25(145)S50=3,625 \begin{array}{lrcl} &a_n&=&3n-4\\[4pt] \text{Find }a_1\text{ by substituting }n=1. & a_1&=&3\cdot1-4\\[4pt] &a_1&=&-1\\[4pt] \text{Find }a_n\text{ by substituting }n=50. & a_{50}&=&3\cdot50-4\\[4pt] \text{Simplify.} & a_{50}&=&146\\[4pt] \text{Use the sum formula.} & S_n&=&\tfrac n2(a_1+a_n)\\[4pt] \text{Substitute in the values.} & S_{50}&=&\tfrac{50}{2}(-1+146)\\[4pt] \text{Simplify.} & S_{50}&=&25(145)\\[4pt] &S_{50}&=&3{,}625 \end{array}

Find the sum of the first 5050 terms of the arithmetic sequence whose general term is an=2n−5a_n=2n-5.

Find the sum of the first 5050 terms of the arithmetic sequence whose general term is an=4n+3a_n=4n+3.

In the next example we are given the sum in summation notation. To add all the terms would be tedious, so we extract the information needed to use the formula to find the sum of the first nn terms.

Example 12.20. Find the sum:

∑i=125(4i+7).\sum_{i=1}^{25}(4i+7).

Solution. To find the sum, we will use the formula Sn=n2(a1+an)S_n=\tfrac n2(a_1+a_n). We know n=25n=25, but we need to find a1a_1 and ana_n in order to use the sum formula.

Expand the summation notation.

∑i=125(4i+7)=(4⋅1+7)+(4⋅2+7)+(4⋅3+7)+⋯+(4⋅25+7). \sum_{i=1}^{25}(4i+7) =(4\cdot1+7)+(4\cdot2+7)+(4\cdot3+7)+\cdots+(4\cdot25+7).

Simplify.

∑i=125(4i+7)=11+15+19+⋯+107.\sum_{i=1}^{25}(4i+7)=11+15+19+\cdots+107.

Identify a1=11a_1=11 and a25=107a_{25}=107. Knowing n=25n=25, a1=11a_1=11, and a25=107a_{25}=107, use the sum formula.

Use the sum formula.Sn=n2(a1+an)Substitute in the values.S25=252(11+107)Simplify.S25=252(118)S25=1,475 \begin{array}{lrcl} \text{Use the sum formula.} & S_n&=&\tfrac n2(a_1+a_n)\\[4pt] \text{Substitute in the values.} & S_{25}&=&\tfrac{25}{2}(11+107)\\[4pt] \text{Simplify.} & S_{25}&=&\tfrac{25}{2}(118)\\[4pt] &S_{25}&=&1{,}475 \end{array}

Find the sum ∑i=130(6i−4)\sum_{i=1}^{30}(6i-4).

Find the sum ∑i=135(5i−3)\sum_{i=1}^{35}(5i-3).

Key terms. An arithmetic sequence is a sequence in which the difference between consecutive terms is always the same. This constant difference dd is the common difference.

This section is adapted from Intermediate Algebra 2e, Section 12.2 by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at OpenStax. Changes: omitted the readiness quiz, media links, and section exercise sets; converted Try It exercises into interactive questions and adapted formatting for the web.

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