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Study Guides > College Algebra

Using the Formula for Arithmetic Series

Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, dd. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first nn terms of an arithmetic series as:

Sn=a1+(a1+d)+(a1+2d)+...+(and)+an{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}.
We can also reverse the order of the terms and write the sum as
Sn=an+(and)+(an2d)+...+(a1+d)+a1{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}.
If we add these two expressions for the sum of the first nn terms of an arithmetic series, we can derive a formula for the sum of the first nn terms of any arithmetic series.
Sn=a1+(a1+d)+(a1+2d)+...+(and)+an+Sn=an+(and)+(an2d)+...+(a1+d)+a12Sn=(a1+an)+(a1+an)+...+(a1+an)\frac{\begin{array}{l}{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}\hfill \\ +{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}\hfill \end{array}}{2{S}_{n}=\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+...+\left({a}_{1}+{a}_{n}\right)}
Because there are nn terms in the series, we can simplify this sum to
2Sn=n(a1+an)2{S}_{n}=n\left({a}_{1}+{a}_{n}\right).
We divide by 2 to find the formula for the sum of the first nn terms of an arithmetic series.
Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}

A General Note: Formula for the Sum of the First n Terms of an Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first nn terms of an arithmetic sequence is
Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}

How To: Given terms of an arithmetic series, find the sum of the first nn terms.

  1. Identify a1{a}_{1} and an{a}_{n}.
  2. Determine nn.
  3. Substitute values for a1an{a}_{1}\text{, }{a}_{n}, and nn into the formula Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}.
  4. Simplify to find Sn{S}_{n}.

Example 2: Finding the First n Terms of an Arithmetic Series

Find the sum of each arithmetic series.
  1. 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}
  2. 20 + 15 + 10 ++50\text{20 + 15 + 10 +}\ldots{ + -50}
  3. k=1123k8\sum _{k=1}^{12}3k - 8

Solution

  1. We are given a1=5{a}_{1}=5 and an=32{a}_{n}=32.Count the number of terms in the sequence to find n=10n=10. Substitute values for a1,an,{a}_{1},{a}_{n}\text{\hspace{0.17em},} and nn into the formula and simplify.
    Sn=n(a1+an)2S10=10(5+32)2=185\begin{array}{l}\begin{array}{l}\hfill \\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \end{array}\hfill \\ {S}_{10}=\frac{10\left(5+32\right)}{2}=185\hfill \end{array}
  2. We are given a1=20{a}_{1}=20 and an=50{a}_{n}=-50.Use the formula for the general term of an arithmetic sequence to find nn.
    an=a1+(n1)d50=20+(n1)(5)70=(n1)(5)14=n115=n\begin{array}{l}{a}_{n}={a}_{1}+\left(n - 1\right)d\hfill \\ -50=20+\left(n - 1\right)\left(-5\right)\hfill \\ -70=\left(n - 1\right)\left(-5\right)\hfill \\ 14=n - 1\hfill \\ 15=n\hfill \end{array}
    Substitute values for a1,an,n{a}_{1},{a}_{n}\text{,}n into the formula and simplify.
    Sn=n(a1+an)2S15=15(2050)2=225\begin{array}{l}\begin{array}{l}\\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\end{array}\hfill \\ {S}_{15}=\frac{15\left(20 - 50\right)}{2}=-225\hfill \end{array}
  3. To find a1{a}_{1}, substitute k=1k=1 into the given explicit formula.
    ak=3k8 a1=3(1)8=5\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array}
    We are given that n=12n=12. To find a12{a}_{12}, substitute k=12k=12 into the given explicit formula.
     ak=3k8a12=3(12)8=28\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array}
    Substitute values for a1,an{a}_{1},{a}_{n}, and nn into the formula and simplify.
     Sn=n(a1+an)2S12=12(5+28)2=138\begin{array}{l}\text{ }{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ {S}_{12}=\frac{12\left(-5+28\right)}{2}=138\hfill \end{array}
Use the formula to find the sum of each arithmetic series.

Try It 2

1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4\text{1}\text{.4 + 1}\text{.6 + 1}\text{.8 + 2}\text{.0 + 2}\text{.2 + 2}\text{.4 + 2}\text{.6 + 2}\text{.8 + 3}\text{.0 + 3}\text{.2 + 3}\text{.4} Solution

Try It 3

13 + 21 + 29 + + 69\text{13 + 21 + 29 + }\dots \text{+ 69} Solution

Try It 4

k=11056k\sum _{k=1}^{10}5 - 6k Solution

Example 3: Solving Application Problems with Arithmetic Series

On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?

Solution

This problem can be modeled by an arithmetic series with a1=12{a}_{1}=\frac{1}{2} and d=14d=\frac{1}{4}. We are looking for the total number of miles walked after 8 weeks, so we know that n=8n=8, and we are looking for S8{S}_{8}. To find a8{a}_{8}, we can use the explicit formula for an arithmetic sequence.
an=a1+d(n1)a8=12+14(81)=94\begin{array}{l}\begin{array}{l}\\ {a}_{n}={a}_{1}+d\left(n - 1\right)\end{array}\hfill \\ {a}_{8}=\frac{1}{2}+\frac{1}{4}\left(8 - 1\right)=\frac{9}{4}\hfill \end{array}
We can now use the formula for arithmetic series.
Sn=n(a1+an)2 S8=8(12+94)2=11\begin{array}{l} {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ \text{ }{S}_{8}=\frac{8\left(\frac{1}{2}+\frac{9}{4}\right)}{2}=11\hfill \end{array}
She will have walked a total of 11 miles.

Try It 5

A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned? Solution

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