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, [latex]d[/latex]. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first [latex]n[/latex] terms of an arithmetic series as:
[latex]{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}[/latex].
We can also reverse the order of the terms and write the sum as
[latex]{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}[/latex].
If we add these two expressions for the sum of the first [latex]n[/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[/latex] terms of any arithmetic series.
[latex]\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)}[/latex]
Because there are [latex]n[/latex] terms in the series, we can simplify this sum to
[latex]2{S}_{n}=n\left({a}_{1}+{a}_{n}\right)[/latex].
We divide by 2 to find the formula for the sum of the first [latex]n[/latex] terms of an arithmetic series.
[latex]{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}[/latex]
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 [latex]n[/latex] terms of an arithmetic sequence is[latex]{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}[/latex]
How To: Given terms of an arithmetic series, find the sum of the first [latex]n[/latex] terms.
- Identify [latex]{a}_{1}[/latex] and [latex]{a}_{n}[/latex].
- Determine [latex]n[/latex].
- Substitute values for [latex]{a}_{1}\text{, }{a}_{n}[/latex], and [latex]n[/latex] into the formula [latex]{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}[/latex].
- Simplify to find [latex]{S}_{n}[/latex].
Example 2: Finding the First n Terms of an Arithmetic Series
Find the sum of each arithmetic series.- [latex]\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[/latex]
- [latex]\text{20 + 15 + 10 +}\ldots{ + -50}[/latex]
- [latex]\sum _{k=1}^{12}3k - 8[/latex]
Solution
- We are given [latex]{a}_{1}=5[/latex] and [latex]{a}_{n}=32[/latex].Count the number of terms in the sequence to find [latex]n=10[/latex].
Substitute values for [latex]{a}_{1},{a}_{n}\text{\hspace{0.17em},}[/latex] and [latex]n[/latex] into the formula and simplify.
[latex]\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}[/latex]
- We are given [latex]{a}_{1}=20[/latex] and [latex]{a}_{n}=-50[/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[/latex].
[latex]\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}[/latex]Substitute values for [latex]{a}_{1},{a}_{n}\text{,}n[/latex] into the formula and simplify.[latex]\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}[/latex]
- To find [latex]{a}_{1}[/latex], substitute [latex]k=1[/latex] into the given explicit formula.
[latex]\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array}[/latex]We are given that [latex]n=12[/latex]. To find [latex]{a}_{12}[/latex], substitute [latex]k=12[/latex] into the given explicit formula.[latex]\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array}[/latex]Substitute values for [latex]{a}_{1},{a}_{n}[/latex], and [latex]n[/latex] into the formula and simplify.[latex]\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}[/latex]
Try It 2
[latex-display]\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}[/latex-display] SolutionTry It 3
[latex-display]\text{13 + 21 + 29 + }\dots \text{+ 69}[/latex-display] SolutionTry It 4
[latex-display]\sum _{k=1}^{10}5 - 6k[/latex-display] SolutionExample 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 [latex]{a}_{1}=\frac{1}{2}[/latex] and [latex]d=\frac{1}{4}[/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[/latex], and we are looking for [latex]{S}_{8}[/latex]. To find [latex]{a}_{8}[/latex], we can use the explicit formula for an arithmetic sequence.[latex]\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}[/latex]
We can now use the formula for arithmetic series.
[latex]\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}[/latex]
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? SolutionLicenses & Attributions
CC licensed content, Specific attribution
- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
