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, $d$ . The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first $n$ terms of an arithmetic series as:

S_{n} = a_{1} + (a_{1} + d) + (a_{1} + 2 d) + . . . + (a_{n} - d) + a_{n}

${S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}$ .

We can also reverse the order of the terms and write the sum as

S_{n} = a_{n} + (a_{n} - d) + (a_{n} - 2 d) + . . . + (a_{1} + d) + a_{1}

${S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}$ .

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

\frac{\begin{matrix} S_{n} = a_{1} + (a_{1} + d) + (a_{1} + 2 d) + . . . + (a_{n} - d) + a_{n} + S_{n} = a_{n} + (a_{n} - d) + (a_{n} - 2 d) + . . . + (a_{1} + d) + a_{1} \end{matrix}}{2 S_{n} = (a_{1} + a_{n}) + (a_{1} + a_{n}) + . . . + (a_{1} + a_{n})}

$\frac{\begin{array}{l}{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}\hfill \\ +{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}\hfill \end{array}}{2{S}_{n}=\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+...+\left({a}_{1}+{a}_{n}\right)}$

Because there are $n$ terms in the series, we can simplify this sum to

2 S_{n} = n (a_{1} + a_{n})

$2{S}_{n}=n\left({a}_{1}+{a}_{n}\right)$ .

We divide by 2 to find the formula for the sum of the first $n$ terms of an arithmetic series.

S_{n} = \frac{n (a_{1} + a_{n})}{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 $n$ terms of an arithmetic sequence is

S_{n} = \frac{n (a_{1} + a_{n})}{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 $n$ terms.

Identify ${a}_{1}$ and ${a}_{n}$ .
Determine $n$ .
Substitute values for ${a}_{1}\text{, }{a}_{n}$ , and $n$ into the formula ${S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}$ .
Simplify to find ${S}_{n}$ .

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

Find the sum of each arithmetic series.

$\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}$
$\text{20 + 15 + 10 +}\ldots{ + -50}$
$\sum _{k=1}^{12}3k - 8$

Solution

We are given ${a}_{1}=5$ and ${a}_{n}=32$ .Count the number of terms in the sequence to find $n=10$ .
Substitute values for ${a}_{1},{a}_{n}\text{\hspace{0.17em},}$ and $n$ into the formula and simplify.

$\begin{array}{l}\begin{array}{l}\hfill \\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \end{array}\hfill \\ {S}_{10}=\frac{10\left(5+32\right)}{2}=185\hfill \end{array}$
We are given ${a}_{1}=20$ and ${a}_{n}=-50$ .Use the formula for the general term of an arithmetic sequence to find $n$ .
$\begin{array}{l}{a}_{n}={a}_{1}+\left(n - 1\right)d\hfill \\ -50=20+\left(n - 1\right)\left(-5\right)\hfill \\ -70=\left(n - 1\right)\left(-5\right)\hfill \\ 14=n - 1\hfill \\ 15=n\hfill \end{array}$

Substitute values for ${a}_{1},{a}_{n}\text{,}n$ into the formula and simplify.

$\begin{array}{l}\begin{array}{l}\\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\end{array}\hfill \\ {S}_{15}=\frac{15\left(20 - 50\right)}{2}=-225\hfill \end{array}$
To find ${a}_{1}$ , substitute $k=1$ into the given explicit formula.
$\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array}$

We are given that $n=12$ . To find ${a}_{12}$ , substitute $k=12$ into the given explicit formula.

$\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array}$

Substitute values for ${a}_{1},{a}_{n}$ , and $n$ into the formula and simplify.

$\begin{array}{l}\text{ }{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ {S}_{12}=\frac{12\left(-5+28\right)}{2}=138\hfill \end{array}$

Use the formula to find the sum of each arithmetic series.

Try It 2

$\text{1}\text{.4 + 1}\text{.6 + 1}\text{.8 + 2}\text{.0 + 2}\text{.2 + 2}\text{.4 + 2}\text{.6 + 2}\text{.8 + 3}\text{.0 + 3}\text{.2 + 3}\text{.4}$

Solution

Try It 3

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

Solution

Try It 4

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

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 ${a}_{1}=\frac{1}{2}$ and $d=\frac{1}{4}$ . We are looking for the total number of miles walked after 8 weeks, so we know that $n=8$ , and we are looking for ${S}_{8}$ . To find ${a}_{8}$ , we can use the explicit formula for an arithmetic sequence.

\begin{matrix} \begin{matrix} a_{n} = a_{1} + d (n - 1) \end{matrix} a_{8} = \frac{1}{2} + \frac{1}{4} (8 - 1) = \frac{9}{4} \end{matrix}

$\begin{array}{l}\begin{array}{l}\\ {a}_{n}={a}_{1}+d\left(n - 1\right)\end{array}\hfill \\ {a}_{8}=\frac{1}{2}+\frac{1}{4}\left(8 - 1\right)=\frac{9}{4}\hfill \end{array}$

We can now use the formula for arithmetic series.

\begin{matrix} S_{n} = \frac{n (a_{1} + a_{n})}{2} S_{8} = \frac{8 (\frac{1}{2} + \frac{9}{4})}{2} = 11 \end{matrix}

$\begin{array}{l} {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ \text{ }{S}_{8}=\frac{8\left(\frac{1}{2}+\frac{9}{4}\right)}{2}=11\hfill \end{array}$

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

Series and Their Notations