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Arithmetic Sequences

Find the common difference for an arithmetic sequence.

Arithmetic Sequences

Determine if a sequence is arithmetic (IA 12.2.1)
Find the general term (nth term) of an arithmetic sequence (IA 12.2.2)

In the last section, 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 ${a}_{1}$ and the common difference is d. We will then look for a pattern.

As we look for a pattern we see that each term starts with ${a}_{1}$ .

The first term adds 0d to the ${a}_{1}$ , the second term adds 1d, the third term adds 2d, the fourth term adds 3d, and the fifth term adds 4d. The number of ds that were added to ${a}_{1}$ is one less than the number of the term. We then have the formula for the general term of an arithmetic sequence.

Example

Find the general term (nth term) of an arithmetic sequence.

Try it.

ⓐ Find the twenty-first term of a sequence where the first term is three and the common difference is eight.
ⓑ Find the eleventh term of a sequence where the third term is 19 and the common difference is five. Give the formula for the general term.

Solution

ⓐ To find the 21st term, use the formula with ${a}_{1}=3$, $d=8$, and $n=21$	${a}_{n}={a}_{1}+(n-1)d$
Substitute	${a}_{21}=3+(21-1)(8)$
Simplify	$\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}$

ⓑ Let's first find ${a}_{1}$. Use the formula with ${a}_{3}=19$, $d=5$, and $n=3$. Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{3}={a}_{1}+(3-1)(5) \\ 19={a}_{1}+(2)(5) \\ 19={a}_{1}+10 \\ {a}_{1=9}\end{array}$
To find the 11th term, use the formula with ${a}_{3}=9$, $d=5$, and $n=11$ Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59 \\ \end{array}$
To find the general term, substitute $a=9$ and $d=5$ into the formula.	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59\end{array}$

Find the general term (nth term) of an arithmetic sequence.

Try it.

Find the sixteenth term of a sequence where the first term is 11 and the common difference is −6.

	${a}_{n}={a}_{1}+(n-1)d$

Try it.

Find the 19th term of a sequence where the 5th term is 1 and the common difference is -4. Give the formula for the general term.

Condensed: the full section is in OpenStax College Algebra 2e.

Finding Common Differences

The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is –3,400.

The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term.

Example

Try it.

Is each sequence arithmetic? If so, find the common difference.

ⓐ $\{1,2,4,8,16,...\}$
ⓑ $\{-3,1,5,9,13,...\}$

Solution

Subtract each term from the subsequent term to determine whether a common difference exists.

ⓐThe sequence is not arithmetic because there is no common difference.
ⓑThe sequence is arithmetic because there is a common difference. The common difference is 4.

Writing Terms of Arithmetic Sequences

Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $n$ and $d$ into formula below.

\[{a}_{n}={a}_{1}+(n-1)d\]

Example

Try it.

Write the first five terms of the arithmetic sequence with ${a}_{1}=17$ and $d=-3$ .

Solution

Adding $-3$ is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.

The first five terms are $\{17,\ 14,\ 11,\ 8,\ 5\}$

Example

Try it.

Given ${a}_{1}=8$ and ${a}_{4}=14$ , find ${a}_{5}$ .

Solution

The sequence can be written in terms of the initial term 8 and the common difference $d$ .

\[\{8,8+d,8+2d,8+3d\}\]

We know the fourth term equals 14; we know the fourth term has the form ${a}_{1}+3d=8+3d$ .

We can find the common difference $d$ .

\[\begin{array}{ll}{a}_{n}={a}_{1}+(n-1)d & \\ {a}_{4}={a}_{1}+3d & \\ {a}_{4}=8+3d & \text{Write the fourth term of the sequence in terms of}{a}_{1}\text{and}d. \\ 14=8+3d & \text{Substitute}14\text{for}{a}_{4}. \\ \ d=2 & \text{Solve for the common difference}.\end{array}\]

Find the fifth term by adding the common difference to the fourth term.

\[{a}_{5}={a}_{4}+2=16\]

Using Recursive Formulas for Arithmetic Sequences

Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given.

\[\begin{array}{lllll}{a}_{n}={a}_{n-1}+d & & & & n\ge 2\end{array}\]

Example

Try it.

Write a recursive formula for the arithmetic sequence.

\[\{-18\text{, }-7\text{, }4\text{, }15\text{, }26\text{, }\ldots \}\]

Solution

The first term is given as $-18$ . The common difference can be found by subtracting the first term from the second term.

\[d=-7-(-18)=11\]

Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.

\[\begin{array}{l}{a}_{1}=-18 \\ {a}_{n}={a}_{n-1}+11,\ \text{for }n\ge 2\end{array}\]

Using Explicit Formulas for Arithmetic Sequences

We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.

\[{a}_{n}={a}_{1}+d(n-1)\]

To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.

The common difference is $-50$ , so the sequence represents a linear function with a slope of $-50$ . To find the $y$ -intercept, we subtract $-50$ from $200:\ 200-(-50)=200+50=250$ . You can also find the $y$ -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is

Recall the slope-intercept form of a line is $y=mx+b.$ When dealing with sequences, we use ${a}_{n}$ in place of $y$ and $n$ in place of $x.$ If we know the slope and vertical intercept of the function, we can substitute them for $m$ and $b$ in the slope-intercept form of a line. Substituting $-50$ for the slope and $250$ for the vertical intercept, we get the following equation:

\[{a}_{n}=-50n+250\]

We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is ${a}_{n}=200-50(n-1)$ , which simplifies to ${a}_{n}=-50n+250.$

Example

Try it.

Write an explicit formula for the arithmetic sequence.

\[\{2\text{, }12\text{, }22\text{, }32\text{, }42\text{, }\ldots \}\]

Solution

The common difference can be found by subtracting the first term from the second term.

\[\begin{array}{ll}d & ={a}_{2}-{a}_{1} \\ & =12-2 \\ & =10\end{array}\]

The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.

\[\begin{array}{l}{a}_{n}=2+10(n-1) \\ {a}_{n}=10n-8\end{array}\]

Condensed: the full section is in OpenStax College Algebra 2e.

Key Equations

recursive formula for nth term of an arithmetic sequence	${a}_{n}={a}_{n-1}+d,n\ge 2$
explicit formula for nth term of an arithmetic sequence	$\begin{array}{l}{a}_{n}={a}_{1}+d(n-1)\end{array}$

Key Concepts

An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
The constant between two consecutive terms is called the common difference.
The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term.
The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly.
A recursive formula for an arithmetic sequence with common difference $d$ is given by ${a}_{n}={a}_{n-1}+d,n\ge 2.$
As with any recursive formula, the initial term of the sequence must be given.
An explicit formula for an arithmetic sequence with common difference $d$ is given by ${a}_{n}={a}_{1}+d(n-1).$
An explicit formula can be used to find the number of terms in a sequence.
In application problems, we sometimes alter the explicit formula slightly to ${a}_{n}={a}_{0}+dn.$

Arithmetic Sequences

Determine if a sequence is arithmetic (IA 12.2.1)
Find the general term (nth term) of an arithmetic sequence (IA 12.2.2)

In the last section, 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 ${a}_{1}$ and the common difference is d. We will then look for a pattern.

As we look for a pattern we see that each term starts with ${a}_{1}$ .

Example

Find the general term (nth term) of an arithmetic sequence.

Try it.

ⓐ Find the twenty-first term of a sequence where the first term is three and the common difference is eight.
ⓑ Find the eleventh term of a sequence where the third term is 19 and the common difference is five. Give the formula for the general term.

Solution

ⓐ To find the 21st term, use the formula with ${a}_{1}=3$, $d=8$, and $n=21$	${a}_{n}={a}_{1}+(n-1)d$
Substitute	${a}_{21}=3+(21-1)(8)$
Simplify	$\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}$

ⓑ Let's first find ${a}_{1}$. Use the formula with ${a}_{3}=19$, $d=5$, and $n=3$. Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{3}={a}_{1}+(3-1)(5) \\ 19={a}_{1}+(2)(5) \\ 19={a}_{1}+10 \\ {a}_{1=9}\end{array}$
To find the 11th term, use the formula with ${a}_{3}=9$, $d=5$, and $n=11$ Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59 \\ \end{array}$
To find the general term, substitute $a=9$ and $d=5$ into the formula.	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59\end{array}$

Find the general term (nth term) of an arithmetic sequence.

Try it.

Find the sixteenth term of a sequence where the first term is 11 and the common difference is −6.

	${a}_{n}={a}_{1}+(n-1)d$

Try it.

Find the 19th term of a sequence where the 5th term is 1 and the common difference is -4. Give the formula for the general term.

Condensed: the full section is in OpenStax Precalculus 2e.

Finding Common Differences

The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term.

Example

Try it.

Is each sequence arithmetic? If so, find the common difference.

ⓐ $\{1,2,4,8,16,...\}$
ⓑ $\{-3,1,5,9,13,...\}$

Solution

Subtract each term from the subsequent term to determine whether a common difference exists.

ⓐThe sequence is not arithmetic because there is no common difference.
ⓑThe sequence is arithmetic because there is a common difference. The common difference is 4.

Writing Terms of Arithmetic Sequences

\[{a}_{n}={a}_{1}+(n-1)d\]

Example

Try it.

Write the first five terms of the arithmetic sequence with ${a}_{1}=17$ and $d=-3$ .

Solution

Adding $-3$ is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.

The first five terms are $\{17,\ 14,\ 11,\ 8,\ 5\}$

Example

Try it.

Given ${a}_{1}=8$ and ${a}_{4}=14$ , find ${a}_{5}$ .

Solution

The sequence can be written in terms of the initial term 8 and the common difference $d$ .

\[\{8,8+d,8+2d,8+3d\}\]

We know the fourth term equals 14; we know the fourth term has the form ${a}_{1}+3d=8+3d$ .

We can find the common difference $d$ .

Find the fifth term by adding the common difference to the fourth term.

\[{a}_{5}={a}_{4}+2=16\]

Using Recursive Formulas for Arithmetic Sequences

\[\begin{array}{lllll}{a}_{n}={a}_{n-1}+d & & & & n\ge 2\end{array}\]

Example

Try it.

Write a recursive formula for the arithmetic sequence.

\[\{-18\text{, }-7\text{, }4\text{, }15\text{, }26\text{, }\ldots \}\]

Solution

The first term is given as $-18$ . The common difference can be found by subtracting the first term from the second term.

\[d=-7-(-18)=11\]

Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.

\[\begin{array}{l}{a}_{1}=-18 \\ {a}_{n}={a}_{n-1}+11,\ \text{for }n\ge 2\end{array}\]

Using Explicit Formulas for Arithmetic Sequences

\[{a}_{n}={a}_{1}+d(n-1)\]

To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.

\[{a}_{n}=-50n+250\]

Example

Try it.

Write an explicit formula for the arithmetic sequence.

\[\{2\text{, }12\text{, }22\text{, }32\text{, }42\text{, }\ldots \}\]

Solution

The common difference can be found by subtracting the first term from the second term.

\[\begin{array}{ll}d & ={a}_{2}-{a}_{1} \\ & =12-2 \\ & =10\end{array}\]

The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.

\[\begin{array}{l}{a}_{n}=2+10(n-1) \\ {a}_{n}=10n-8\end{array}\]

Condensed: the full section is in OpenStax Precalculus 2e.

Key Equations

recursive formula for nth term of an arithmetic sequence	${a}_{n}={a}_{n-1}+d,n\ge 2$
explicit formula for nth term of an arithmetic sequence	$\begin{array}{l}{a}_{n}={a}_{1}+d(n-1)\end{array}$

Key Concepts

An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
The constant between two consecutive terms is called the common difference.
The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term.
The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly.
A recursive formula for an arithmetic sequence with common difference $d$ is given by ${a}_{n}={a}_{n-1}+d,n\ge 2.$
As with any recursive formula, the initial term of the sequence must be given.
An explicit formula for an arithmetic sequence with common difference $d$ is given by ${a}_{n}={a}_{1}+d(n-1).$
An explicit formula can be used to find the number of terms in a sequence.
In application problems, we sometimes alter the explicit formula slightly to ${a}_{n}={a}_{0}+dn.$

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 in an arithmetic sequence, ${a}_{n}-{a}_{n-1},$ is d, the common difference, for n greater than or equal to two.

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

Example

Try it.

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

ⓐ $5,9,13,17,21,25\text{,}\ \ldots$

ⓑ $4,9,12,17,20,25\text{,}\ \ldots$

Solution

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

ⓐ

Find the difference of the consecutive terms.	$\begin{array}{lllllllllll}5, & & \ 9, & & \ 13, & & \ 17 & & \ 21, & \ 25,\ldots \\ \\ \\ 9-5 & & \ 13-9 & & \ 17-13 & & \ 21-17 & & \ 25-21 & & \\ 4 & & \ 4 & & \ 4 & & \ 4 & & \ 4 & & \end{array}$
	The sequence is arithmetic. The common difference is $d=4\text{.}$

ⓑ

Find the difference of the consecutive terms.	$\begin{array}{lllllllllll}4, & & \ 9, & & \ 12, & & \ 17 & & \ 20, & \ 25,\ldots \\ \\ \\ 9-4 & & \ 12-9 & & \ 17-12 & & \ 20-17 & & \ 25-20 & & \\ 5 & & \ 3 & & \ 5 & & \ 3 & & \ 5 & & \end{array}$
	The sequence is not arithmetic as all the differences between the consecutive terms are not the same. There is no common difference.

ⓒ

Find the difference of the consecutive terms.	$\begin{array}{lllllllllll}10, & & \ 3, & & \ -4, & & \ -11 & & \ -18, & \ -25,\ldots & \\ \\ \\ 3-10 & & \ -4-3 & & \ -11-(-4) & & \ -18-(-11) & & \ -25-(-18) & & \\ -7 & & \ -7 & & \ -7 & & \ -7 & & \ -7 & & \end{array}$
	The sequence is arithmetic. The common difference is $d=-7\text{.}$

If we know the first term, ${a}_{1},$ and the common difference, d, we can list a finite number of terms of the sequence.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Find the General Term (

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 ${a}_{1}$ and the common difference is d. We will then look for a pattern.

As we look for a pattern we see that each term starts with ${a}_{1}$.

The first term adds 0d to the ${a}_{1}$, the second term adds 1d, the third term adds 2d, the fourth term adds 3d, and the fifth term adds 4d. The number of ds that were added to ${a}_{1}$ is one less than the number of the term. This leads us to the following

\[{a}_{n}={a}_{1}+(n-1)d\]

We will use this formula in the next example to find the 15^th term of a sequence.

Example

Try it.

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, ${a}_{15},$ use the formula with ${a}_{1}=3\ \text{and}\ d=6\text{.}$	$\ {a}_{n}={a}_{1}+(n-1)d$
Substitute in the values.	$\ {a}_{15}=3+(15-1)6$
Simplify.	$\ {a}_{15}=3+(14)6$
	$\ {a}_{15}=87$

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

Try it.

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

Solution

To first find the first term, ${a}_{1},$ use the formula with ${a}_{7}=10,n=7,\ \text{and}\ d=-2.$	$\ {a}_{n}={a}_{1}+(n-1)d$
Substitute in the values.	$\ 10={a}_{1}+(7-1)(-2)$
Simplify.	$\ 10={a}_{1}+(6)(-2)$
	$\ 10={a}_{1}-12$
	$\ {a}_{1}=22$
Find the twelfth term, ${a}_{12},$ using the formula with ${a}_{1}=22,n=12,\ \text{and}\ d=-2\text{.}$	$\ {a}_{n}={a}_{1}+(n-1)d$
Substitute in the values.	$\ {a}_{12}=22+(12-1)(-2)$
Simplify.	$\ {a}_{12}=22+(11)(-2)$
	$\ {a}_{12}=0$
	The twelfth term of the sequence is 0, ${a}_{12}=0.$
To find the general term, substitute the values into the formula.	$\ {a}_{n}={a}_{1}+(n-1)d$
	$\ {a}_{n}=22+(n-1)(-2)$
	$\ {a}_{n}=22-2n+2$
	The general term is ${a}_{n}=-2n+24.$

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Find the Sum of the First

As with the general sequences, it is often useful to find the sum of an arithmetic sequence. The sum, ${S}_{n},$ of the first $n$ terms of any arithmetic sequence is written as ${S}_{n}={a}_{1}+{a}_{2}+{a}_{3}+...+{a}_{n}.$ 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, ${a}_{1},$ and keep adding a d to get the next term as:

\[{S}_{n}={a}_{1}+({a}_{1}+d)+({a}_{1}+2d)+\ldots +{a}_{n}.\]

We can also reverse the order of the terms and write the sum by starting with ${a}_{n}$ and keep subtracting d to get the next term as

\[{S}_{n}={a}_{n}+({a}_{n}-d)+({a}_{n}-2d)+\ldots +{a}_{1}.\]

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

\[\begin{array}{lllllllllllllllllllll}\underline{\begin{array}{lllllllllll}{S}_{n} & = & {a}_{1} & \ + & ({a}_{1}+d) & + & ({a}_{1}+2d) & + & \ldots & + & {a}_{n} \\ +{S}_{n} & = & {a}_{n} & \ + & ({a}_{n}-d) & + & ({a}_{n}-2d) & + & \ldots & + & {a}_{1}\end{array}} \\ \\ 2{S}_{n}\ =\ ({a}_{1}+{a}_{n})+\ ({a}_{1}+{a}_{n})\ +\ ({a}_{1}+{a}_{n})\ +\ \ldots \ +\ ({a}_{1}+{a}_{n})\end{array}\]

Because there are n sums of $({a}_{1}+{a}_{n})$ on the right side of the equation, we rewrite the right side as $n({a}_{1}+{a}_{n}).$

\[2{S}_{n}=n({a}_{1}+{a}_{n})\]

We divide by two to solve for ${S}_{n}.$

\[{S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\]

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

Example

Try it.

Find the sum of the first 30 terms of the arithmetic sequence: 8, 13, 18, 23, 28, …

Solution

To find the sum, we will use the formula ${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n}).$ We know ${a}_{1}=8$, $d=5$ and $n=30,$ but we need to find ${a}_{n}$ in order to use the sum formula.

	$\ {a}_{n}={a}_{1}+(n-1)d$
$\text{Find}\ {a}_{n}\ \text{where}\ {a}_{1}=8,d=5\ \text{and}\ n=30.$	${a}_{30}=8+(30-1)5$
Simplify.	${a}_{30}=8+(29)5$
	${a}_{30}=153$
$\text{Knowing}\ {a}_{1}=8,n=30,\ \text{and}\ {a}_{30}=153,$ use the sum formula.	$\ {S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$
Substitute in the values.	${S}_{30}=\frac{30}{2}(8+153)$
Simplify.	${S}_{30}=15(161)$
Simplify.	${S}_{30}=2,415$

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

General Term (nth term) of an Arithmetic Sequence
The general term of an arithmetic sequence with first term ${a}_{1}$ and the common difference d is
\[{a}_{n}={a}_{1}+(n-1)d\]
Sum of the First n Terms of an Arithmetic Sequence
The sum, ${S}_{n},$ of the first n terms of an arithmetic sequence, where ${a}_{1}$ is the first term and ${a}_{n}$ is the nth term is
\[{S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\]

Arithmetic Sequences

Determine if a Sequence is Arithmetic

In the following exercises, determine if each sequence is arithmetic, and if so, indicate the common difference.

Try it.

$4,12,20,28,36,44\text{,}\ \ldots$

Solution

The sequence is arithmetic with common difference $d=8.$

Try it.

$-7,-2,3,8,13,18\text{,}\ \ldots$

Try it.

$-15,-16,3,12,21,30\text{,}\ \ldots$

Solution

The sequence is not arithmetic.

Try it.

$11,5,-1,-7-13,-19\text{,}\ \ldots$

Try it.

$8,5,2,-1,-4,-7\text{,}\ \ldots$

Solution

The sequence is arithmetic with common difference $d=-3.$

Try it.

$15,5,-5,-15,-25,-35\text{,}\ \ldots$

In the following exercises, write the first five terms of each sequence with the given first term and common difference.

Try it.

${a}_{1}=11$ and $d=7$

Solution

$11,18,25,32,39$

Try it.

${a}_{1}=18$ and $d=9$

Try it.

${a}_{1}=-7$ and $d=4$

Solution

$-7,-3,1,5,9$

Try it.

${a}_{1}=-8$ and $d=5$

Try it.

${a}_{1}=14$ and $d=-9$

Solution

$14,5,-4,-13,-22$

Try it.

${a}_{1}=-3$ and $d=-3$

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

In the following exercises, find the term described using the information provided.

Try it.

Find the twenty-first term of a sequence where the first term is three and the common difference is eight.

Solution

$163$

Try it.

Find the twenty-third term of a sequence where the first term is six and the common difference is four.

Try it.

Find the thirtieth term of a sequence where the first term is $-14$ and the common difference is five.

Solution

$131$

Try it.

Find the fortieth term of a sequence where the first term is $-19$ and the common difference is seven.

Try it.

Find the sixteenth term of a sequence where the first term is 11 and the common difference is $-6.$

Solution

$-79$

Try it.

Find the fourteenth term of a sequence where the first term is eight and the common difference is $-3.$

Try it.

Find the twentieth term of a sequence where the fifth term is $-4$ and the common difference is $-2.$ Give the formula for the general term.

Solution

${a}_{20}=-34.$ The general term is ${a}_{n}=-2n+6.$

Try it.

Find the thirteenth term of a sequence where the sixth term is $-1$ and the common difference is $-4.$ Give the formula for the general term.

Try it.

Find the eleventh term of a sequence where the third term is 19 and the common difference is five. Give the formula for the general term.

Solution

${a}_{11}=59.$ The general term is ${a}_{n}=5n+4.$

Try it.

Find the fifteenth term of a sequence where the tenth term is 17 and the common difference is seven. Give the formula for the general term.

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Find the eighth term of a sequence where the seventh term is $-8$ and the common difference is $-5.$ Give the formula for the general term.

Solution

${a}_{8}=-13.$ The general term is ${a}_{n}=-5n+27.$

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Find the fifteenth term of a sequence where the tenth term is $-11$ and the common difference is $-3.$ Give the formula for the general term.

In the following exercises, find the first term and common difference of the sequence with the given terms. Give the formula for the general term.

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The second term is 14 and the thirteenth term is 47.

Solution

${a}_{1}=11,$$d=3.$ The general term is ${a}_{n}=3n+8.$

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The third term is 18 and the fourteenth term is 73.

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The second term is 13 and the tenth term is $-51.$

Solution

${a}_{1}=21,$$d=-8.$ The general term is ${a}_{n}=-8n+29.$

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The third term is four and the tenth term is $-38$.

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The fourth term is $-6$ and the fifteenth term is 27.

Solution

${a}_{1}=-15,$$d=3.$ The general term is ${a}_{n}=3n-18.$

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The third term is $-13$ and the seventeenth term is 15.

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

In the following exercises, find the sum of the first 30 terms of each arithmetic sequence.

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$11,14,17,20,23\text{,}\ \ldots$

Solution

1,635

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$12,18,24,30,36\text{,}\ \ldots$

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$8,5,2,-1,-4\text{,}\ \ldots$

Solution

$-1,065$

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$16,10,4,-2,-8\text{,}\ \ldots$

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$-17,-15,-13,-11,-9\text{,}\ \ldots$

Solution

360

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$-15,-12,-9,-6,-3\text{,}\ \ldots$

In the following exercises, find the sum of the first 50 terms of the arithmetic sequence whose general term is given.

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${a}_{n}=5n-1$

Solution

6,325

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${a}_{n}=2n+7$

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${a}_{n}=-3n+5$

Solution

–3,575

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${a}_{n}=-4n+3$

In the following exercises, find each sum.

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$\sum _{i=1}^{40}(8i-7)$

Solution

6,280

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$\sum _{i=1}^{45}(7i-5)$

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$\sum _{i=1}^{50}(3i+6)$

Solution

4,125

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$\sum _{i=1}^{25}(4i+3)$

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$\sum _{i=1}^{35}(-6i-2)$

Solution

$-3,850$

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$\sum _{i=1}^{30}(-5i+1)$

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Determine if each sequence is arithmetic. If so, indicate the common difference.
ⓐ $5,9,13,17,21,25\text{,}\ \ldots$
ⓑ $4,9,12,17,20,25\text{,}\ \ldots$

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To determine if the sequence is arithmetic, we find the difference of the consecutive terms shown.
ⓐ
$\begin{array}{lllllllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the difference of} \\ \text{the consecutive terms.}\end{array} & & & & & \begin{array}{lllllllllll}5, & & \ 9, & & \ 13, & & \ 17 & & \ 21, & \ 25,\ldots \\ \\ \\ 9-5 & & \ 13-9 & & \ 17-13 & & \ 21-17 & & \ 25-21 & & \\ 4 & & \ 4 & & \ 4 & & \ 4 & & \ 4 & & \end{array} \\ \\ \\ & & & & & \text{The sequence is arithmetic. The common difference is}\ d=4\text{.}\end{array}$
ⓑ
$\begin{array}{lllllllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the difference of} \\ \text{the consecutive terms.}\end{array} & & & & & \begin{array}{lllllllllll}4, & & \ 9, & & \ 12, & & \ 17 & & \ 20, & \ 25,\ldots \\ \\ \\ 9-4 & & \ 12-9 & & \ 17-12 & & \ 20-17 & & \ 25-20 & & \\ 2 & & \ 3 & & \ 5 & & \ 3 & & \ 5 & & \end{array} \\ \\ \\ & & & & & \begin{array}{l}\text{The sequence is not arithmetic as all the differences between} \\ \text{the consecutive terms are not the same.} \\ \text{There is no common difference.}\end{array}\end{array}$
-4, 4, 2, 10, 8, 16, …
Find the difference of consecutive terms.
-3, -1, 1, 3, 5, 7, …
Find the difference of consecutive terms.
Write the first five terms of the sequence where the first term is 5 and the common difference is $d=-6.$

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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.
\[\begin{array}{lllllllllllllll}{a}_{1} & & & {a}_{2} & & & {a}_{3} & & & {a}_{4} & & & {a}_{5} & & \\ 5 & & & 5+(-6) & & & -1+(-6) & & & -7+(-6) & & & -13+(-6) \\ & & & -1 & & & -7 & & & -13 & & & -19\end{array}\]
The sequence is $5,-1,-7,-13,-19\text{,}\ \ldots$
Write the first five terms of the sequence where the first term is –4 and the common difference is $d=7.$
\[\begin{array}{lllllllllllllll}{a}_{1} & & & {a}_{2} & & & {a}_{3} & & & {a}_{4} & & & {a}_{5} & & \\ -4\end{array}\]
The sequence is: ________________________________________

ⓐ Find the twenty-first term of a sequence where the first term is three and the common difference is eight.
ⓑ Find the eleventh term of a sequence where the third term is 19 and the common difference is five. Give the formula for the general term.

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ⓐ To find the 21st term, use the formula with ${a}_{1}=3$, $d=8$, and $n=21$	${a}_{n}={a}_{1}+(n-1)d$
Substitute	${a}_{21}=3+(21-1)(8)$
Simplify	$\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}$

ⓑ Let's first find ${a}_{1}$. Use the formula with ${a}_{3}=19$, $d=5$, and $n=3$. Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{3}={a}_{1}+(3-1)(5) \\ 19={a}_{1}+(2)(5) \\ 19={a}_{1}+10 \\ {a}_{1=9}\end{array}$
To find the 11th term, use the formula with ${a}_{3}=9$, $d=5$, and $n=11$ Substitute these values and simplify	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59 \\ \end{array}$
To find the general term, substitute $a=9$ and $d=5$ into the formula.	$\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59\end{array}$

Find the sixteenth term of a sequence where the first term is 11 and the common difference is −6.
${a}_{n}={a}_{1}+(n-1)d$
Find the 19th term of a sequence where the 5th term is 1 and the common difference is -4. Give the formula for the general term.
Is each sequence arithmetic? If so, find the common difference.
1. ⓐ $\{1,2,4,8,16,...\}$
2. ⓑ $\{-3,1,5,9,13,...\}$
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Subtract each term from the subsequent term to determine whether a common difference exists.
1. ⓐThe sequence is not arithmetic because there is no common difference.
2. ⓑThe sequence is arithmetic because there is a common difference. The common difference is 4.
Is the given sequence arithmetic? If so, find the common difference.
\[\{18,\ 16,\ 14,\ 12,\ 10,\ldots \}\]

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The sequence is arithmetic. The common difference is $-2.$
Is the given sequence arithmetic? If so, find the common difference.
\[\{1,\ 3,\ 6,\ 10,\ 15,\ldots \}\]

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The sequence is not arithmetic because $3-1\ne 6-3.$
Write the first five terms of the arithmetic sequence with ${a}_{1}=17$ and $d=-3$ .

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Adding $-3$ is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.
The first five terms are $\{17,\ 14,\ 11,\ 8,\ 5\}$
List the first five terms of the arithmetic sequence with ${a}_{1}=1$ and $d=5$ .

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$\{1,6,11,16,21\}$
Given ${a}_{1}=8$ and ${a}_{4}=14$ , find ${a}_{5}$ .

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The sequence can be written in terms of the initial term 8 and the common difference $d$ .
\[\{8,8+d,8+2d,8+3d\}\]
We know the fourth term equals 14; we know the fourth term has the form ${a}_{1}+3d=8+3d$ .
We can find the common difference $d$ .
\[\begin{array}{ll}{a}_{n}={a}_{1}+(n-1)d & \\ {a}_{4}={a}_{1}+3d & \\ {a}_{4}=8+3d & \text{Write the fourth term of the sequence in terms of}{a}_{1}\text{and}d. \\ 14=8+3d & \text{Substitute}14\text{for}{a}_{4}. \\ \ d=2 & \text{Solve for the common difference}.\end{array}\]
Find the fifth term by adding the common difference to the fourth term.
\[{a}_{5}={a}_{4}+2=16\]
Given ${a}_{3}=7$ and ${a}_{5}=17$ , find ${a}_{2}$ .

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${a}_{2}=2$
Write a recursive formula for the arithmetic sequence.
\[\{-18\text{, }-7\text{, }4\text{, }15\text{, }26\text{, }\ldots \}\]

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The first term is given as $-18$ . The common difference can be found by subtracting the first term from the second term.
\[d=-7-(-18)=11\]
Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.
\[\begin{array}{l}{a}_{1}=-18 \\ {a}_{n}={a}_{n-1}+11,\ \text{for }n\ge 2\end{array}\]
Write a recursive formula for the arithmetic sequence.
\[\{25\text{, }37\text{, }49\text{, }61\text{, }\ldots \}\]

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$\begin{array}{l}{a}_{1}=25 \\ {a}_{n}={a}_{n-1}+12,\ \text{for }n\ge 2\end{array}$
Write an explicit formula for the arithmetic sequence.
\[\{2\text{, }12\text{, }22\text{, }32\text{, }42\text{, }\ldots \}\]

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The common difference can be found by subtracting the first term from the second term.
\[\begin{array}{ll}d & ={a}_{2}-{a}_{1} \\ & =12-2 \\ & =10\end{array}\]
The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.
\[\begin{array}{l}{a}_{n}=2+10(n-1) \\ {a}_{n}=10n-8\end{array}\]
Write an explicit formula for the following arithmetic sequence.
\[\{50,47,44,41,\ldots \}\]

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${a}_{n}=53-3n$
Find the number of terms in the finite arithmetic sequence.
\[\{8\text{, }1\text{, }-6\text{, }...\text{, }-41\}\]

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The common difference can be found by subtracting the first term from the second term.
\[1-8=-7\]
The common difference is $-7$ . Substitute the common difference and the initial term of the sequence into the $n\text{th}$ term formula and simplify.
\[\begin{array}{l}{a}_{n}={a}_{1}+d(n-1) \\ {a}_{n}=8+(-7)(n-1) \\ {a}_{n}=15-7n\end{array}\]
Substitute $-41$ for ${a}_{n}$ and solve for $n$
\[\begin{array}{l}-41=15-7n \\ \ \ \ \ 8=n\end{array}\]
There are eight terms in the sequence.
Find the number of terms in the finite arithmetic sequence.
\[\{6\text{, }11\text{, }16\text{, }...\text{, }56\}\]

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There are 11 terms in the sequence.
A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week.
1. ⓐWrite a formula for the child’s weekly allowance in a given year.
2. ⓑWhat will the child’s allowance be when he is 16 years old?
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1. ⓐ
  The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2.
  
  Let $A$ be the amount of the allowance and $n$ be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get:
  \[{A}_{n}=1+2n\]
2. ⓑ
  We can find the number of years since age 5 by subtracting.
  \[16-5=11\]
  We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16.
  \[{A}_{11}=1+2(11)=23\]
  The child’s allowance at age 16 will be $23 per week.
A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today?

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The formula is ${T}_{n}=10+4n,$ and it will take her 42 minutes.
What is an arithmetic sequence?

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A sequence where each successive term of the sequence increases (or decreases) by a constant value.
How is the common difference of an arithmetic sequence found?
How do we determine whether a sequence is arithmetic?

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We find whether the difference between all consecutive terms is the same. This is the same as saying that the sequence has a common difference.
What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence?
Describe how linear functions and arithmetic sequences are similar. How are they different?

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Both arithmetic sequences and linear functions have a constant rate of change. They are different because their domains are not the same; linear functions are defined for all real numbers, and arithmetic sequences are defined for natural numbers or a subset of the natural numbers.
$\{5,11,17,23,29,...\}$
$\{0,\frac{1}{2},1,\frac{3}{2},2,...\}$

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The common difference is $\frac{1}{2}$
$\{11.4,9.3,7.2,5.1,3,...\}$
$\{4,16,64,256,1024,...\}$

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The sequence is not arithmetic because $16-4\ne 64-16.$
${a}_{1}=-25$ , $d=-9$
${a}_{1}=0$ , $d=\frac{2}{3}$

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$0,\ \frac{2}{3},\ \frac{4}{3},\ 2,\ \frac{8}{3}$
${a}_{1}=17,\ {a}_{7}=-31$
${a}_{13}=-60,\ {a}_{33}=-160$

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$0,-5,-10,-15,-20$
First term is 3, common difference is 4, find the 5^th term.
First term is 4, common difference is 5, find the 4^th term.

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${a}_{4}=19$
First term is 5, common difference is 6, find the 8^th term.
First term is 6, common difference is 7, find the 6^th term.

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${a}_{6}=41$

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How to: Arithmetic Sequences

Find the common difference for an arithmetic sequence.
Write terms of an arithmetic sequence.
Use a recursive formula for an arithmetic sequence.
Use an explicit formula for an arithmetic sequence.
Determine if a sequence is arithmetic (IA 12.2.1)
Find the general term (nth term) of an arithmetic sequence (IA 12.2.2)
Add the common difference to the first term to find the second term.
Add the common difference to the second term to find the third term.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours. Because its source is licensed for non-commercial use, this lesson is free, sells nothing, and is shared under the same licence.

More in Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

ⓐ To find the 21st term, use the formula with \({a}_{1}=3\), \(d=8\), and \(n=21\)	\({a}_{n}={a}_{1}+(n-1)d\)
Substitute	\({a}_{21}=3+(21-1)(8)\)
Simplify	\(\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}\)

ⓑ Let's first find \({a}_{1}\). Use the formula with \({a}_{3}=19\), \(d=5\), and \(n=3\). Substitute these values and simplify	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{3}={a}_{1}+(3-1)(5) \\ 19={a}_{1}+(2)(5) \\ 19={a}_{1}+10 \\ {a}_{1=9}\end{array}\)
To find the 11th term, use the formula with \({a}_{3}=9\), \(d=5\), and \(n=11\) Substitute these values and simplify	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59 \\ \end{array}\)
To find the general term, substitute \(a=9\) and \(d=5\) into the formula.	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59\end{array}\)

ⓐ To find the 21st term, use the formula with \({a}_{1}=3\), \(d=8\), and \(n=21\)	\({a}_{n}={a}_{1}+(n-1)d\)
Substitute	\({a}_{21}=3+(21-1)(8)\)
Simplify	\(\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}\)

ⓑ Let's first find \({a}_{1}\). Use the formula with \({a}_{3}=19\), \(d=5\), and \(n=3\). Substitute these values and simplify	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{3}={a}_{1}+(3-1)(5) \\ 19={a}_{1}+(2)(5) \\ 19={a}_{1}+10 \\ {a}_{1=9}\end{array}\)
To find the 11th term, use the formula with \({a}_{3}=9\), \(d=5\), and \(n=11\) Substitute these values and simplify	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59 \\ \end{array}\)
To find the general term, substitute \(a=9\) and \(d=5\) into the formula.	\(\begin{array}{l}{a}_{n}={a}_{1}+(n-1)d \\ {a}_{11}=9+(11-1)(5) \\ {a}_{11}=9+(10)(5) \\ {a}_{11}=59\end{array}\)

To find the fifteenth term, \({a}_{15},\) use the formula with \({a}_{1}=3\ \text{and}\ d=6\text{.}\)	\(\ {a}_{n}={a}_{1}+(n-1)d\)
Substitute in the values.	\(\ {a}_{15}=3+(15-1)6\)
Simplify.	\(\ {a}_{15}=3+(14)6\)
	\(\ {a}_{15}=87\)

recursive formula for nth term of an arithmetic sequence	\({a}_{n}={a}_{n-1}+d,n\ge 2\)
explicit formula for nth term of an arithmetic sequence	\(\begin{array}{l}{a}_{n}={a}_{1}+d(n-1)\end{array}\)

Find the difference of the consecutive terms.	\(\begin{array}{lllllllllll}5, & & \ 9, & & \ 13, & & \ 17 & & \ 21, & \ 25,\ldots \\ \\ \\ 9-5 & & \ 13-9 & & \ 17-13 & & \ 21-17 & & \ 25-21 & & \\ 4 & & \ 4 & & \ 4 & & \ 4 & & \ 4 & & \end{array}\)
	The sequence is arithmetic. The common difference is \(d=4\text{.}\)

Find the difference of the consecutive terms.	\(\begin{array}{lllllllllll}4, & & \ 9, & & \ 12, & & \ 17 & & \ 20, & \ 25,\ldots \\ \\ \\ 9-4 & & \ 12-9 & & \ 17-12 & & \ 20-17 & & \ 25-20 & & \\ 5 & & \ 3 & & \ 5 & & \ 3 & & \ 5 & & \end{array}\)
	The sequence is not arithmetic as all the differences between the consecutive terms are not the same. There is no common difference.

Find the difference of the consecutive terms.	\(\begin{array}{lllllllllll}10, & & \ 3, & & \ -4, & & \ -11 & & \ -18, & \ -25,\ldots & \\ \\ \\ 3-10 & & \ -4-3 & & \ -11-(-4) & & \ -18-(-11) & & \ -25-(-18) & & \\ -7 & & \ -7 & & \ -7 & & \ -7 & & \ -7 & & \end{array}\)
	The sequence is arithmetic. The common difference is \(d=-7\text{.}\)

To first find the first term, \({a}_{1},\) use the formula with \({a}_{7}=10,n=7,\ \text{and}\ d=-2.\)	\(\ {a}_{n}={a}_{1}+(n-1)d\)
Substitute in the values.	\(\ 10={a}_{1}+(7-1)(-2)\)
Simplify.	\(\ 10={a}_{1}+(6)(-2)\)
	\(\ 10={a}_{1}-12\)
	\(\ {a}_{1}=22\)
Find the twelfth term, \({a}_{12},\) using the formula with \({a}_{1}=22,n=12,\ \text{and}\ d=-2\text{.}\)	\(\ {a}_{n}={a}_{1}+(n-1)d\)
Substitute in the values.	\(\ {a}_{12}=22+(12-1)(-2)\)
Simplify.	\(\ {a}_{12}=22+(11)(-2)\)
	\(\ {a}_{12}=0\)
	\(\)The twelfth term of the sequence is 0, \({a}_{12}=0.\)
To find the general term, substitute the values into the formula.	\(\ {a}_{n}={a}_{1}+(n-1)d\)
	\(\ {a}_{n}=22+(n-1)(-2)\)
	\(\ {a}_{n}=22-2n+2\)
	\(\)The general term is \({a}_{n}=-2n+24.\)

	\(\ {a}_{n}={a}_{1}+(n-1)d\)
\(\text{Find}\ {a}_{n}\ \text{where}\ {a}_{1}=8,d=5\ \text{and}\ n=30.\)	\({a}_{30}=8+(30-1)5\)
Simplify.	\({a}_{30}=8+(29)5\)
	\({a}_{30}=153\)
\(\text{Knowing}\ {a}_{1}=8,n=30,\ \text{and}\ {a}_{30}=153,\) use the sum formula.	\(\ {S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\)
Substitute in the values.	\({S}_{30}=\frac{30}{2}(8+153)\)
Simplify.	\({S}_{30}=15(161)\)
Simplify.	\({S}_{30}=2,415\)