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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 shown in .

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

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

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.
ⓐ $5,9,13,17,21,25\text{,}\ \text{\ldots }$
ⓑ $4,9,12,17,20,25\text{,}\ \text{\ldots }$

උත්තරය හෙළි කරන්න

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 $d=-6.$

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

ⓐ 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. ⓑ $\{-3,1,5,9,13,...\}$
උත්තරය හෙළි කරන්න
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.
\[\{18,\ 16,\ 14,\ 12,\ 10,\ldots \}\]

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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 $d=-3$ .

උත්තරය හෙළි කරන්න

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 $d=5$ .

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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 ${a}_{2}$ .

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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.
\[\{25\text{, }37\text{, }49\text{, }61\text{, }\text{\ldots }\}\]

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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.
\[\{50,47,44,41,\ldots \}\]

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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.
\[\{6\text{, }11\text{, }16\text{, }...\text{, }56\}\]

උත්තරය හෙළි කරන්න

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?
උත්තරය හෙළි කරන්න
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?

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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?

උත්තරය හෙළි කරන්න

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?

උත්තරය හෙළි කරන්න

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.
$\{5,11,17,23,29,...\}$
$\{0,\frac{1}{2},1,\frac{3}{2},2,...\}$

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

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

උත්තරය හෙළි කරන්න

$0,-5,-10,-15,-20$
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.

උත්තරය හෙළි කරන්න

${a}_{4}=19$
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.

උත්තරය හෙළි කරන්න

${a}_{6}=41$

Symbols used here

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

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.

ඔයාගේම උත්සහ කරන්න

Parts of this page are adapted from OpenStax College 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.

තවත් 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 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	\(\begin{array}{l}{a}_{21}=3+(20)(8) \\ {a}_{21}=3+160 \\ {a}_{21}=163\end{array}\)

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	\(\begin{array}{l}{a}_{n}={a}_{1}+d(n-1)\end{array}\)