The Central Idea
Consider the sequence
\[ 5,\ 8,\ 11,\ 14,\ 17,\ldots \]Each term is \(3\) greater than the previous term:
\[ 8-5=3, \] \[ 11-8=3, \] \[ 14-11=3. \]Because the difference between consecutive terms is constant, this is an arithmetic sequence.
The constant difference is
\[ \boxed{d=3}. \]What Is an Arithmetic Sequence?
An arithmetic sequence is a sequence in which the difference between consecutive terms is constant.
If the common difference is \(d\), then
\[ a_n-a_{n-1}=d. \]Equivalently,
\[ \boxed{a_n=a_{n-1}+d}. \]This is the recursive structure of every arithmetic sequence.
The Common Difference
The constant amount added from one term to the next is called the common difference.
It is usually represented by
\[ d. \]To find it, subtract any term from the term immediately after it:
\[ \boxed{d=a_n-a_{n-1}}. \]For an arithmetic sequence, this calculation gives the same result for every pair of consecutive terms.
Example: Finding the Common Difference
Consider
\[ 12,\ 19,\ 26,\ 33,\ 40,\ldots \]Subtract consecutive terms:
\[ 19-12=7, \] \[ 26-19=7, \] \[ 33-26=7. \]Therefore,
\[ \boxed{d=7}. \]The Common Difference Can Be Negative
An arithmetic sequence does not have to increase.
Consider
\[ 20,\ 16,\ 12,\ 8,\ 4,\ldots \]Then
\[ 16-20=-4, \] \[ 12-16=-4. \]Thus,
\[ \boxed{d=-4}. \]A negative common difference produces a decreasing arithmetic sequence.
The Common Difference Can Be Zero
A constant sequence is also arithmetic.
For
\[ 6,\ 6,\ 6,\ 6,\ldots \]we have
\[ 6-6=0. \]Therefore,
\[ \boxed{d=0}. \]Every term remains unchanged.
Recognizing an Arithmetic Sequence
To determine whether a listed sequence is arithmetic:
- subtract each term from the term after it;
- compare the differences;
- if every difference is the same, the sequence is arithmetic.
For example,
\[ 4,\ 9,\ 14,\ 19,\ldots \]has differences
\[ 5,\ 5,\ 5,\ldots \]so it is arithmetic.
Example: A Sequence That Is Not Arithmetic
Consider
\[ 2,\ 4,\ 8,\ 16,\ldots \]The differences are
\[ 4-2=2, \] \[ 8-4=4, \] \[ 16-8=8. \]The differences are not constant.
Therefore, the sequence is
\[ \boxed{\text{not arithmetic}}. \]Later we will recognize this as a geometric sequence.
The Recursive Definition
An arithmetic sequence is naturally described recursively.
If the first term is \(a_1\) and the common difference is \(d\), then
\[ \boxed{ a_1=\text{given},\qquad a_n=a_{n-1}+d\quad(n\geq2) }. \]The initial condition supplies the starting point, and the recurrence adds the same amount at every step.
Example: Writing a Recursive Definition
Consider
\[ 7,\ 12,\ 17,\ 22,\ldots \]The first term is
\[ a_1=7, \]and the common difference is
\[ d=5. \]Therefore,
\[ \boxed{ a_1=7,\qquad a_n=a_{n-1}+5\quad(n\geq2) }. \]Building the Explicit Formula
Suppose the first term is \(a_1\) and the common difference is \(d\).
Then
\[ a_2=a_1+d. \]The third term is
\[ a_3=a_1+2d. \]The fourth term is
\[ a_4=a_1+3d. \]The \(n\)th term is reached after \(n-1\) equal steps.
Therefore,
\[ \boxed{a_n=a_1+(n-1)d}. \]The Explicit Formula
The standard explicit formula for an arithmetic sequence indexed from \(1\) is
\[ \boxed{a_n=a_1+(n-1)d}. \]Here:
- \(a_n\) is the \(n\)th term;
- \(a_1\) is the first term;
- \(n\) is the index;
- \(d\) is the common difference.
Why the Formula Uses \(n-1\)
The first term already exists before any common differences are added.
To reach \(a_2\), add \(d\) once.
To reach \(a_3\), add \(d\) twice.
To reach \(a_n\), add \(d\)
\[ n-1 \]times.
That is why the formula is
\[ a_n=a_1+(n-1)d \]rather than \(a_1+nd\).
Example: Writing an Explicit Formula
Consider
\[ 10,\ 14,\ 18,\ 22,\ldots \]We have
\[ a_1=10 \]and
\[ d=4. \]Substitute into
\[ a_n=a_1+(n-1)d. \]Then
\[ a_n=10+4(n-1). \]Simplify:
\[ a_n=10+4n-4 \] \[ \boxed{a_n=4n+6}. \]Equivalent Explicit Forms
The formulas
\[ a_n=10+4(n-1) \]and
\[ a_n=4n+6 \]define the same sequence.
The first form makes the first term and common difference visible.
The second resembles the familiar slope-intercept form of a linear function.
Both representations are useful.
Finding a Distant Term
Suppose
\[ a_1=6,\qquad d=8. \]Find \(a_{25}\).
Use
\[ a_n=a_1+(n-1)d. \]Then
\[ a_{25}=6+(25-1)(8) \] \[ =6+24(8) \] \[ =6+192 \] \[ =\boxed{198}. \]Finding the Index of a Known Term
Suppose
\[ a_n=5+3(n-1). \]At what index does the value \(47\) occur?
Set
\[ 47=5+3(n-1). \]Subtract \(5\):
\[ 42=3(n-1). \]Divide by \(3\):
\[ 14=n-1. \]Therefore,
\[ \boxed{n=15}. \]Determining Whether a Value Is a Term
Let
\[ a_n=4n+1,\qquad n\geq1. \]Is \(30\) a term of the sequence?
Solve
\[ 4n+1=30. \]Then
\[ 4n=29 \]and
\[ n=\frac{29}{4}. \]Since the allowed indices are positive integers, \(30\) is not a term.
\[ \boxed{30\notin\{a_n:n\geq1\}}. \]Finding a Missing Term
Suppose
\[ 8,\ 13,\ \square,\ 23,\ 28 \]is arithmetic.
The common difference is
\[ 13-8=5. \]Therefore, the missing term is
\[ 13+5=\boxed{18}. \]Finding Several Missing Terms
Suppose
\[ 4,\ \square,\ \square,\ 19 \]is arithmetic.
There are three equal steps from \(4\) to \(19\).
The total change is
\[ 19-4=15. \]Therefore,
\[ d=\frac{15}{3}=5. \]The sequence is
\[ \boxed{4,\ 9,\ 14,\ 19}. \]Arithmetic Means
A number inserted between two values so that the resulting terms form an arithmetic sequence is called an arithmetic mean.
To insert one arithmetic mean between \(6\) and \(14\), find the midpoint:
\[ \frac{6+14}{2}=10. \]Thus,
\[ \boxed{6,\ 10,\ 14} \]is an arithmetic sequence.
Several Arithmetic Means
Suppose we want to insert three arithmetic means between \(2\) and \(22\).
The completed sequence has five terms:
\[ 2,\ \square,\ \square,\ \square,\ 22. \]There are four equal steps.
Therefore,
\[ d=\frac{22-2}{4}=5. \]The sequence is
\[ \boxed{2,\ 7,\ 12,\ 17,\ 22}. \]Finding the Common Difference from Two Terms
Suppose
\[ a_4=17 \]and
\[ a_{10}=41. \]From index \(4\) to index \(10\) there are
\[ 10-4=6 \]equal steps.
The total change in value is
\[ 41-17=24. \]Therefore,
\[ d=\frac{24}{6} \] \[ =\boxed{4}. \]A General Two-Term Formula
For any two terms \(a_m\) and \(a_n\) of an arithmetic sequence,
\[ a_n-a_m=(n-m)d. \]Therefore,
\[ \boxed{a_n=a_m+(n-m)d}. \]This form is useful when a term other than \(a_1\) is already known.
Example: Using a Known Later Term
Suppose
\[ a_6=25 \]and
\[ d=3. \]Find \(a_{14}\).
Use
\[ a_{14}=a_6+(14-6)d. \]Then
\[ a_{14}=25+8(3) \] \[ =25+24 \] \[ =\boxed{49}. \]Finding the First Term from Another Term
Suppose
\[ a_8=31 \]and
\[ d=4. \]Use
\[ a_8=a_1+(8-1)d. \]Then
\[ 31=a_1+7(4). \]Therefore,
\[ 31=a_1+28 \]and
\[ \boxed{a_1=3}. \]Finding an Explicit Formula from Two Terms
Suppose
\[ a_3=11 \]and
\[ a_7=27. \]First find the common difference:
\[ d=\frac{27-11}{7-3} \] \[ =\frac{16}{4} \] \[ =4. \]Now use
\[ a_3=a_1+2d. \]Thus,
\[ 11=a_1+8, \]so
\[ a_1=3. \]Therefore,
\[ a_n=3+4(n-1) \] \[ \boxed{a_n=4n-1}. \]Arithmetic Sequences and Linear Functions
The explicit formula
\[ a_n=a_1+(n-1)d \]can be expanded:
\[ a_n=a_1+dn-d. \]Rearranging gives
\[ \boxed{a_n=dn+(a_1-d)}. \]This has the same algebraic form as
\[ y=mx+b. \]The common difference \(d\) plays the role of slope.
The Common Difference Is a Discrete Rate of Change
In an arithmetic sequence,
\[ \frac{a_n-a_{n-1}}{n-(n-1)} = \frac{a_n-a_{n-1}}{1} = d. \]Thus the common difference is the rate of change in the sequence value for each one-unit increase in the index.
This is the discrete analogue of constant slope.
A Sequence Table
Consider
\[ a_n=3n+2. \]| Index \(n\) | Term \(a_n\) |
|---|---|
| \(1\) | \(5\) |
| \(2\) | \(8\) |
| \(3\) | \(11\) |
| \(4\) | \(14\) |
| \(5\) | \(17\) |
The input increases by \(1\), while the output increases by \(3\).
The Discrete Graph
The sequence
\[ a_n=3n+2 \]produces the points
\[ (1,5),\ (2,8),\ (3,11),\ (4,14),\ldots \]These points lie on the line
\[ y=3x+2. \]But the sequence itself contains only the specified indexed points.
The line describes a continuous extension; the sequence remains discrete.
Constant First Differences
Arithmetic sequences are characterized by constant first differences.
Consider
\[ 4,\ 10,\ 16,\ 22,\ 28. \]The first differences are
\[ 6,\ 6,\ 6,\ 6. \]Since the first differences are constant, the sequence is arithmetic.
Later, finite differences will help us identify polynomial sequences whose first differences are not constant.
Indexing from Zero
If an arithmetic sequence begins at \(n=0\), the standard formula changes slightly.
If \(a_0\) is the starting term, then
\[ \boxed{a_n=a_0+nd}. \]This occurs because moving from index \(0\) to index \(n\) requires exactly \(n\) steps.
Example: Arithmetic Sequence Beginning at Zero
Let
\[ a_0=7 \]and
\[ d=4. \]Then
\[ a_n=7+4n. \]The sequence begins
\[ a_0=7, \] \[ a_1=11, \] \[ a_2=15, \] \[ a_3=19. \]Thus,
\[ \boxed{7,\ 11,\ 15,\ 19,\ldots}. \]A More General Indexing Formula
Suppose the sequence begins at some index \(k\), with known value \(a_k\).
Then
\[ \boxed{a_n=a_k+(n-k)d}. \]This formula simply counts how many equal index steps separate \(k\) and \(n\).
The familiar formula
\[ a_n=a_1+(n-1)d \]is the special case \(k=1\).
Arithmetic Sequences in Applications
Arithmetic sequences appear whenever a quantity changes by the same amount at each discrete step.
Examples include:
- adding the same amount to an account every month;
- increasing production by a fixed number of units each cycle;
- moving a fixed distance at each step;
- reducing inventory by a constant number each period;
- arranging objects in rows whose sizes increase by a fixed amount;
- following a linear depreciation schedule over discrete periods.
Example: A Savings Pattern
Suppose a person begins a savings schedule by depositing \(50\) dollars in the first month and increases the monthly deposit by \(10\) dollars each month.
The monthly deposit amounts form the arithmetic sequence
\[ 50,\ 60,\ 70,\ 80,\ldots \]Here,
\[ a_1=50 \]and
\[ d=10. \]The deposit in month \(n\) is
\[ a_n=50+10(n-1). \]Example: Fixed Decrease
A storage tank contains \(900\) liters at the first recorded time and loses \(35\) liters during each equal time interval.
The recorded amounts are
\[ 900,\ 865,\ 830,\ 795,\ldots \]so
\[ d=-35. \]The \(n\)th recorded amount is
\[ a_n=900-35(n-1). \]Common Error: Using a Ratio Instead of a Difference
Arithmetic sequences are identified by constant differences, not constant ratios.
For
\[ 3,\ 6,\ 9,\ 12,\ldots \]the differences are
\[ 3,\ 3,\ 3. \]The ratios are not constant:
\[ \frac63=2, \] \[ \frac96=\frac32. \]The sequence is arithmetic because of the constant difference.
Common Error: Reversing the Subtraction
The common difference is found by subtracting the earlier term from the later term:
\[ d=a_n-a_{n-1}. \]For
\[ 20,\ 15,\ 10,\ 5,\ldots \]we have
\[ d=15-20=-5. \]Writing \(20-15=5\) would reverse the direction of change.
Common Error: Using \(nd\) Instead of \((n-1)d\)
Suppose
\[ a_1=5,\qquad d=3. \]The correct formula is
\[ a_n=5+3(n-1). \]If we incorrectly use
\[ 5+3n, \]then at \(n=1\) we get
\[ 8 \]instead of the required first term \(5\).
Common Error: Assuming Every Linear-Looking List Is Arithmetic
A visual impression is not enough.
Consider
\[ 2,\ 5,\ 9,\ 12,\ldots \]The differences are
\[ 3,\ 4,\ 3. \]Since they are not constant, the sequence is not arithmetic.
Always calculate the differences.
Common Error: Forgetting the Starting Index
The formula
\[ a_n=a_1+(n-1)d \]assumes indexing begins at \(1\).
If indexing begins at \(0\), use
\[ a_n=a_0+nd. \]More generally, if the known term is \(a_k\), use
\[ a_n=a_k+(n-k)d. \]A Reliable Procedure for Analyzing an Arithmetic Sequence
- Identify the indexing convention.
- Calculate consecutive differences.
- Verify that the difference is constant.
- Record the common difference \(d\).
- Identify a known term such as \(a_1\).
- Write the recursive rule if needed.
- Write the explicit formula if needed.
- Substitute known information to solve for missing quantities.
- Check the result against the original sequence.
Example: Full Arithmetic-Sequence Workflow
Consider
\[ 18,\ 13,\ 8,\ 3,\ -2,\ldots \]First find the common difference:
\[ 13-18=-5. \]Thus,
\[ d=-5. \]The recursive definition is
\[ a_1=18,\qquad a_n=a_{n-1}-5. \]The explicit definition is
\[ a_n=18-5(n-1). \]Simplify:
\[ a_n=18-5n+5 \] \[ \boxed{a_n=23-5n}. \]Example: Recovering a Sequence from Two Terms
Suppose
\[ a_5=24 \]and
\[ a_{12}=52. \]First find the common difference:
\[ d=\frac{52-24}{12-5} \] \[ =\frac{28}{7} \] \[ =4. \]Now use
\[ a_5=a_1+4d. \]Therefore,
\[ 24=a_1+16 \]and
\[ a_1=8. \]The explicit rule is
\[ a_n=8+4(n-1) \] \[ \boxed{a_n=4n+4}. \]Practice 1: Find the Common Difference
Find the common difference of
\[ 9,\ 15,\ 21,\ 27,\ldots \]Practice 2: Decreasing Arithmetic Sequence
Find the common difference of
\[ 30,\ 24,\ 18,\ 12,\ldots \]Practice 3: Write a Recursive Definition
Write a recursive definition for
\[ 11,\ 16,\ 21,\ 26,\ldots \]assuming indexing begins at \(1\).
Practice 4: Write an Explicit Formula
Write an explicit formula for
\[ 4,\ 10,\ 16,\ 22,\ldots \]assuming indexing begins at \(1\).
Practice 5: Find a Distant Term
Let
\[ a_1=7,\qquad d=5. \]Find
\[ a_{20}. \]Practice 6: Find the Index
Let
\[ a_n=3n+4. \]At what index does the value \(49\) occur?
Practice 7: Find the Missing Terms
Complete the arithmetic sequence:
\[ 5,\ \square,\ \square,\ 26. \]Practice 8: Find the Common Difference from Two Terms
Suppose
\[ a_4=14 \]and
\[ a_{10}=38. \]Find \(d\).
Practice 9: Find the Explicit Formula from Two Terms
Suppose
\[ a_2=9 \]and
\[ a_6=25. \]Find an explicit formula for \(a_n\).
Practice 10: Identify the Error
A student says that
\[ 2,\ 6,\ 12,\ 20,\ldots \]is arithmetic because the terms are increasing regularly.
Explain the error.
Practice Answers
Practice 1
\[ d=15-9 \] \[ =\boxed{6}. \]Practice 2
\[ d=24-30 \] \[ =\boxed{-6}. \]Practice 3
The first term is \(11\), and the common difference is \(5\).
\[ \boxed{ a_1=11,\qquad a_n=a_{n-1}+5\quad(n\geq2) }. \]Practice 4
Here,
\[ a_1=4,\qquad d=6. \]Therefore,
\[ a_n=4+6(n-1) \] \[ =4+6n-6 \] \[ \boxed{a_n=6n-2}. \]Practice 5
\[ a_{20}=7+(20-1)(5) \] \[ =7+95 \] \[ =\boxed{102}. \]Practice 6
Solve
\[ 3n+4=49. \]Then
\[ 3n=45 \]and
\[ \boxed{n=15}. \]Practice 7
There are three equal steps from \(5\) to \(26\).
\[ d=\frac{26-5}{3} \] \[ =7. \]Therefore,
\[ \boxed{5,\ 12,\ 19,\ 26}. \]Practice 8
\[ d=\frac{38-14}{10-4} \] \[ =\frac{24}{6} \] \[ =\boxed{4}. \]Practice 9
First find the common difference:
\[ d=\frac{25-9}{6-2} \] \[ =\frac{16}{4} \] \[ =4. \]Since
\[ a_2=a_1+d, \]we have
\[ 9=a_1+4, \]so
\[ a_1=5. \]Therefore,
\[ a_n=5+4(n-1) \] \[ \boxed{a_n=4n+1}. \]Practice 10
Increasing terms do not automatically form an arithmetic sequence.
Calculate the differences:
\[ 6-2=4, \] \[ 12-6=6, \] \[ 20-12=8. \]Because
\[ 4,\ 6,\ 8 \]are not equal, the sequence is
\[ \boxed{\text{not arithmetic}}. \]What to Remember
An arithmetic sequence changes by a constant difference:
\[ \boxed{d=a_n-a_{n-1}}. \]Its recursive form is
\[ \boxed{ a_1=\text{given},\qquad a_n=a_{n-1}+d }. \]Its standard explicit form is
\[ \boxed{a_n=a_1+(n-1)d}. \]More generally,
\[ \boxed{a_n=a_k+(n-k)d}. \]- The common difference may be positive, negative, or zero.
- Constant first differences identify arithmetic sequences.
- The common difference is a discrete rate of change.
- Arithmetic sequences correspond to linear relationships on discrete domains.
- The starting index determines the correct explicit formula.
- Known terms can be used to recover \(d\), \(a_1\), or the general term.
Where We Go Next
Arithmetic sequences are generated by repeated addition or subtraction. Another fundamental family is generated by repeated multiplication.
The next lesson, Geometric Sequences, introduces the common ratio, develops recursive and explicit formulas for geometric sequences, and connects repeated multiplication with exponential relationships.