02.8 Arithmetic Sequences
Estimated study time: 45–55 minutes
Prerequisites: Basic algebra, substitution into formulas, and solving linear equations.
Learning outcome: Students can identify arithmetic sequences, calculate their terms, and solve problems involving common differences.
1. What is a numerical sequence?
A numerical sequence is an ordered list of numbers.
For example:
\[ 3,\ 6,\ 9,\ 12,\ 15,\ldots \]
Each number in the sequence is called a term.
The order matters. The sequences:
\[ 2,\ 4,\ 6 \]
and:
\[ 6,\ 4,\ 2 \]
have the same numbers but are different sequences because the terms appear in a different order.
Finite and infinite sequences
A finite sequence has a fixed number of terms.
Example:
\[ 4,\ 8,\ 12,\ 16,\ 20. \]
This sequence has five terms.
An infinite sequence continues indefinitely.
Example:
\[ 4,\ 8,\ 12,\ 16,\ 20,\ldots \]
The dots indicate that the sequence continues.
Important: A few displayed terms do not uniquely determine a sequence. When continuing a pattern, use the rule stated or clearly intended in the question.
2. Terms, indices, and sequence notation
We usually label the terms:
\[ a_1,\ a_2,\ a_3,\ldots,\ a_n,\ldots \]
Here:
- \(a_1\) is the first term.
- \(a_2\) is the second term.
- \(a_n\) is the term in position \(n\).
- \(n\) is the index, or position number.
In this lesson, indexing begins at:
\[ n=1. \]
Example 1: Reading sequence notation
For the sequence:
\[ 5,\ 9,\ 13,\ 17,\ 21,\ldots \]
we have:
\[ a_1=5, \]\[ a_2=9, \]\[ a_3=13, \]\[ a_4=17, \]\[ a_5=21. \]
The statement:
\[ a_4=17 \]
means:
The fourth term has the value \(17\).
The index \(4\) and the term value \(17\) represent different things.
| Notation | Meaning | Example |
|---|---|---|
| \(a_1\) | First term | \(a_1=5\) |
| \(a_n\) | Term in position \(n\) | \(a_4=17\) |
| \(n\) | Index or position number | \(1,2,3,\ldots\) |
| \((a_n)\) | The sequence as a whole | \((a_n)=(5,9,13,17,\ldots)\) |
A sequence can be described by a formula
Suppose:
\[ a_n=3n+2, \qquad n=1,2,3,\ldots \]
Substituting consecutive positive integers gives:
\[ a_1=3(1)+2=5, \]\[ a_2=3(2)+2=8, \]\[ a_3=3(3)+2=11. \]
Therefore, the sequence begins:
\[ 5,\ 8,\ 11,\ 14,\ldots \]
Unlike a function whose inputs may include all real numbers, a sequence uses discrete index values.
3. What is an arithmetic sequence?
An arithmetic sequence is a sequence in which the difference between every pair of consecutive terms is constant.
This fixed difference is called the common difference, written \(d\).
\[ d=a_{n+1}-a_n. \]
Equivalently:
\[ a_{n+1}=a_n+d. \]
In words:
Add the same number each time to obtain the next term.
Example 2: Positive common difference
Consider:
\[ 4,\ 7,\ 10,\ 13,\ldots \]
The consecutive differences are:
\[ 7-4=3, \]\[ 10-7=3, \]\[ 13-10=3. \]
Therefore, this is an arithmetic sequence with:
\[ d=3. \]
Example 3: Negative common difference
Consider:
\[ 12,\ 7,\ 2,\ -3,\ldots \]
The differences are:
\[ 7-12=-5, \]\[ 2-7=-5, \]\[ -3-2=-5. \]
Therefore:
\[ d=-5. \]
Adding a negative common difference makes the terms decrease.
Example 4: Zero common difference
Consider:
\[ 6,\ 6,\ 6,\ 6,\ldots \]
Each difference is zero:
\[ d=0. \]
A constant sequence is an arithmetic sequence.
Direction of an arithmetic sequence
- If \(d>0\), the sequence is strictly increasing.
- If \(d<0\), the sequence is strictly decreasing.
- If \(d=0\), the sequence is constant.
Arithmetic sequences may contain negative numbers, fractions, or decimals.
For example:
\[ \frac12,\ \frac34,\ 1,\ \frac54,\ldots \]
has common difference:
\[ d=\frac14. \]
4. Recognizing arithmetic patterns
To test whether a displayed numerical pattern is arithmetic, subtract each term from the next.
Always use the same subtraction order:
\[ \text{next term}-\text{previous term}. \]
Worked example 1: An arithmetic pattern
Determine whether the following terms form an arithmetic sequence:
\[ -2,\ 3,\ 8,\ 13,\ 18. \]
Calculate:
\[ 3-(-2)=5, \]\[ 8-3=5, \]\[ 13-8=5, \]\[ 18-13=5. \]
All consecutive differences are equal, so the finite sequence is arithmetic:
\[ d=5. \]
Worked example 2: A non-arithmetic pattern
Consider:
\[ 2,\ 4,\ 8,\ 16. \]
The differences are:
\[ 4-2=2, \]\[ 8-4=4, \]\[ 16-8=8. \]
The differences are not constant, so the sequence is not arithmetic.
Although each term is twice the previous term, a constant multiplication factor does not make a sequence arithmetic.
| Sequence | Consecutive differences | Arithmetic? |
|---|---|---|
| \(3,7,11,15\) | \(4,4,4\) | Yes; \(d=4\) |
| \(10,7,4,1\) | \(-3,-3,-3\) | Yes; \(d=-3\) |
| \(5,5,5,5\) | \(0,0,0\) | Yes; \(d=0\) |
| \(1,4,9,16\) | \(3,5,7\) | No |
5. The general term formula
For an arithmetic sequence with first term \(a_1\) and common difference \(d\):
\[ a_n=a_1+(n-1)d. \]
This formula calculates the term in any specified position directly.
Why does the formula contain \(n-1\)?
Starting from the first term:
\[ a_1=a_1, \]\[ a_2=a_1+d, \]\[ a_3=a_1+2d, \]\[ a_4=a_1+3d. \]
To reach the second term, take one step.
To reach the fourth term, take three steps.
To reach the \(n\)th term, take:
\[ n-1 \]
steps.
Each step adds \(d\), giving:
\[ a_n=a_1+(n-1)d. \]
Worked example 3: Finding a distant term
Find the twentieth term of:
\[ 5,\ 9,\ 13,\ 17,\ldots \]
Identify:
\[ a_1=5, \qquad d=4. \]
Use:
\[ a_n=5+(n-1)4. \]
For \(n=20\):
\[ a_{20}=5+(20-1)4. \]\[ a_{20}=5+76=81. \]
The twentieth term is:
\[ 81. \]
Worked example 4: A decreasing sequence
An arithmetic sequence has:
\[ a_1=18, \qquad d=-3. \]
Find \(a_8\).
\[ a_8=18+(8-1)(-3). \]\[ a_8=18-21=-3. \]
The eighth term is:
\[ -3. \]
Writing a simplified general formula
For:
\[ a_1=5, \qquad d=4, \]
we can simplify:
\[ a_n=5+4(n-1). \]\[ a_n=4n+1. \]
Both formulas describe the same sequence.
6. Finding a missing term
6.1 One missing term between two consecutive neighbors
Suppose three consecutive terms of an arithmetic sequence are:
\[ p,\ q,\ r. \]
Their differences must be equal:
\[ q-p=r-q. \]
Therefore:
\[ 2q=p+r, \]
so:
\[ q=\frac{p+r}{2}. \]
The middle term is the arithmetic mean of its neighbors.
Worked example 5
Find the missing term:
\[ 7,\ \square,\ 19. \]
Because these are consecutive terms of an arithmetic sequence:
\[ \text{Missing term}=\frac{7+19}{2}=13. \]
The sequence is:
\[ 7,\ 13,\ 19. \]
Checking:
\[ 13-7=6, \qquad 19-13=6. \]
6.2 Several missing terms
Count the number of steps between the known terms.
Worked example 6
Insert three terms between \(4\) and \(20\) to form an arithmetic sequence:
\[ 4,\ \square,\ \square,\ \square,\ 20. \]
There are five terms, so there are four steps from the first to the last.
Therefore:
\[ d=\frac{20-4}{4}=4. \]
The completed sequence is:
\[ 4,\ 8,\ 12,\ 16,\ 20. \]
Do not divide by the number of missing terms. Divide by the number of steps between the known terms.
7. Finding the common difference
7.1 From consecutive terms
Use:
\[ d=a_{n+1}-a_n. \]
For example, if:
\[ a_5=17, \qquad a_6=12, \]
then:
\[ d=12-17=-5. \]
7.2 From two non-consecutive terms
Suppose we know \(a_p\) and \(a_q\), where \(q>p\).
There are:
\[ q-p \]
steps between them.
Therefore:
\[ a_q-a_p=(q-p)d. \]
Hence:
\[ d=\frac{a_q-a_p}{q-p}. \]
Worked example 7
An arithmetic sequence satisfies:
\[ a_3=11, \qquad a_8=31. \]
Find the common difference and the first term.
Step 1: Find the common difference.
\[ d=\frac{31-11}{8-3}. \]\[ d=\frac{20}{5}=4. \]
Step 2: Find the first term.
Since:
\[ a_3=a_1+2d, \]
we have:
\[ 11=a_1+2(4). \]
Therefore:
\[ a_1=3. \]
The general term is:
\[ a_n=3+4(n-1). \]
Simplifying:
\[ a_n=4n-1. \]
Finding a term from another known term
You do not always need to find \(a_1\).
Starting from any known term \(a_p\):
\[ a_n=a_p+(n-p)d. \]
For the sequence above:
\[ a_{12}=a_8+(12-8)d. \]\[ a_{12}=31+4(4)=47. \]
8. Determining the number of terms
For a finite arithmetic sequence with \(n\) terms:
\[ a_n=a_1+(n-1)d. \]
When \(d\neq 0\), rearrange:
\[ a_n-a_1=(n-1)d, \]\[ \frac{a_n-a_1}{d}=n-1. \]
Therefore:
\[ n=\frac{a_n-a_1}{d}+1. \]
The result must be a positive integer.
Worked example 8: Positive common difference
How many terms are in:
\[ 7,\ 11,\ 15,\ldots,\ 63? \]
Identify:
\[ a_1=7, \qquad d=4, \qquad a_n=63. \]
Then:
\[ n=\frac{63-7}{4}+1. \]\[ n=\frac{56}{4}+1=14+1=15. \]
There are:
\[ 15\text{ terms}. \]
Worked example 9: Negative common difference
How many terms are in:
\[ 30,\ 26,\ 22,\ldots,\ -10? \]
Here:
\[ a_1=30, \qquad d=-4, \qquad a_n=-10. \]
Therefore:
\[ n=\frac{-10-30}{-4}+1. \]\[ n=\frac{-40}{-4}+1=11. \]
There are:
\[ 11\text{ terms}. \]
What if the common difference is zero?
The formula for \(n\) above divides by \(d\), so it cannot be used when:
\[ d=0. \]
In a constant sequence, the first and last term values alone do not determine the number of terms.
For example, both sequences:
\[ 5,\ 5,\ 5 \]
and:
\[ 5,\ 5,\ 5,\ 5,\ 5 \]
begin and end with \(5\), but have different lengths.
9. Checking whether a number belongs to a sequence
To check whether a number is a term:
- Set the general term equal to that number.
- Solve for \(n\).
- Check that \(n\) is a positive integer.
- For a finite sequence, also check that \(n\) does not exceed its length.
Worked example 10
Does \(50\) belong to:
\[ 4,\ 10,\ 16,\ 22,\ldots? \]
The general term is:
\[ a_n=4+6(n-1). \]
Set:
\[ 4+6(n-1)=50. \]
Then:
\[ 6(n-1)=46, \]\[ n=1+\frac{46}{6}=\frac{26}{3}. \]
This is not an integer.
Therefore, \(50\) is not a term of the sequence.
Comparison
Check whether \(52\) belongs:
\[ 4+6(n-1)=52. \]\[ 6(n-1)=48. \]\[ n=9. \]
Therefore:
\[ a_9=52. \]
10. Simple applications
Arithmetic sequences model situations involving the same added or subtracted amount at each step.
Worked example 11: Seats in an auditorium
The first row has \(18\) seats. Each following row has \(3\) more seats than the previous row.
How many seats are in the twelfth row?
Let \(a_n\) be the number of seats in row \(n\).
Then:
\[ a_1=18, \qquad d=3. \]
Therefore:
\[ a_{12}=18+(12-1)3. \]\[ a_{12}=18+33=51. \]
The twelfth row has:
\[ 51\text{ seats}. \]
This is the number of seats in that row, not the total across all twelve rows.
Worked example 12: Weekly saving amounts
A student saves \(50\) RMB in week \(1\). Each week, the student saves \(10\) RMB more than in the previous week.
How much does the student save during week \(8\)?
Let \(a_n\) be the amount saved during week \(n\).
\[ a_1=50, \qquad d=10. \]
Then:
\[ a_8=50+(8-1)10. \]\[ a_8=120. \]
The student saves:
\[ 120\text{ RMB during week }8. \]
This does not represent the cumulative savings after eight weeks.
Worked example 13: A decreasing quantity
A tank contains \(120\) liters of water initially. Exactly \(8\) liters are removed each minute.
How much remains after \(10\) minutes?
The initial reading corresponds to time zero. If we label it as \(a_1\), then:
\[ a_1=120. \]
After one minute:
\[ a_2=112. \]
After ten minutes, we need the eleventh reading:
\[ a_{11}=120+(11-1)(-8). \]\[ a_{11}=120-80=40. \]
There are:
\[ 40\text{ liters} \]
remaining after ten minutes.
Indexing matters: The first listed reading may represent time zero rather than one elapsed time period. The model applies only while the tank still contains water.
| Purpose | Formula | Condition or reminder |
|---|---|---|
| Find a term from the first term | \(a_n=a_1+(n-1)d\) | Indexing starts at \(1\). |
| Find the common difference | \(d=a_{n+1}-a_n\) | Use consecutive terms. |
| Use two known terms | \(d=\frac{a_q-a_p}{q-p}\) | \(q>p\) |
| Find a term from another term | \(a_n=a_p+(n-p)d\) | Count the steps between indices. |
| Find a middle term | \(a_n=\frac{a_{n-1}+a_{n+1}}{2}\) | The three terms must be consecutive. |
| Determine the number of terms | \(n=\frac{a_n-a_1}{d}+1\) | Requires \(d\neq0\) and a positive integer result. |
12. Common mistakes
Mistake 1: Using \(n\) instead of \(n-1\)
Incorrect:
\[ a_n=a_1+nd. \]
Correct:
\[ a_n=a_1+(n-1)d. \]
There are \(n-1\) steps from the first term to the \(n\)th term.
Mistake 2: Reversing the subtraction order
For:
\[ 10,\ 7,\ 4,\ldots \]
the common difference is:
\[ d=7-10=-3, \]
not \(3\).
Mistake 3: Dividing by the number of terms rather than the number of steps
Between the third and eighth terms, there are:
\[ 8-3=5 \]
steps.
Use:
\[ a_8-a_3=5d. \]
Mistake 4: Accepting a non-integer index
A term cannot occupy position:
\[ n=4.5. \]
If solving for \(n\) does not give a positive integer, the proposed value is not a term.
Mistake 5: Confusing a term with a total
The eighth weekly saving amount and the total saved over eight weeks are different quantities.
The formula:
\[ a_n=a_1+(n-1)d \]
finds an individual term.
Mistake 6: Treating percentage growth as arithmetic growth
Adding \(10\) each time produces an arithmetic sequence.
Increasing by \(10\%\) each time generally does not, because the amount added changes.
13. Practice quiz
Try these questions before reading the answers.
Question 1: Terms and indices
For:
\[ -3,\ 2,\ 7,\ 12,\ldots \]
state:
\[ a_1,\qquad a_3,\qquad d. \]
Question 2: Recognizing arithmetic sequences
Which of the following finite sequences are arithmetic?
A.
\[ 2,\ 5,\ 8,\ 11 \]
B.
\[ 2,\ 4,\ 8,\ 16 \]
C.
\[ 9,\ 9,\ 9,\ 9 \]
D.
\[ 10,\ 6,\ 2,\ -2 \]
Question 3: Finding a term
An arithmetic sequence has:
\[ a_1=7, \qquad d=3. \]
Find:
\[ a_{15}. \]
Question 4: General term
Find a simplified formula for the \(n\)th term of:
\[ 12,\ 8,\ 4,\ 0,\ldots \]
Question 5: One missing term
Find the missing term in this arithmetic sequence:
\[ -4,\ \square,\ 10. \]
Question 6: Several missing terms
Insert two terms between \(5\) and \(20\) to form an arithmetic sequence:
\[ 5,\ \square,\ \square,\ 20. \]
Question 7: Two known terms
An arithmetic sequence satisfies:
\[ a_4=14, \qquad a_{10}=38. \]
Find:
\[ d \quad\text{and}\quad a_1. \]
Question 8: Number of terms
How many terms are in:
\[ 6,\ 11,\ 16,\ldots,\ 81? \]
Question 9: Membership
Is \(100\) a term of:
\[ 3,\ 7,\ 11,\ 15,\ldots? \]
Explain your answer.
Question 10: Application
The first row of a theater has \(16\) seats. Each following row has \(2\) more seats than the preceding row.
How many seats are in row \(18\)?
Question 11: Negative common difference
An arithmetic sequence has:
\[ a_1=40, \qquad d=-6. \]
Find the first term that is negative.
Question 12: Formula interpretation
A sequence is defined by:
\[ a_n=5n-2, \qquad n=1,2,3,\ldots \]
Find its first term and common difference.
14. Answers and explanations
Answer 1
\[ a_1=-3, \]\[ a_3=7, \]\[ d=2-(-3)=5. \]
Answer 2
A, C, and D are arithmetic.
For A:
\[ d=3. \]
For C:
\[ d=0. \]
For D:
\[ d=-4. \]
Sequence B has differences:
\[ 2,\ 4,\ 8, \]
so it is not arithmetic.
Answer 3
\[ a_{15}=7+(15-1)3. \]\[ a_{15}=7+42=49. \]
Answer 4
Here:
\[ a_1=12, \qquad d=-4. \]
Therefore:
\[ a_n=12-4(n-1). \]
Simplifying:
\[ a_n=16-4n. \]
Answer 5
The middle term is:
\[ \frac{-4+10}{2}=3. \]
The sequence is:
\[ -4,\ 3,\ 10. \]
Answer 6
There are three steps between \(5\) and \(20\):
\[ d=\frac{20-5}{3}=5. \]
The completed sequence is:
\[ 5,\ 10,\ 15,\ 20. \]
Answer 7
\[ d=\frac{38-14}{10-4}. \]\[ d=\frac{24}{6}=4. \]
Using the fourth term:
\[ 14=a_1+3(4). \]
Therefore:
\[ a_1=2. \]
Answer 8
\[ a_1=6, \qquad d=5, \qquad a_n=81. \]
Then:
\[ n=\frac{81-6}{5}+1. \]\[ n=15+1=16. \]
There are:
\[ 16\text{ terms}. \]
Answer 9
Set:
\[ 3+4(n-1)=100. \]
Then:
\[ 4(n-1)=97, \]\[ n=\frac{101}{4}=25.25. \]
The index is not an integer.
Therefore, \(100\) is not a term.
Answer 10
\[ a_{18}=16+(18-1)2. \]\[ a_{18}=16+34=50. \]
Row \(18\) has:
\[ 50\text{ seats}. \]
Answer 11
We need:
\[ a_n=40-6(n-1)<0. \]
Simplify:
\[ 46-6n<0. \]\[ n>\frac{46}{6}=\frac{23}{3}. \]
The smallest positive integer satisfying this is:
\[ n=8. \]
Then:
\[ a_8=40-6(7)=-2. \]
The first negative term is:
\[ a_8=-2. \]
Answer 12
Substitute \(n=1\):
\[ a_1=5(1)-2=3. \]
Find the difference:
\[ a_{n+1}-a_n =\bigl(5(n+1)-2\bigr)-(5n-2). \]\[ a_{n+1}-a_n=5. \]
Therefore:
\[ a_1=3, \qquad d=5. \]
15. Lesson review
You should now be able to:
- Interpret sequence notation and distinguish an index from a term value.
- Recognize an arithmetic sequence by its constant consecutive differences.
- Calculate terms using:
\[ a_n=a_1+(n-1)d. \]
- Find the common difference from two known terms:
\[ d=\frac{a_q-a_p}{q-p}. \]
- Find missing terms by counting equal steps.
- Determine the number of terms when \(d\neq0\):
\[ n=\frac{a_n-a_1}{d}+1. \]
- Check that an index is a positive integer.
- Model situations involving a fixed added or subtracted amount per step.