Develops the sequences whose consecutive terms differ by a constant, connecting them to the linear functions of chapter 2. Derives the term formula and explains its off-by-one, derives the sum formula by pairing terms from the ends, and applies both to modelling.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.2 Arithmetic Sequences, pp. 1307-1319
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1307-1319 — the pages these objectives are drawn from
Warm-up
Some lists have an obvious next term and some do not.
Discussion prompt
In the list 5, 8, 11, 14, what makes the next term obvious?
Hint: Compare consecutive terms.
Answer:
Each term exceeds the previous by exactly three. The step never changes.
So the next term is 17, and the pattern continues indefinitely with the same step.
A constant step is the defining feature of this family. And a constant rate of change should sound familiar — it is exactly what slope measured in chapter 2.
Concept
An arithmetic sequence adds the same amount at each step. Because the change is constant, the explicit formula is linear in the position.
common difference — the constant amount added to each term of an arithmetic sequence to get the next
\[ a_n=a_1+(n-1)d \]
The common difference plays the role of slope and the first term plays the role of an intercept, though at position one rather than zero. That off-by-one is the source of the formula's minus one and of most errors with it.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1307-1310
Section
Section 1
Concept
Testing whether a sequence is arithmetic means computing the differences between consecutive terms and confirming they all agree.
Subtracting in the right order matters: later minus earlier gives the difference that is added going forwards. Reversing it gives the negative, which then produces a formula describing the sequence backwards.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1307-1311
Picture it
Each arrow adds the same amount.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
The bottom line is the connection worth carrying. A constant rate of change is a slope, so everything from chapter 2 about lines applies to these sequences.
Worked example
Compute all the differences.
\[ \text{Is } 7,\;3,\;-1,\;-5 \text{ arithmetic?} \]
First difference
Why: Second minus first.
\[ -4 \]
Second difference
Why: Third minus second.
\[ -4 \]
Third difference
Why: Fourth minus third.
\[ -4 \]
Compare
Why: All equal.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
\[ d=-4 \]
Verify: check the direction
Why: A negative common difference means the sequence decreases, which the terms confirm. Subtracting in the other order would have given positive four and described the sequence read backwards, which is a different sequence.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1308-1310
Prediction
A sequence has terms 10, 7, 4.
Predict first
What comes next?
Correct: One.
Why: The common difference is negative three, so the next term is 4 minus 3. A negative difference gives a decreasing sequence, which the terms already show.
Worked example
One differing pair settles it.
\[ \text{Is } 2,\;4,\;8,\;16 \text{ arithmetic?} \]
First difference
Why: Two.
\[ 2 \]
Second difference
Why: Four.
\[ 4 \]
Compare
Why: They differ.
Note what it is instead
Why: Each term doubles.
Figure (svg): The solution to Worked example a sequence that is not arithmetic shown as a ladder of expressions, one row per legal move
\[ \text{not arithmetic} \]
Verify: check what pattern it does have
Why: Each term is twice the previous, so the ratios are constant rather than the differences. That is the geometric family, which §11.3 develops — and testing both differences and ratios is how the two are distinguished.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1310-1311
Trap
\[ 7,\;3,\;-1: \quad d=7-3=4 \]
Subtract the later term from the earlier
Why: The order is chosen to avoid a negative.
The formula then describes the sequence increasing, when it decreases.
Later minus earlier. The common difference is what gets added going forwards.
A decreasing sequence genuinely has a negative common difference, and avoiding the minus sign misdescribes it.
Checking the direction is the test: a negative difference must go with terms that decrease.
Faded example
From two consecutive terms.
Fill in the blanks
d=3-7=-4, \textdecreases___
Why: Later minus earlier gives the amount added going forwards, which is negative here. Its sign tells you immediately whether the sequence rises or falls.
Sorting
The differences must be constant.
Sort into buckets
Sort each sequence.
Step zero
You are given a list and asked to identify the pattern.
Discussion prompt
What do you compute first?
Hint: Two tests, quickly.
Answer:
The differences between consecutive terms. If they are constant, the sequence is arithmetic and everything in this section applies.
If not, compute the ratios. If those are constant, it is geometric and §11.3 applies instead.
Both tests take seconds and between them cover the two families this chapter develops. Trying to spot a formula before running them is guessing where a test would settle it.
Section
Section 2
Concept
An arithmetic sequence's terms lie on a straight line when plotted against position, with the common difference as the slope.
The sequence is the dots and not the line. Points between them have no meaning, because there is no term at position one and a half — which is exactly what restricting the domain to counting numbers means.
Figure (svg): A contrast between a linear function and an arithmetic sequence, showing that they are the same object with different domains
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1311-1314
Picture it
The same structure, different domains.
Figure (svg): A contrast between a linear function and an arithmetic sequence, showing that they are the same object with different domains
The last row on each side records the one genuine difference. The sequence's graph is dots on a line rather than the line itself, because only whole positions exist.
Worked example
The common difference is the slope.
\[ \text{A sequence has } a_n=4n+3. \text{ Find its common difference and first term.} \]
Identify the coefficient
Why: Of the index.
\[ 4 \]
Interpret it
Why: The change per step.
\[ d = 4 \]
Substitute the first index
Why: For the first term.
\[ 4 + 3 \]
State both
Why: Difference and first term.
\[ d = 4, a _{1} = 7 \]
Figure (svg): A contrast between a linear function and an arithmetic sequence, showing that they are the same object with different domains
\[ d=4,\; a_1=7 \]
Verify: check the second term
Why: The formula gives 11 at the second position, which is 7 plus 4 — consistent with the common difference. The coefficient of the index is always the common difference, exactly as a slope is.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1312-1313
Prediction
An arithmetic sequence is plotted against position.
Predict first
What do the points do?
Correct: Lie on a straight line.
Why: A constant change per step is exactly a constant slope, so the points fall on a line. They are dots rather than the line itself, since only whole positions exist.
Worked example
The intercept sits at position zero.
\[ \text{Why is the constant } 3 \text{ not the first term of } a_n=4n+3? \]
Note where the constant applies
Why: At index zero.
Note where the sequence starts
Why: At index one.
Compute the first term
Why: Substitute one.
\[ 7 \]
Compare
Why: They differ by the common difference.
Figure (svg): The solution to Worked example why the constant is not the first term shown as a ladder of expressions, one row per legal move
\[ a_1=7\ne 3 \]
Verify: check the relationship
Why: The first term exceeds the constant by exactly the common difference, since going from position zero to position one takes one step. That relationship holds for every arithmetic sequence written this way.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1313-1314
Error analysis
A student describes a sequence from its formula.
Annotate
On: \( a_n=4n+3 \;\Longrightarrow\; a_1=3 \)
The analogy with slope-intercept form is close but the intercept sits at zero, where the sequence has no term. Substituting the first index rather than reading a constant avoids relying on the analogy where it breaks.
Faded example
From the explicit formula.
Fill in the blanks
a_n=4n+3 \;\Longrightarrow\; d=4
Why: The coefficient of the index is the amount added per step, which is the common difference. It plays exactly the role slope plays for a line.
Sorting
The common difference behaves like a slope.
Sort into buckets
Sort each case.
Explain it
Arithmetic sequences and linear functions are closely related.
Discussion prompt
Explain the relationship and the one difference.
Hint: What is allowed as an input?
Answer:
They have the same structure: a constant rate of change, so the explicit formula is linear and the plotted points lie on a line.
The common difference is the slope, and everything chapter 2 said about interpreting a slope transfers unchanged.
The one difference is the domain: a sequence has terms only at whole positions, so its graph is dots rather than a continuous line. A good explanation notes that points between the dots have no meaning, since there is no term at position one and a half.
Section
Section 3
Concept
Reaching the nth term from the first takes one fewer step than the index, because the first term takes no steps at all.
The check is one substitution: putting the first index into the formula must give the first term back. If it gives the first term plus one difference, the minus one was omitted.
Figure (svg): A diagram counting the steps from the first term to the nth, showing that there is one fewer step than the index
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1314-1317
Picture it
Four terms and only three arrows.
Figure (svg): A diagram counting the steps from the first term to the nth, showing that there is one fewer step than the index
Counting the arrows makes the minus one unmistakable. The formula adds one difference per arrow, and there is always one fewer arrow than box.
Worked example
One substitution.
\[ \text{Find the fiftieth term of } 5,\;8,\;11,\;\ldots \]
Identify the first term
Why: From the list.
\[ 5 \]
Identify the common difference
Why: Later minus earlier.
\[ 3 \]
Count the steps
Why: One fewer than fifty.
\[ 49 \]
Compute
Why: First term plus 49 differences.
\[ 5 + 147 \]
Figure (svg): A diagram counting the steps from the first term to the nth, showing that there is one fewer step than the index
\[ a_{50}=152 \]
Verify: check the first term
Why: Substituting index one gives 5 plus zero differences, which is 5 — the correct first term. That single substitution confirms the minus one is in the right place.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1315-1316
Prediction
Reaching the twentieth term from the first.
Predict first
How many differences are added?
Correct: Nineteen.
Why: The first term takes no steps, so reaching the twentieth takes nineteen. Counting the gaps between terms rather than the terms themselves makes this unmistakable.
Worked example
Solve rather than substitute.
\[ \text{Which term of } 5,\;8,\;11,\;\ldots \text{ equals } 98? \]
Write the formula
Why: First term and difference.
\[ 5 + 3(n - 1) = 98 \]
Isolate the bracket
Why: Subtract and divide.
\[ n - 1 = 31 \]
Solve
Why: Add one.
\[ n = 32 \]
Check it is a counting number
Why: It is.
Figure (svg): The solution to Worked example find the position of a term shown as a ladder of expressions, one row per legal move
\[ n=32 \]
Verify: substitute back
Why: The thirty-second term is 5 plus 31 times 3, which is 5 plus 93, or 98 — matching. Had the solution come out fractional, the value would simply not appear in the sequence.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1316-1317
Trap
\[ a_n=a_1+nd \]
Add one difference per term
Why: The steps are counted as though there were one per term.
Substituting the first index gives the first term plus one difference, which is the second term.
There is one fewer step than terms. The first term is reached without adding anything.
So the difference is added n minus one times, not n times.
Substituting the first index is the check. It must return the first term exactly, and the wrong formula returns the second.
Faded example
First term 5, difference 3, fiftieth term.
Fill in the blanks
a_1=5+3(50-147)=5+___=152
Why: Forty-nine steps of three each add 147 to the first term. The minus one is what converts a term count into a step count.
Sorting
Substitute the first index.
Sort into buckets
Sort each formula for a sequence starting at 5 with difference 3.
Explain it to yourself
The formula subtracts one from the index.
Discussion prompt
Explain where it comes from.
Hint: Count the gaps.
Answer:
The formula starts at the first term and adds one difference per step. The first term is already there, so reaching it takes no steps at all.
Between n terms there are n minus one gaps, so that many differences get added. Drawing the terms as boxes and the steps as arrows makes the count visible.
And the check is immediate: substituting the first index must return the first term. A good explanation notes that this one substitution catches the error every time, which is why it is worth doing automatically.
Section
Section 4
Concept
Pairing the first term with the last, the second with the second-last, and so on gives pairs that all have the same total — so the sum is the number of terms times the average of the first and last.
The odd case still works, because the middle term equals the average of the ends and so contributes exactly what half a pair would. Writing the formula as a count times an average rather than as a count of pairs covers both cases without a special rule.
Figure (svg): A diagram pairing the first and last terms, the second and second-last, and so on, showing that each pair has the same total
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1317-1319
Picture it
Every arc joins a pair with the same total.
Figure (svg): A diagram pairing the first and last terms, the second and second-last, and so on, showing that each pair has the same total
As one end rises by the common difference the other falls by it, so every pair's total is the same. That is the whole derivation.
Worked example
Count, average, multiply.
\[ \text{Sum the first } 20 \text{ terms of } 5,\;8,\;11,\;\ldots \]
Find the last term
Why: By the term formula.
\[ 5 + 19(3) = 62 \]
Average the ends
Why: First and last.
\[ \frac{5 + 62}{2} = 33.5 \]
Multiply by the count
Why: Twenty terms.
\[ 20 \times 33.5 \]
Compute
Why: The sum.
\[ 670 \]
Figure (svg): A diagram pairing the first and last terms, the second and second-last, and so on, showing that each pair has the same total
\[ S_{20}=670 \]
Verify: check with a small case
Why: The first four terms are 5, 8, 11, 14, summing to 38 — and the formula gives four times the average of 5 and 14, which is four times 9.5, or 38. The method works on a case small enough to check by hand.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1317-1318
Prediction
Pairing an arithmetic sequence's terms from the ends inward.
Predict first
What do the pair totals do?
Correct: They are all the same.
Why: Moving one step inward from each end raises one term by the common difference and lowers the other by the same amount, so the total is unchanged. That constancy is what makes the sum formula work.
Worked example
The first hundred counting numbers.
\[ \text{Sum } 1+2+3+\cdots+100. \]
Identify the ends
Why: First and last.
\[ 1\text{ and } 100 \]
Average them
Why: Their mean.
\[ 50.5 \]
Multiply by the count
Why: One hundred terms.
\[ 100 \times 50.5 \]
Compute
Why: The sum.
\[ 5050 \]
Figure (svg): The solution to Worked example the classic sum shown as a ladder of expressions, one row per legal move
\[ 5050 \]
Verify: check the pairing directly
Why: Pairing 1 with 100, 2 with 99 and so on gives fifty pairs each totalling 101, and fifty times 101 is 5050 — the same answer by the same reasoning stated differently. This is the sum famously computed by pairing rather than by adding term by term.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1318-1319
Error analysis
A student sums twenty terms.
Annotate
On: \( S_{20}=20\cdot\frac{5+65}{2} \)
The sum formula depends on the term formula, so an off-by-one there propagates. Checking the last term by substituting into the term formula, and checking that by substituting the first index, catches both at once.
Faded example
Twenty terms from 5 to 62.
Fill in the blanks
S=20\cdot\frac2670}=20\cdot 33.5=___
Why: The sum is the number of terms times the average of the first and last. Writing it as a count times an average rather than as pairs handles odd and even counts without a special case.
Sorting
Three quantities, and one is often computed first.
Sort into buckets
Sort each item.
Explain it
The sum formula comes from a clever rearrangement.
Discussion prompt
Explain the derivation to a classmate.
Hint: What happens as you move inward from both ends?
Answer:
Pair the first term with the last, the second with the second-last, and so on. Moving one step inward raises one term by the common difference and lowers the other by exactly the same amount.
So every pair has the same total, and the sum is that total times the number of pairs — which is half the number of terms.
Writing it as the count times the average of the ends covers the odd case too, since the middle term equals that average. A good explanation notes this is the reasoning Gauss is said to have used on the first hundred counting numbers.
Section
Section 5
Concept
Arithmetic sequences model situations where a quantity changes by the same amount each period — simple interest, fixed depreciation, regular savings.
The indexing decision is the modelling step most often glossed over. Whether the initial value is term one or term zero changes every subsequent computation, so it should be stated explicitly rather than assumed.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1313-1319
Picture it
The pattern every application shares.
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
Whenever a quantity changes by the same amount each period, this is the model. Constant proportional change is a different family, developed in the next section.
Worked example
A fixed amount added each month.
\[ \text{You start with } \$200 \text{ and add } \$50 \text{ monthly. What is the balance after } 12 \text{ months?} \]
Set the first term
Why: The starting balance.
\[ 200 \]
Set the common difference
Why: The monthly addition.
\[ 50 \]
Decide the indexing
Why: Month zero is the start.
\[ \text{after } 12\text{ months is } 12\text{ steps} \]
Compute
Why: Starting value plus twelve additions.
\[ 200 + 600 \]
Figure (svg): A sequence of terms with the constant difference between consecutive pairs marked by arrows
\[ \$800 \]
Verify: state the indexing explicitly
Why: Twelve months of deposits means twelve additions, so the answer uses twelve steps rather than eleven. Treating the starting balance as term one and asking for term twelve would have given only eleven deposits — which is why the indexing has to be stated rather than assumed.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1314-1317
Prediction
A question asks how many seats there are altogether.
Predict first
Which formula applies?
Correct: The sum formula.
Why: Altogether asks for a total across all the rows, which is a sum. The term formula would give the count in one particular row, which answers a different question entirely.
Worked example
The sum formula, not the term formula.
\[ \text{A theatre has } 20 \text{ seats in row one and two more in each later row. How many seats in } 15 \text{ rows?} \]
Identify the sequence
Why: Seats per row.
\[ \text{first term } 20,\text{ difference } 2 \]
Find the last row's seats
Why: By the term formula.
\[ 20 + 14(2) = 48 \]
Average the ends
Why: First and last rows.
\[ 34 \]
Multiply by the count
Why: Fifteen rows.
\[ 510 \]
Figure (svg): The solution to Worked example a total rather than a value shown as a ladder of expressions, one row per legal move
\[ 510 \]
Verify: check the question asked for a total
Why: The question wants all the seats, not the seats in the fifteenth row — so the sum formula applies rather than the term formula. Answering 48 would have answered a different question, which is the commonest slip in these problems.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1317-1319
Trap
\[ \text{fifteen rows, so the answer is }a_{15}=48 \]
Compute the fifteenth term
Why: The question is read as asking for a single value.
That gives the seats in the last row, not the seats in the building.
A total needs the sum formula, not the term formula. The term formula gives one value at one position.
Reading the question for the words 'total', 'altogether' or 'in all' identifies which is wanted.
The two answers differ enormously — 48 against 510 here — so the reading matters more than the arithmetic.
Sorting
Single value or running total.
Sort into buckets
Sort each question.
Faded example
Starting at 200 with 50 added monthly.
Fill in the blanks
a_1=200, \quad d=50
Why: The starting value is the first term and the fixed periodic change is the common difference. Stating which period counts as the first term is the remaining modelling decision.
Explain it to yourself
Whether the start is term zero or term one matters.
Discussion prompt
Explain why, and how to settle it.
Hint: How many steps have happened?
Answer:
The index counts positions, and the formula adds one difference per step from term one. So whether the starting value is term one or term zero changes how many steps a given time corresponds to.
After twelve monthly deposits, twelve additions have happened. If the starting balance is term one, then twelve additions reaches term thirteen, not term twelve.
The fix is to state the correspondence explicitly — which real period is term one — before writing any formula. A good explanation notes that this is a modelling decision rather than a mathematical one, which is why it must be recorded rather than deduced.
Comparison
Fill the blanks from memory. Two formulas answering two questions.
Comparison matrix
| term formula | sum formula | |
|---|---|---|
| answers | the value at one position | the total of many terms |
| needs | first term, difference, index | count, first term, last term |
| the off-by-one | one fewer step than the index | inherited from the term formula |
| question words | at, in the nth | total, altogether, in all |
The last row is what decides between them in an application. Reading the question for those words settles which formula before any computation.
Pattern
Five steps, and the last one catches the section's recurring error.
Step 5 costs one substitution and catches the off-by-one, which propagates into the sum formula as well as appearing in the term formula.
OpenStax Algebra and Trigonometry 2e, §13.2 Arithmetic Sequences §13.2
Check
The term formula.
Check your understanding
How many common differences are added to reach the twentieth term from the first?
Answer: A
Why: The first term is reached without adding anything, so there is one fewer step than the index. Counting the gaps between terms rather than the terms themselves makes this immediate.
Check
The sum formula.
Check your understanding
What is the sum of an arithmetic sequence's first n terms?
Answer: A
Why: Pairing terms from the ends gives pairs of equal total, so the sum is that total times the number of pairs — which is the count times the average of the ends. Writing it that way handles odd and even counts alike.
Check
Identifying the family.
Check your understanding
What makes a sequence arithmetic?
Answer: A
Why: A constant difference is the defining property, and it makes the explicit formula linear. A constant ratio defines the geometric family instead, which the next section develops.
Real world
Simple interest and straight-line depreciation are both arithmetic.
Discussion prompt
An asset loses the same amount of value each year. What kind of sequence is its value?
Hint: Constant change, in which direction?
Answer:
Arithmetic, with a negative common difference — the annual depreciation amount. The starting value is the first term.
This is called straight-line depreciation precisely because the values fall on a straight line, which is the connection to chapter 2 made concrete in accounting practice.
The alternative, where the asset loses a fixed percentage each year, is geometric instead and gives a curve. Which method is used is an accounting choice with real tax consequences, and the two produce noticeably different values in the middle years.
Commit first
State your confidence along with your answer.
Predict first
Why does the term formula subtract one from the index?
Correct: There is one fewer step than there are terms.
Why: The first term is reached without adding any differences, so between n terms there are n minus one gaps. Substituting the first index and checking that it returns the first term confirms the count is right.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate's term formula gives the second term when they substitute one. Explain what went wrong.
Hint: How many steps did they count?
Answer:
They added one difference per term rather than one per step, so the formula is one step ahead throughout.
The first term takes no steps to reach, so between n terms there are n minus one gaps — which is where the minus one comes from.
The check is the fix: substituting the first index must return the first term. A good explanation notes that doing that substitution automatically catches this error before it propagates into a sum computation, where it is much harder to spot.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the recurring error lives, and it propagates into the fourth. The second is what makes the whole family feel familiar rather than new.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw four terms as boxes with arrows between them, and count both to show where the minus one comes from. Beside it, write the term formula and the sum formula, noting what each answers. Underneath, derive the sum formula by pairing terms from the ends, and work one application labelling which period is term one.
If your box-and-arrow count shows one fewer arrow than box, and your application states the indexing explicitly, the section's two recurring errors are both accounted for.
Recap
Five things, and the third is the one to check every time.
| if you remember one thing | it should be this |
|---|---|
| about the family | constant difference, so a linear formula |
| about the term formula | one fewer step than the index |
| about the sum | count times the average of the two end terms |
| about applications | read for total or value, and state the indexing |
Section 11.3 changes addition to multiplication: a constant ratio rather than a constant difference, which gives exponential growth and — surprisingly — a finite total even when the terms never stop.
OpenStax, Precalculus, §11.2 Arithmetic Sequences §11.2, pp. 1307-1319 — everything on these slides traces back here
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