The common difference, the rule for the nth term of an arithmetic sequence, recovering that rule from a term and d or from two terms, the linear graph, and the formula for the sum of a finite arithmetic series.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 12 — Sequences and Series
Analyze Arithmetic Sequences and Series
Objectives
Five outcomes. One kind of pattern, and a formula for its total.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-807 — the lesson these objectives are drawn from
Warm-up
Lesson 12.1 gave you rules for sequences found by rewriting terms until the position appeared.
Discussion prompt
Look at 4, 9, 14, 19. What happens between one term and the next? Can you use that to reach the 15th term without listing all fifteen?
Hint: Subtract consecutive terms.
Answer:
Each term is 5 more than the one before. To reach the 15th term you take 14 steps of 5 from the first term.
\[ a_{15} = 4 + 14(5) = 4 + 70 = 74 \]
That is the whole lesson in one line. A sequence with a constant difference is called arithmetic, and its rule is always the first term plus however many steps you have taken, times the step size.
Concept
An arithmetic sequence has a constant difference d between consecutive terms. Its nth term is the first term plus n minus 1 steps of size d, and the sum of its first n terms is the mean of the first and last, times how many there are.
common difference — The constant amount d by which each term of an arithmetic sequence exceeds the one before it. It is found as a_2 minus a_1, and the same value must appear for every consecutive pair.
\[ a_n = a_1 + (n-1)d; \qquad S_n = n\!\left(\frac{a_1+a_n}{2}\right) \]
A negative d means the terms decrease. A d of zero gives a constant sequence, which is still arithmetic.
Figure (svg): Two columns comparing the sequence rule with the series rule
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-804
Section
Section 1
Concept
Subtract each term from the one after it. If every one of those differences is the same number, the sequence is arithmetic and that number is d.
\[ d = a_2-a_1 = a_3-a_2 = a_4-a_3 = \cdots \]
Checking one difference proves nothing, since any two terms differ by something. Several must agree.
Figure (svg): Two sequences with the differences of consecutive terms written underneath
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-802 — Identify arithmetic sequences
Picture it
Example 1, both parts.
Figure (svg): Two sequences with the differences of consecutive terms written underneath
On the left every difference is 5. On the right they grow 2, 4, 6, 8, so the sequence has a pattern but not a constant one.
Worked example
Example 1, both parts.
\[ \text{Is } -4,1,6,11,16,\dots \text{ arithmetic? Is } 3,5,9,15,23,\dots? \]
First: subtract consecutive terms
Why: One minus negative 4, six minus 1, and so on.
\[ 5, 5, 5, 5 \]
First: conclude
Why: Every difference is the same.
Second: subtract consecutive terms
Why: Five minus 3, nine minus 5, and so on.
\[ 2, 4, 6, 8 \]
Second: conclude
Why: The differences are not constant.
Figure (svg): Two sequences with the differences of consecutive terms written underneath
\[ d = 5; \qquad \text{not arithmetic} \]
Verify: look at what the second one is doing
Why: Its differences 2, 4, 6, 8 are themselves arithmetic, so the sequence is quadratic rather than linear — the second differences are constant at 2. A sequence can be perfectly regular without being arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-802
Fill the middle
Guided Practice 1.
Fill in the blanks
17, 14, 11, 8, 5 \;\Longrightarrow\; d = 14 - 17 = -3
Why: Subtracting in the order later minus earlier gives negative 3, and the negative sign correctly records that the terms are falling.
Worked example
Guided Practice 1.
\[ \text{Is } 17,14,11,8,5,\dots \text{ arithmetic? Explain.} \]
Subtract consecutive terms
Why: Fourteen minus 17, eleven minus 14, and so on.
\[ -3, -3, -3, -3 \]
Check they all agree
Why: All four differences are negative 3.
State the conclusion
Why: The difference is the same throughout.
Note the sign
Why: A negative d makes the terms decrease.
Figure (svg): The solution to Worked example a decreasing sequence shown as a ladder of expressions, one row per algebraic move
\[ d = -3 \]
Verify: check the direction against the sign
Why: The terms fall by 3 each step and d is negative 3, which agree. Subtracting in the wrong order gives positive 3 and would predict a rising sequence, so the sign is a built-in check on the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-802
Trap
\[ 3, 5, 9, 15, 23, \dots \]
Compute a sub 2 minus a sub 1
Why: Five minus 3 is 2, so d is taken to be 2.
\[ d = 2 \quad \text{(wrong)} \]
The next difference is 9 minus 5, which is 4. One difference can be computed for any sequence whatever; it proves nothing.
\[ 5-3 = 2, \; 9-5 = 4, \; 15-9 = 6 \]
Compute several differences and compare
Why: Arithmetic requires them all to be equal.
\[ 2 \neq 4 \;\Longrightarrow\; \text{not arithmetic} \]
Three or four differences is enough in practice. If they agree, the pattern is almost certainly the intended one.
Sorting
Test the differences.
Sort into buckets
Sort each sequence.
The fourth is the important non-example: it is perfectly regular, but the regularity is a constant RATIO, which is Lesson 12.3's topic rather than this one's.
Prediction
Commit before reasoning.
Predict first
Is the sequence 7, 7, 7, 7, ... arithmetic?
Correct: Yes, with d equal to 0.
\[ a_n = 7+(n-1)(0) = 7 \]
Why: The definition asks for a constant difference, and zero is a constant. The rule becomes a_n equals 7 plus n minus 1 times 0, which is just 7, and the graph is a horizontal row of points — a line of slope zero, consistent with d being the slope. It happens to be geometric too, with a common ratio of 1, which makes it the one sequence that belongs to both families.
Two truths and a lie
Two of these are true. Keep the one that is not.
Eliminate the wrong options
Which statement about common differences is FALSE?
Survives elimination: B
Why: Any two consecutive terms differ by something, so a single difference carries no information about whether the pattern continues. The other two are both consequences of the sequence being a linear function of its position.
Section
Section 2
Concept
Once you have the first term and the common difference, the rule for the nth term is the first term plus n minus 1 steps of size d. Simplifying puts it in slope-intercept form.
\[ a_n = a_1+(n-1)d \]
The d is multiplied by n minus 1, not by n, because reaching the first term takes no steps at all.
Figure (svg): The arithmetic sequence rule with each symbol labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-803 — Rule for an Arithmetic Sequence
Picture it
Every symbol in the rule.
Figure (svg): The arithmetic sequence rule with each symbol labelled
Reaching term one takes zero steps, term two takes one, term n takes n minus 1. That count is what multiplies d.
Worked example
Example 2, both parts.
\[ \text{Write a rule and find } a_{15} \text{ for } 4,9,14,19,\dots \text{ and } 60,52,44,36,\dots \]
First: find a_1 and d
Why: The first term is 4 and 9 minus 4 is 5.
\[ a _{1} = 4, d = 5 \]
First: substitute and simplify
Why: Four plus n minus 1 times 5.
\[ a _{n} = -1 + 5 n \]
First: evaluate at 15
Why: Negative 1 plus 75.
\[ a _{15} = 74 \]
Second: repeat
Why: Sixty and 52 minus 60 is negative 8.
\[ a _{n} = 68 - 8 n, a _{15} = -52 \]
Figure (svg): The arithmetic sequence rule with each symbol labelled
\[ a_n = -1+5n, \; a_{15} = 74; \qquad a_n = 68-8n, \; a_{15} = -52 \]
Verify: check each rule at n equal to 1
Why: The first gives negative 1 plus 5, which is 4 — the stated first term. The second gives 68 minus 8, which is 60. Testing at n equal to 1 catches the n against n minus 1 error immediately, because that error shifts every term by exactly one d.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 803-803
Fill the middle
Example 2a.
Fill in the blanks
a_n = 4+(n-1)5 = 4+5n-5 = -1+5n
Why: Four minus 5 is negative 1, so the simplified rule is negative 1 plus 5n. The constant is always a_1 minus d.
Worked example
Guided Practice 2.
\[ \text{Write a rule for } 17,14,11,8,\dots \text{ and find } a_{20}. \]
Find a_1 and d
Why: The first term is 17 and the difference is negative 3.
\[ a _{1} = 17, d = -3 \]
Substitute
Why: Seventeen plus n minus 1 times negative 3.
\[ a _{n} = 17 - 3 n + 3 \]
Simplify
Why: Collect the constants.
\[ a _{n} = 20 - 3 n \]
Evaluate at 20
Why: Twenty minus 60.
\[ a _{20} = -40 \]
Figure (svg): The solution to Worked example a rule and its 20th term shown as a ladder of expressions, one row per algebraic move
\[ a_n = 20-3n, \; a_{20} = -40 \]
Verify: read the simplified rule as a line
Why: In the form 20 minus 3n the coefficient of n is negative 3, which is d, and the constant 20 is what the rule would give at n equal to 0 — one step BEFORE the first term. That is why the constant is not a_1 but a_1 minus d.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 804-804
Error analysis
A student writes a rule for 37, 24, 11, -2, -15, using a_1 = 37 and d = -13.
Annotate
On: \( a_n = a_1+nd = 37+n(-13) = 37-13n \)
The whole sequence has been shifted one place. Substituting n equal to 1 into any proposed rule catches this in a single line.
Matching
Find a_1 and d, then substitute.
Match the pairs
Why: In every simplified rule the coefficient of n is d and the constant is a_1 minus d. Reading those two numbers off a rule is the fastest way to recover the sequence it describes.
Comparison
Fill the blanks. The unsimplified form and the simplified one.
Comparison matrix
| Question | a_n = 4 + (n - 1)5 | a_n = -1 + 5n |
|---|---|---|
| What the constant is | the first term | the first term minus d |
| Value at n = 1 | 4 | 4 |
| Where d appears | multiplying (n - 1) | the coefficient of n |
| Value at n = 15 | 74 | 74 |
Both forms are correct and give identical values. The unsimplified one shows where the numbers came from; the simplified one shows the line.
Prediction
Commit before reasoning.
Predict first
In the simplified rule a_n = 20 - 3n, what does the 20 represent?
Correct: The value the rule would give at n equal to 0, one step before the first term.
\[ a_1 = 20-3 = 17; \quad \text{constant} = a_1-d \]
Why: The first term is 17, not 20, because the rule at n equal to 1 gives 20 minus 3. In line language the constant is the y-intercept, and the sequence's graph starts at n equal to 1 rather than 0, so the intercept sits one step outside the sequence. Expecting the constant to be a_1 is the same off-by-one that the error analysis above punished.
Section
Section 3
Concept
If you are given some term other than the first, substitute it into the general rule and solve for a_1. If you are given two terms, substitute both and solve the resulting system for a_1 and d.
\[ a_{19} = a_1+18d \;\Longrightarrow\; a_1 = a_{19}-18d \]
Graphing the result confirms the work: the points of any arithmetic sequence lie on a line whose slope is d.
Figure (svg): The first six terms of an arithmetic sequence plotted as collinear points
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 803-804 — Write a rule given a term and common difference
Picture it
Example 3b: the sequence with a19 equal to 48 and d equal to 3.
Figure (svg): The first six terms of an arithmetic sequence plotted as collinear points
The dashed line is drawn only to show the pattern. The sequence itself is the six dots, and the line's slope is d.
Worked example
Example 3.
\[ \text{One term is } a_{19} = 48 \text{ and } d = 3. \text{ Write a rule.} \]
Substitute n equal to 19
Why: Into the general rule.
\[ 48 = a _{1} + 18(3) \]
Simplify
Why: Eighteen times 3 is 54.
\[ 48 = a _{1} + 54 \]
Solve for a_1
Why: Subtract 54.
\[ a _{1} = -6 \]
Write the rule
Why: Negative 6 plus n minus 1 times 3.
\[ a _{n} = -9 + 3 n \]
Figure (svg): The first six terms of an arithmetic sequence plotted as collinear points
\[ a_n = -9+3n \]
Verify: check at n equal to 19
Why: Negative 9 plus 3 times 19 is negative 9 plus 57, which is 48 — the given term. Checking at the GIVEN position rather than at n equal to 1 is the right test here, because that is the only value the problem actually supplied.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 803-803
Fill the middle
Example 3a.
Fill in the blanks
48 = a_1+18(3) = a_1+54 \;\Longrightarrow\; a_1 = -6
Why: Forty-eight minus 54 is negative 6. The nineteenth term sits eighteen steps of 3 above the first, so the first is well below it.
Worked example
Example 4.
\[ \text{Two terms are } a_8 = 21 \text{ and } a_{27} = 97. \text{ Write a rule.} \]
Write two equations
Why: Substituting 27 and then 8 for n.
\[ 97 = a _{1} + 26 d; 21 = a _{1} + 7 d \]
Subtract to eliminate a_1
Why: Ninety-seven minus 21, and 26d minus 7d.
\[ 76 = 19 d \]
Solve for d
Why: Seventy-six over 19.
\[ d = 4 \]
Back-substitute and simplify
Why: Ninety-seven equals a_1 plus 104.
\[ a _{1} = -7, a _{n} = -11 + 4 n \]
Figure (svg): The solution to Worked example from two terms shown as a ladder of expressions, one row per algebraic move
\[ a_n = -11+4n \]
Verify: check both given terms
Why: At n equal to 8 the rule gives negative 11 plus 32, which is 21; at n equal to 27 it gives negative 11 plus 108, which is 97. Both match, which is the point of using both equations — a rule fitted to one term alone could be anything.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 804-804
Trap
\[ a_{19} = 48, \; d = 3 \]
Use 48 as a_1
Why: The given value is the only term in sight, so it is taken as the start.
\[ a_n = 48+(n-1)3 = 45+3n \quad \text{(wrong)} \]
That rule gives 51 at n equal to 2, but it should give 48 at n equal to 19. The subscript says which position the value belongs to.
\[ 48 = a_1+18(3) \;\Longrightarrow\; a_1 = -6 \]
Substitute the position, then solve for a_1
Why: The rule is written for a_1, so a_1 must be found first.
\[ a_n = -6+(n-1)3 = -9+3n \]
Eighteen steps of 3 separate the first term from the nineteenth, so a_1 sits 54 below 48.
Fill the middle
Example 4, Step 2.
Fill in the blanks
97-21 = 76 = (26-7)d = 19d \;\Longrightarrow\; d = 4
Why: Seventy-six over 19 is 4. Subtracting the two equations removes a_1 entirely, which is why this is faster than substitution.
Matching
Solve for a_1 first in each case.
Match the pairs
Why: The first two needed one equation each; the last two needed a system. In the third, subtracting gives 45 equals 9d, so d is 5 and a_1 is negative 4.
Prediction
Commit before reasoning.
Predict first
Why does every arithmetic sequence graph as points on a line?
Correct: Because the simplified rule has the form constant plus dn, which is a linear function of n.
\[ a_n = (a_1-d)+dn; \quad \text{slope } d \]
Why: Writing the rule as a_1 minus d plus dn shows it is a line in disguise, with slope d and intercept a_1 minus d. The only difference from an ordinary line is that n is restricted to the whole numbers, which turns the line into a row of dots on it. This is exactly why a constant first difference signals a linear rule, the same test you used for shapes of data in Lesson 11.5.
Section
Section 4
Concept
Adding the terms of an arithmetic sequence gives an arithmetic series. Its sum is the average of the first and last terms, multiplied by how many terms there are.
\[ S_n = n\!\left(\frac{a_1+a_n}{2}\right) \]
The formula follows from writing the sum forwards and backwards and adding: the d terms cancel, leaving n copies of a_1 plus a_n.
Figure (svg): The reverse-and-add derivation of the arithmetic series sum formula
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 804-804 — The Sum of a Finite Arithmetic Series
Picture it
Where the formula comes from.
Figure (svg): The reverse-and-add derivation of the arithmetic series sum formula
Each column of the two lines adds to a_1 plus a_n, and there are n columns, so twice the sum is n times that.
Worked example
Example 5.
\[ \text{Find } \sum_{i=1}^{20}(4+3i). \]
Identify the first term
Why: Substitute i equal to 1.
\[ a _{1} = 4 + 3 = 7 \]
Identify the last term
Why: Substitute i equal to 20.
\[ a _{20} = 4 + 60 = 64 \]
Apply the sum formula
Why: Twenty times the mean of 7 and 64.
\[ S _{20} = 20(\frac{71}{2}) \]
Compute
Why: Twenty times 35.5.
\[ 710 \]
Figure (svg): The reverse-and-add derivation of the arithmetic series sum formula
\[ S_{20} = 20\!\left(\tfrac{7+64}{2}\right) = 710 \]
Verify: confirm the series is arithmetic
Why: Evaluating 4 plus 3i at i equal to 1, 2, 3, 4 gives 7, 10, 13, 16, whose differences are all 3 — so the formula applies. Any term rule that is linear in the index produces an arithmetic series, with d equal to the coefficient of i.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 805-805
Fill the middle
Example 5.
Fill in the blanks
a_64 = 4+3(20) = ___
Why: Substituting the upper limit into the term rule gives 64. The sum formula needs the last term's VALUE, not its position.
Worked example
Guided Practice 5.
\[ \text{Find } \sum_{i=1}^{12}(2+7i). \]
Find a_1
Why: Two plus 7.
\[ 9 \]
Find a_12
Why: Two plus 84.
\[ 86 \]
Apply the formula
Why: Twelve times the mean of 9 and 86.
\[ S _{12} = 12(\frac{95}{2}) \]
Compute
Why: Six times 95.
\[ 570 \]
Figure (svg): The solution to Worked example another sigma series shown as a ladder of expressions, one row per algebraic move
\[ S_{12} = 570 \]
Verify: check with the special formulas of Lesson 12.1
Why: Splitting gives twelve 2s plus 7 times the sum of 1 to 12, which is 24 plus 7 times 78, or 24 plus 546, giving 570. The two routes agree, and the split route is the reason this formula is not strictly necessary — only much faster.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 805-805
Error analysis
A student evaluates the sum of 4 plus 3i for i from 1 to 20.
Annotate
On: \( S_{20} = 20\!\left(\frac{7+20}{2}\right) = 270 \)
The formula needs three numbers with different jobs: how many terms, what the first is, and what the last is. Confusing n with a_n is the commonest slip.
Sorting
In the sum of 4 plus 3i from i equal to 1 to 20.
Sort into buckets
Sort each number by its role in the sum formula.
Notice that d is absent from the sum formula. It is needed to FIND the last term when that term is not given, but not in the formula itself.
Matching
Find the first and last terms, then apply the formula.
Match the pairs
Why: Every one of these is arithmetic because the term rule is linear in the index. The last two are the same series stopped at different points, and stopping six terms earlier cut the total by roughly two thirds.
Prediction
Commit before reasoning.
Predict first
In the reverse-and-add derivation, why does every column of the two lines add to the same value?
Correct: Because moving one step forward in the top line moves one step backward in the bottom, so the two d changes cancel.
\[ (a_1+kd)+(a_n-kd) = a_1+a_n \]
Why: The top line reads a_1, a_1 plus d, a_1 plus 2d and so on while the bottom reads a_n, a_n minus d, a_n minus 2d. Column k contributes a_1 plus k d plus a_n minus k d, and the kd terms cancel exactly, leaving a_1 plus a_n. That happens in every column regardless of n's parity, which is why the formula has no even-or-odd cases.
Section
Section 5
Concept
In an applied problem, count the first few terms to find a_1 and d, write the sequence rule, use it to find the last term, and only then apply the sum formula.
\[ a_n = 3n; \quad S_{14} = 14\!\left(\frac{3+42}{2}\right) = 315 \]
The sum formula needs the last term, and the sequence rule is usually the only way to get it.
Figure (svg): A house of cards with the number of cards per row labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 805-805 — Use an arithmetic sequence and series in real life
Picture it
Example 6: rows counted from the top.
Figure (svg): A house of cards with the number of cards per row labelled
Each row adds two leaning cards and one flat card, so the counts are 3, 6, 9, 12 and the rule is 3n.
Worked example
Example 6, both parts.
\[ \text{Rows hold } 3,6,9,12,\dots \text{ cards. Write a rule, then total } 14 \text{ rows.} \]
Find a_1 and d
Why: Three cards in the top row, three more each time.
\[ a _{1} = 3, d = 3 \]
Write the rule
Why: Three plus n minus 1 times 3.
\[ a _{n} = 3 n \]
Find the last row
Why: Three times 14.
\[ a _{14} = 42 \]
Apply the sum formula
Why: Fourteen times the mean of 3 and 42.
\[ S _{14} = 315 \]
Figure (svg): A house of cards with the number of cards per row labelled
\[ a_n = 3n; \quad S_{14} = 315 \]
Verify: sanity-check the size of the total
Why: Fourteen rows averaging 22.5 cards each gives 315, and the middle rows really do hold around 21 to 24 cards. A total below 42 or above 588 would have to be wrong, since 42 is the largest single row and 588 is fourteen of them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 805-805
Fill the middle
Example 6b.
Fill in the blanks
S_315 = 14\!\left(\frac______\right) = 14(22.5) = ___
Why: Fourteen times 22.5 is 315 cards. The mean of the first and last rows, 22.5, is the average row size across the whole house.
Worked example
Guided Practice 6.
\[ \text{How many cards are in a house with } 8 \text{ rows?} \]
Find the last row
Why: Three times 8.
\[ a _{8} = 24 \]
Apply the sum formula
Why: Eight times the mean of 3 and 24.
\[ S _{8} = 8(\frac{27}{2}) \]
Compute
Why: Four times 27.
\[ 108 \]
Compare with 14 rows
Why: Three hundred fifteen against 108.
Figure (svg): The solution to Worked example a smaller house shown as a ladder of expressions, one row per algebraic move
\[ S_8 = 108 \]
Verify: check the growth is faster than the row count
Why: Fourteen rows is 1.75 times eight rows, but 315 is nearly three times 108. The total grows roughly as the square of the number of rows, because both the count of rows and the size of each row are growing together.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 805-805
Trap
\[ a_n = 3n, \; 14 \text{ rows} \]
Substitute 14 as the last term
Why: The number 14 is the only large number in the problem.
\[ S_{14} = 14\!\left(\frac{3+14}{2}\right) = 119 \quad \text{(wrong)} \]
Fourteen is how MANY rows there are. The last row holds 3 times 14, which is 42 cards.
\[ a_{14} = 3(14) = 42 \]
Find the last term with the sequence rule first
Why: Then the sum formula has all three of its inputs.
\[ S_{14} = 14\!\left(\frac{3+42}{2}\right) = 315 \]
The sequence rule and the series formula work as a pair: the first supplies the last term the second needs.
Ranking
Smallest first.
Put in order
Why: The values are 42, 108, 315, 570 and 710. The single row is smallest by far, which is the point of a series: a total dwarfs any one term once there are many terms.
Comparison
Fill the blanks. Two different questions about the same house of cards.
Comparison matrix
| Question | How many cards in row 14? | How many cards in all 14 rows? |
|---|---|---|
| Formula | a_n = a_1 + (n - 1)d | S_n = n(a_1 + a_n)/2 |
| Answer | 42 | 315 |
| What you need | a_1, d and n | a_1, a_n and n |
| Kind of answer | one term | a total |
Reading the question for the words each and total is what picks the formula. The series formula needs the sequence rule's answer as one of its inputs.
Prediction
Commit before reasoning.
Predict first
Does the formula for the sum of a finite arithmetic series contain the common difference?
Correct: No — it uses only n, a_1 and a_n, though d is often needed to find a_n first.
\[ S_n = n\!\left(\frac{a_1+a_n}{2}\right); \quad \text{no } d \]
Why: The derivation cancelled every d, which is exactly why the final formula has none. If a problem hands you the first and last terms and the count directly, you can total the series without ever knowing d. When the last term is not given, though, the sequence rule is the only route to it, and that rule does need d.
Comparison
Fill the blanks. One reaches a term, the other a total.
Comparison matrix
| Question | Sequence rule | Series formula |
|---|---|---|
| Formula | a_n = a_1 + (n - 1)d | S_n = n(a_1 + a_n)/2 |
| Answers | one term | the total of the first n terms |
| Uses d? | yes | no |
| Graph | collinear points | grows quadratically in n |
The sum grows as the square of n because both how many terms there are and how big they are grow together, which is why 14 rows of cards holds nearly three times what 8 rows holds.
Pattern
Test, rule, then total.
The common difference multiplies n minus 1, not n. Substituting n equal to 1 into your rule catches that error every time.
OpenStax Algebra and Trigonometry 2e, §13.2 Arithmetic Sequences §13.2
Check
Find a_1 and d first.
Check your understanding
What is a rule for the nth term of 60, 52, 44, 36, ...?
Answer: A
Why: With a_1 = 60 and d = -8, the rule 60 + (n - 1)(-8) simplifies to 68 - 8n.
Check
Solve for the first term.
Check your understanding
For an arithmetic sequence, a_30 = 57 and d = 4. What is a rule for the nth term?
Answer: A
Why: 57 = a_1 + 29(4) gives a_1 = -59, so a_n = -59 + (n - 1)4 = -63 + 4n.
Check
Find the last term before totalling.
Check your understanding
What is the sum of 4 + 3i for i from 1 to 20?
Answer: A
Why: a_1 = 7 and a_20 = 64, so S_20 = 20 times the mean of 7 and 64, which is 710.
Real world
A marching band forms a triangle. The front row has 3 marchers and each row behind has 2 more than the row in front. There are 12 rows.
Discussion prompt
How many marchers are in the last row, and how many are in the band?
Hint: Find the rule first, then the last term, then the total.
Answer:
\[ a_1 = 3, \; d = 2 \;\Longrightarrow\; a_n = 3+(n-1)2 = 1+2n \]
\[ a_{12} = 1+24 = 25 \]
\[ S_{12} = 12\!\left(\frac{3+25}{2}\right) = 12(14) = 168 \]
The back row holds 25 marchers and the band has 168 in total.
Notice that the terms 3, 5, 7, ... are the odd numbers, so the total is the sum of the first twelve odd numbers — which is 12 squared, or 144, plus the 24 extra that comes from starting at 3 rather than 1. Both routes give 168, and the second is worth knowing: the sum of the first n odd numbers is always n squared, which is one of the prettiest facts in this chapter and follows directly from the formula you just used.
Commit first
Answer, then rate your confidence honestly.
Predict first
You know a_1 = 5, n = 40 and a_40 = 122. Can you find the sum without computing d?
Correct: Yes — the sum formula uses only n, a_1 and a_n, giving 40 times 63.5, or 2540.
\[ S_{40} = 40\!\left(\frac{5+122}{2}\right) = 2540 \]
Why: Every d cancelled in the reverse-and-add derivation, so the final formula contains none. Here the mean of 5 and 122 is 63.5, and forty terms averaging that gives 2540 exactly. The common difference happens to be 3, since 122 minus 5 is 117 and 117 over 39 is 3, but knowing that changes nothing about the total. The rule needs d; the series formula does not.
Explain it
They can find a slope and have never met sequences.
Discussion prompt
In four sentences or fewer, explain why the sum of an arithmetic series is the average of the ends times the count.
Hint: Think about pairing terms from the two ends.
Answer:
Pair the first term with the last, the second with the second-last, and so on. Because the sequence goes up by the same amount each step, every pair adds to the same total.
So the whole series is just that pair total, repeated. Dividing by two turns pair totals into per-term averages, which is why the formula reads count times the average of the two ends.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the n minus 1, always test your rule at n equal to 1. For two given terms, write both equations and subtract; a_1 vanishes and d falls out. For n against a_n, remember n counts terms while a_n is a value, and find a_n with the sequence rule first. For the graph, write the rule as a_1 minus d plus dn and read off the slope.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an arithmetic sequences page. Top left: write two sequences, one arithmetic and one not, and show the consecutive differences under each. Top right: write the rule a_n = a_1 + (n - 1)d in large letters and label all four symbols, then write beside it in one sentence why d multiplies n minus 1. Middle: take a term and a common difference of your own choosing, solve for a_1, write the rule, and plot the first six terms as points on a set of axes, drawing the line through them dashed and labelling its slope. Bottom left: reproduce the reverse-and-add derivation in full and box the sum formula. Bottom right: solve an applied problem of your own invention, showing the sequence rule finding the last term and the series formula finding the total.
If your applied problem used the row count as the last term, recheck it: that is the mistake this lesson punishes most often.
Recap
Five things, and two formulas that work as a pair.
| If you see | Then |
|---|---|
| Constant consecutive differences | The sequence is arithmetic; d is that difference |
| a_1 and d | Write a_n = a_1 + (n - 1)d and simplify |
| a term other than the first | Substitute its position and solve for a_1 |
| Two terms | Write two equations and subtract to get d |
| A total wanted | Find a_n first, then use n times the mean of the ends |
| A simplified rule | The coefficient of n is d; the constant is a_1 minus d |
Lesson 12.3 replaces the constant difference with a constant ratio, which changes lines into exponential curves and changes both formulas.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.2 Analyze Arithmetic Sequences and Series §12.2, pp. 802-807 — everything on these slides traces back here
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