The common ratio, the rule for the nth term of a geometric sequence, recovering that rule from a term and r or from two terms, the exponential graph, the sum of a finite geometric series, and percent growth as a geometric model.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 12 — Sequences and Series
Analyze Geometric Sequences and Series
Objectives
Five outcomes. The same five as last lesson, with multiplying in place of adding.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-815 — the lesson these objectives are drawn from
Warm-up
Lesson 12.2 tested sequences by subtracting consecutive terms.
Discussion prompt
Try that on 625, 125, 25, 5, 1. What do the differences do? Now try dividing instead.
Hint: The differences are -500, -100, -20, -4.
Answer:
The differences shrink fast and are nowhere near constant, so the sequence is not arithmetic.
\[ \frac{125}{625} = \frac{1}{5}, \quad \frac{25}{125} = \frac{1}{5}, \quad \frac{5}{25} = \frac{1}{5} \]
But every RATIO is one fifth. A sequence with a constant ratio is called geometric, and everything from the last lesson has a matching version here with multiplication in place of addition.
Concept
A geometric sequence has a constant ratio r between consecutive terms. Its nth term is the first term multiplied by r some n minus 1 times, and the sum of its first n terms has a closed formula that works for any r other than 1.
common ratio — The constant value r obtained by dividing any term of a geometric sequence by the term before it. The same value must appear for every consecutive pair.
\[ a_n = a_1 r^{\,n-1}; \qquad S_n = a_1\!\left(\frac{1-r^{\,n}}{1-r}\right) \]
A ratio between negative one and one shrinks the terms toward zero. A negative ratio makes the signs alternate.
Figure (svg): Two columns comparing arithmetic with geometric sequences
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-812
Section
Section 1
Concept
Divide each term by the one before it. If every one of those ratios is the same number, the sequence is geometric and that number is r.
\[ r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \cdots \]
As with the difference test, one ratio proves nothing: any two nonzero terms have a ratio. Several must agree.
Figure (svg): Two sequences with the ratios of consecutive terms written underneath
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-810 — Identify geometric sequences
Picture it
Example 1, both parts.
Figure (svg): Two sequences with the ratios of consecutive terms written underneath
On the left the ratios are five halves, nine fifths, fourteen ninths — all different. On the right every ratio is one fifth.
Worked example
Example 1, both parts.
\[ \text{Is } 4,10,18,28,40,\dots \text{ geometric? Is } 625,125,25,5,1,\dots? \]
First: divide consecutive terms
Why: Ten over 4, eighteen over 10, and so on.
\[ \frac{5}{2}, \frac{9}{5}, \frac{14}{9}, \frac{10}{7} \]
First: conclude
Why: The ratios are all different.
Second: divide consecutive terms
Why: One twenty-five over 625, twenty-five over 125.
\[ \frac{1}{5}, \frac{1}{5}, \frac{1}{5}, \frac{1}{5} \]
Second: conclude
Why: Every ratio is the same.
Figure (svg): Two sequences with the ratios of consecutive terms written underneath
\[ \text{not geometric}; \qquad r = \tfrac{1}{5} \]
Verify: check what the first one is instead
Why: Its differences are 6, 8, 10, 12, which are themselves arithmetic — so it is quadratic rather than exponential. A sequence can fail both the difference test and the ratio test and still be perfectly regular.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-810
Fill the middle
Guided Practice 3.
Fill in the blanks
-4, 8, -16, 32 \;\Longrightarrow\; r = \frac-2___ = ___
Why: Eight over negative 4 is negative 2, and the negative sign is what makes the terms alternate in sign.
Worked example
Guided Practice 1 to 3.
\[ \text{Are } 81,27,9,3,1; \; 1,2,6,24,120; \; -4,8,-16,32,-64 \text{ geometric?} \]
First: divide consecutive terms
Why: Twenty-seven over 81, nine over 27.
\[ \frac{1}{3}\text{ each time} \]
Second: divide consecutive terms
Why: Two over 1, six over 2, twenty-four over 6.
\[ 2, 3, 4, 5 \]
Third: divide consecutive terms
Why: Eight over negative 4, negative 16 over 8.
\[ -2\text{ each time} \]
Conclude
Why: The first and third pass, the second fails.
Figure (svg): The solution to Worked example three more sequences shown as a ladder of expressions, one row per algebraic move
\[ r = \tfrac{1}{3}; \quad \text{not geometric}; \quad r = -2 \]
Verify: read the sign pattern of the third
Why: Its terms alternate negative, positive, negative, positive, and r is negative 2. A negative ratio always produces alternating signs, which is a fast visual check: alternating signs mean r is negative, and no alternation means r is positive.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-810
Trap
\[ 625, 125, 25, 5, 1 \]
Subtract consecutive terms
Why: The habit from the last lesson is applied automatically.
\[ -500, -100, -20, -4 \quad \text{(not constant)} \]
The student concludes there is no pattern at all, when in fact there is a perfectly clean one.
\[ \frac{125}{625} = \frac{1}{5} = \frac{25}{125} = \frac{5}{25} \]
Try dividing as well as subtracting
Why: A sequence can be regular in either way.
\[ \text{geometric with } r = \tfrac{1}{5} \]
Run both tests before deciding a sequence has no pattern. Subtract for arithmetic, divide for geometric.
Sorting
Run both tests.
Sort into buckets
Sort each sequence.
The fifth item is the interesting one: 0.75, 1.5 looks like doubling until the third term arrives, which is precisely why one ratio is never enough.
Prediction
Commit before reasoning.
Predict first
In 625, 125, 25, 5, 1, ... with r = 1/5, what happens to the terms as n grows?
Correct: They shrink toward zero without ever reaching it.
\[ \left|r\right| < 1 \;\Longrightarrow\; a_n \to 0 \]
Why: Multiplying by one fifth repeatedly makes the terms 1/5, 1/25, 1/125 and so on — always positive, always smaller, never zero. That shrinking is what will make an INFINITE geometric series have a finite total in Lesson 12.4, something no arithmetic series can ever do. When the ratio is bigger than 1 the terms grow instead, and the infinite total is unbounded.
Two truths and a lie
Two of these are true. Keep the one that is not.
Eliminate the wrong options
Which statement about common ratios is FALSE?
Survives elimination: B
Why: The sequence 1, 2, 6, 24, 120 fails both tests: its differences change and so do its ratios. Failing one test says nothing about the other, so both must be run before concluding anything.
Section
Section 2
Concept
Once you have the first term and the common ratio, the rule for the nth term is the first term multiplied by r raised to n minus 1.
\[ a_n = a_1 r^{\,n-1} \]
The exponent is n minus 1, not n, because the first term has not been multiplied by r at all.
Figure (svg): The geometric sequence rule with each symbol labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-811 — Rule for a Geometric Sequence
Picture it
Every symbol in the rule.
Figure (svg): The geometric sequence rule with each symbol labelled
The exponent counts multiplications. Term one has had none, term two one, term n has had n minus 1.
Worked example
Example 2, both parts.
\[ \text{Write a rule and find } a_7 \text{ for } 4,20,100,500,\dots \text{ and } 152,-76,38,-19,\dots \]
First: find a_1 and r
Why: The first term is 4 and 20 over 4 is 5.
\[ a _{1} = 4, r = 5 \]
First: write the rule and evaluate
Why: Four times 5 to the sixth.
\[ a _{n} = 4(5) ^{n - 1}, a _{7} = 62, 500 \]
Second: find a_1 and r
Why: One fifty-two, and negative 76 over 152.
\[ a _{1} = 152, r = -\frac{1}{2} \]
Second: write the rule and evaluate
Why: One fifty-two times one half to the sixth.
\[ a _{7} = \frac{152}{64} = \frac{19}{8} \]
Figure (svg): The geometric sequence rule with each symbol labelled
\[ a_n = 4(5)^{\,n-1}, \; a_7 = 62{,}500; \qquad a_n = 152\!\left(-\tfrac{1}{2}\right)^{\,n-1}, \; a_7 = \tfrac{19}{8} \]
Verify: check the sign of the second 7th term
Why: The exponent 6 is even, so the negative ratio's sign cancels and a_7 is positive — matching the pattern that odd-numbered terms are positive here. Tracking the parity of the exponent is the whole of sign management with a negative ratio.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 811-811
Fill the middle
Example 2a.
Fill in the blanks
a_7 = 4(5)^62500 = 4(5)^6 = 4(15625) = ___
Why: Four times 15,625 is 62,500. Six multiplications by 5 took the sequence from 4 past sixty thousand.
Worked example
Guided Practice 4.
\[ \text{Write a rule for } 3,15,75,375,\dots \text{ and find } a_8. \]
Find a_1 and r
Why: The first term is 3 and 15 over 3 is 5.
\[ a _{1} = 3, r = 5 \]
Write the rule
Why: Three times 5 to the n minus 1.
\[ a _{n} = 3(5) ^{n - 1} \]
Evaluate at 8
Why: Three times 5 to the seventh.
\[ 3(78, 125) \]
Compute
Why: Three times 78,125.
\[ 234, 375 \]
Figure (svg): The solution to Worked example a rule and its 8th term shown as a ladder of expressions, one row per algebraic move
\[ a_n = 3(5)^{\,n-1}, \; a_8 = 234{,}375 \]
Verify: compare with an arithmetic sequence of the same start
Why: Starting at 3 and ADDING 12 each time reaches 87 by the eighth term; multiplying by 5 reaches 234,375. That gap after only eight steps is why exponential growth eventually overwhelms linear growth in any model.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 812-812
Error analysis
A student writes a rule for 4, 20, 100, 500, using a_1 = 4 and r = 5.
Annotate
On: \( a_n = a_1 r^{\,n} = 4(5)^{\,n} \)
The whole sequence has been shifted one place, exactly as multiplying d by n did in the arithmetic case. Substituting n equal to 1 catches it at once.
Matching
Find a_1 and r, then substitute.
Match the pairs
Why: In every rule the number in front is a_1 and the number in the parentheses is r. The first and third share a ratio but start in different places, which scales the whole sequence without changing its shape.
Comparison
Fill the blanks. Both start at 3.
Comparison matrix
| Question | a_n = 3 + (n - 1)5 | a_n = 3(5)^(n - 1) |
|---|---|---|
| What repeats | adding 5 | multiplying by 5 |
| Second term | 8 | 15 |
| Eighth term | 38 | 234,375 |
| Graph | collinear points | exponential curve |
Both rules use n minus 1 for the same reason, and both start at the same place. Eight steps later they are six thousand times apart.
Prediction
Commit before reasoning.
Predict first
In a_n = 152(-1/2)^(n-1), for which n is the term positive?
Correct: Odd n, because then the exponent n minus 1 is even.
\[ (-\tfrac{1}{2})^{6} > 0; \quad (-\tfrac{1}{2})^{5} < 0 \]
Why: A negative base raised to an even exponent is positive, and n minus 1 is even exactly when n is odd. So terms 1, 3, 5, 7 are positive and terms 2, 4, 6 are negative, which matches the given sequence 152, negative 76, 38, negative 19. Reading the parity of the exponent rather than of n itself is what keeps the signs straight, and it is another consequence of the exponent being n minus 1.
Section
Section 3
Concept
If you are given a term other than the first, substitute its position into the general rule and solve for a_1. If you are given two terms, solve one for a_1, substitute into the other, and the powers of r collapse to a single equation.
\[ a_4 = a_1 r^3 \;\Longrightarrow\; a_1 = \frac{a_4}{r^3} \]
Graphing confirms the work: the points of a geometric sequence with positive r lie on an exponential curve.
Figure (svg): The first six terms of a geometric sequence plotted on an exponential curve
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 811-812 — Write a rule given a term and common ratio
Picture it
Example 3b: the sequence with a4 equal to 12 and r equal to 2.
Figure (svg): The first six terms of a geometric sequence plotted on an exponential curve
The dashed curve is drawn only to show the pattern. Each step doubles, so the gaps widen as n grows.
Worked example
Example 3.
\[ \text{One term is } a_4 = 12 \text{ and } r = 2. \text{ Write a rule.} \]
Substitute n equal to 4
Why: Into the general rule.
\[ 12 = a _{1}(2) ^{3} \]
Simplify
Why: Two cubed is 8.
\[ 12 = 8 a _{1} \]
Solve for a_1
Why: Twelve over 8.
\[ a _{1} = 1.5 \]
Write the rule
Why: One point five times 2 to the n minus 1.
\[ a _{n} = 1.5(2) ^{n - 1} \]
Figure (svg): The first six terms of a geometric sequence plotted on an exponential curve
\[ a_n = 1.5(2)^{\,n-1} \]
Verify: check at n equal to 4
Why: One point five times 2 cubed is 1.5 times 8, which is 12 — the given term. Listing the first six terms 1.5, 3, 6, 12, 24, 48 also shows 12 sitting in the fourth position, exactly where it was promised.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 811-811
Fill the middle
Example 3a.
Fill in the blanks
12 = a_1(2)^3 = 8a_1 \;\Longrightarrow\; a_1 = 1.5
Why: Twelve over 8 is 1.5. The fourth term sits three doublings above the first, so the first is one eighth of it.
Worked example
Example 4.
\[ \text{Two terms are } a_3 = -48 \text{ and } a_6 = 3072. \text{ Write a rule.} \]
Write two equations
Why: Substituting 3 and then 6 for n.
\[ -48 = a _{1} r ^{2}; 3072 = a _{1} r ^{5} \]
Solve the first for a_1
Why: Negative 48 over r squared.
\[ a _{1} = -48 / r ^{2} \]
Substitute into the second
Why: The r squared cancels three powers.
\[ 3072 = -48 r ^{3} \]
Solve for r, then a_1
Why: R cubed is negative 64.
\[ r = -4, a _{1} = -3 \]
Figure (svg): The solution to Worked example from two terms shown as a ladder of expressions, one row per algebraic move
\[ a_n = -3(-4)^{\,n-1} \]
Verify: check both given terms
Why: At n equal to 3 the rule gives negative 3 times 16, which is negative 48; at n equal to 6 it gives negative 3 times negative 1024, which is 3072. Dividing one equation by the other is the quicker route: 3072 over negative 48 is negative 64, which is r cubed directly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 812-812
Trap
\[ a_4 = 12, \; r = 2 \]
Use 12 as a_1
Why: The given value is the only term in sight, so it is taken as the start.
\[ a_n = 12(2)^{\,n-1} \quad \text{(wrong)} \]
That rule gives 96 at n equal to 4, but it should give 12 there. The subscript names the position, not the start.
\[ 12 = a_1(2)^3 \;\Longrightarrow\; a_1 = 1.5 \]
Substitute the position, then solve for a_1
Why: Three doublings separate the first term from the fourth.
\[ a_n = 1.5(2)^{\,n-1} \]
Undoing three doublings means dividing 12 by 8, which is why a_1 is well below the given term.
Fill the middle
Example 4, Step 2.
Fill in the blanks
\frac-4___ = r^___ = r^3 \;\Longrightarrow\; r = ___
Why: Three thousand seventy-two over negative 48 is negative 64, whose cube root is negative 4. Dividing the two equations removes a_1 entirely.
Matching
Solve for a_1 first in each case.
Match the pairs
Why: The last one has a subtlety: r squared is one quarter, so r could be one half or negative one half, and both give a valid sequence through the two stated terms. The positive choice is the usual convention.
Prediction
Commit before reasoning.
Predict first
From a_2 = -12 and a_4 = -3 you get r squared equal to 1/4. How many rules fit?
Correct: Two — r could be 1/2 or -1/2, giving a_1 = -24 or a_1 = 24.
\[ r^2 = \tfrac{1}{4} \;\Longrightarrow\; r = \pm\tfrac{1}{2} \]
Why: The two given positions are two apart, so only r squared is determined and the square root has two signs. Both resulting sequences pass through negative 12 at position 2 and negative 3 at position 4; they differ at the odd positions. Notice that this never happens in Example 4, where the positions were three apart and a cube root has just one real value. Whether the ambiguity arises depends on the parity of the gap between the two given positions.
Section
Section 4
Concept
Adding the terms of a geometric sequence gives a geometric series. Its sum has a closed formula built from the first term, the ratio and how many terms there are.
\[ S_n = a_1\!\left(\frac{1-r^{\,n}}{1-r}\right), \quad r \neq 1 \]
The derivation subtracts r times the sum from the sum. Every middle term cancels, leaving only the two ends.
Figure (svg): The subtract-r-times-it derivation of the geometric series sum formula
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 812-812 — The Sum of a Finite Geometric Series
Picture it
Where the formula comes from.
Figure (svg): The subtract-r-times-it derivation of the geometric series sum formula
Each middle term appears once with a plus and once with a minus. Only the first term and the new last one survive.
Worked example
Example 5.
\[ \text{Find } \sum_{i=1}^{16} 4(3)^{\,i-1}. \]
Identify the first term
Why: Substitute i equal to 1.
\[ a _{1} = 4 \]
Identify the ratio
Why: The base of the power.
\[ r = 3 \]
Apply the formula
Why: Four times one minus 3 to the 16th, over 1 minus 3.
\[ S _{16} = 4(1 - 3 ^{16}) / (-2) \]
Compute
Why: Three to the 16th is 43,046,721.
\[ 86, 093, 440 \]
Figure (svg): The subtract-r-times-it derivation of the geometric series sum formula
\[ S_{16} = 86{,}093{,}440 \]
Verify: compare with the last term alone
Why: The sixteenth term is 4 times 3 to the fifteenth, which is 57,395,628 — roughly two thirds of the whole total. With a ratio of 3 the last term always dominates, so a total far larger than twice the last term would signal an error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 813-813
Fill the middle
Guided Practice 7.
Fill in the blanks
S_8 = 6\!\left(\frac-510___\right) = 6\!\left(\frac______\right) = ___
Why: Negative 255 over 3 is negative 85, and 6 times that is negative 510. Note that the denominator became 3, not negative 1, because subtracting a negative adds.
Worked example
Guided Practice 7.
\[ \text{Find } \sum_{i=1}^{8} 6(-2)^{\,i-1}. \]
Identify a_1 and r
Why: Six, and negative 2.
\[ a _{1} = 6, r = -2 \]
Substitute
Why: One minus negative 2 to the eighth, over 1 minus negative 2.
\[ S _{8} = 6(1 - 256) / 3 \]
Simplify the numerator
Why: Negative 2 to the eighth is positive 256.
\[ 6(-255) / 3 \]
Compute
Why: Six times negative 85.
\[ -510 \]
Figure (svg): The solution to Worked example a negative ratio shown as a ladder of expressions, one row per algebraic move
\[ S_8 = -510 \]
Verify: check by pairing the terms
Why: The terms are 6, negative 12, 24, negative 48, 96, negative 192, 384, negative 768. Each consecutive pair sums to negative 6, negative 24, negative 96 and negative 384, and those four add to negative 510. A negative total makes sense because the largest term in the list is negative.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 813-813
Error analysis
A student totals the series 7 + 7 + 7 + 7 + 7 with the geometric sum formula.
Annotate
On: \( S_5 = 7\!\left(\frac{1-1^5}{1-1}\right) = \frac{0}{0} \)
This is why the boxed formula carries the condition r not equal to 1. It is the single ratio the derivation cannot handle.
Sorting
In the sum of 4 times 3 to the i minus 1, for i from 1 to 16.
Sort into buckets
Sort each number by its role in the sum formula.
Unlike the arithmetic sum formula, this one does need r — and unlike that formula, it does NOT need the last term.
Comparison
Fill the blanks. Both total n terms.
Comparison matrix
| Question | Arithmetic | Geometric |
|---|---|---|
| Formula | S_n = n(a_1 + a_n)/2 | S_n = a_1(1 - r^n)/(1 - r) |
| Needs the last term? | yes | no |
| Needs d or r? | no | yes |
| Excluded case | none | r = 1 |
The two formulas need opposite inputs, which is a useful memory hook: the arithmetic one wants both ends, the geometric one wants the start and the multiplier.
Prediction
Commit before reasoning.
Predict first
In the derivation, why do all the middle terms disappear when r times S is subtracted from S?
Correct: Because multiplying S by r shifts every term one place, so each middle term appears in both lines.
\[ S-rS = a_1-a_1r^{\,n} \]
Why: The sum contains a_1 r, a_1 r squared and so on; multiplying by r turns those into a_1 r squared, a_1 r cubed and so on — the same list shifted by one. Subtracting therefore cancels everything except the very first term of one line and the very last of the other. This telescoping is the same idea as the reverse-and-add trick of Lesson 12.2, adapted from addition to multiplication.
Section
Section 5
Concept
A quantity that increases by the same percent each period forms a geometric sequence with ratio 1 plus that percent as a decimal. Totalling several periods is a geometric series.
\[ r = 1+0.059 = 1.059 \]
A decrease of p percent gives r equal to 1 minus p instead, and the same formulas apply unchanged.
Figure (svg): Box office revenue rising by a constant percentage each year
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 813-813 — Use a geometric sequence and series in real life
Picture it
Example 6: U.S. box office revenue, 1990 to 2003.
Figure (svg): Box office revenue rising by a constant percentage each year
The bars grow by 5.9 percent each year, so each is 1.059 times the one before, and the total for the period is a geometric series.
Worked example
Example 6, both parts.
\[ \text{Revenue was } \$5.02 \text{ billion in } 1990 \text{ and rose } 5.9\% \text{ a year. Model it and total } 1990\text{-}2003. \]
Find r from the percent
Why: One plus 0.059.
\[ r = 1.059 \]
Write the rule with n = 1 for 1990
Why: Five point zero two times 1.059 to the n minus 1.
\[ a _{n} = 5.02(1.059) ^{n - 1} \]
Count the years
Why: Nineteen ninety through 2003 inclusive.
\[ 14\text{ years} \]
Apply the sum formula
Why: Five point zero two times 1 minus 1.059 to the 14th, over negative 0.059.
\[ \text{about } 105 \]
Figure (svg): Box office revenue rising by a constant percentage each year
\[ a_n = 5.02(1.059)^{\,n-1}; \quad S_{14} \approx 105 \]
Verify: sanity-check the total against the years
Why: Fourteen years averaging about $7.5 billion gives $105 billion, and the yearly figures do run from $5.02 to about $11 billion, so an average near $7.5 billion is right. Counting 2003 minus 1990 as 13 years instead of 14 is the classic slip; both endpoints count.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 813-813
Fill the middle
Example 6a.
Fill in the blanks
\text1.059 5.9\% \;\Longrightarrow\; r = 1+0.059 = ___
Why: The new value keeps all of the old and adds 5.9 percent more, so the multiplier is 1.059 rather than 0.059.
Worked example
Guided Practice 8.
\[ \text{Estimate the box office revenue in } 2000. \]
Find the position of 2000
Why: Two thousand minus 1990, plus 1.
\[ n = 11 \]
Substitute into the rule
Why: Five point zero two times 1.059 to the tenth.
\[ a _{11} = 5.02(1.059) ^{10} \]
Evaluate the power
Why: One point zero five nine to the tenth.
\[ \text{about } 1.774 \]
Multiply
Why: Five point zero two times 1.774.
\[ \text{about } 8.91 \]
Figure (svg): The solution to Worked example one year's revenue shown as a ladder of expressions, one row per algebraic move
\[ a_{11} \approx 8.91 \]
Verify: check the growth over ten years
Why: Revenue rose from $5.02 billion to about $8.91 billion in ten years, a 77 percent increase — which is what compounding 5.9 percent ten times should give, since 1.059 to the tenth is about 1.77. A simple 59 percent, from adding 5.9 ten times, would have been the arithmetic answer and is noticeably too small.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 813-813
Trap
\[ \text{growth of } 5.9\% \;\Longrightarrow\; r = 0.059 \]
Use the percent directly
Why: The number 5.9 percent is read straight into r.
\[ a_2 = 5.02(0.059) = 0.296 \quad \text{(wrong)} \]
A ratio of 0.059 would shrink the revenue to a twentieth of itself, not grow it by six percent.
\[ r = 1+0.059 = 1.059 \]
Add the percent to 1
Why: The new amount is the whole of the old plus the increase.
\[ a_2 = 5.02(1.059) \approx 5.32 \]
For a decrease of p percent the ratio is 1 minus p instead, so a 5.9 percent fall would give 0.941.
Matching
Growth adds to 1; decay subtracts from it.
Match the pairs
Why: A 100 percent increase is a doubling, which is why r is 2 rather than 1. Every decay ratio lands strictly between 0 and 1, and every growth ratio above 1.
Ranking
Smallest first.
Put in order
Why: The values are -510, 5.02, 8.91, 105 and 86,093,440. The negative total comes first precisely because a negative ratio can drag a sum below zero, something no positive-ratio series can do.
Prediction
Commit before reasoning.
Predict first
Growing at 5.9 percent for 13 years, is the total increase 13 times 5.9 percent, or about 77 percent?
Correct: About 116 percent, because each year's increase is computed on a larger base.
\[ 1.059^{13} \approx 2.11; \quad 13(0.059) = 0.767 \]
Why: One point zero five nine raised to the thirteenth is about 2.11, an increase of 111 percent, whereas adding 5.9 percent thirteen times suggests only 76.7 percent. The gap is compounding: the second year's 5.9 percent is taken on a base that already grew, and every later year compounds again. This is exactly the difference between an arithmetic and a geometric model, and it is why the distinction matters far outside this chapter.
Comparison
Fill the blanks. One adds, the other multiplies.
Comparison matrix
| Question | Arithmetic | Geometric |
|---|---|---|
| Test | subtract consecutive terms | divide consecutive terms |
| Rule | a_n = a_1 + (n - 1)d | a_n = a_1 times r to the n - 1 |
| Graph | collinear points | an exponential curve |
| Sum | n(a_1 + a_n)/2 | a_1(1 - r^n)/(1 - r) |
Both rules use n minus 1 for the same reason: the first term has had nothing done to it yet, whether the operation is adding d or multiplying by r.
Pattern
Test, rule, then total.
For percent growth the ratio is 1 plus the percent as a decimal, never the percent alone.
OpenStax Algebra and Trigonometry 2e, §13.3 Geometric Sequences §13.3
Check
Find a_1 and r first.
Check your understanding
What is a rule for the nth term of 4, 20, 100, 500, ...?
Answer: A
Why: The first term is 4 and every ratio is 5, so the rule is 4 times 5 to the n minus 1.
Check
Solve for the first term.
Check your understanding
One term of a geometric sequence is a_6 = -96 and r = 2. What is a rule for the nth term?
Answer: A
Why: -96 = a_1 times 2^5 = 32a_1, so a_1 = -3.
Check
Identify a_1 and r, then substitute.
Check your understanding
What is the sum of 6 times (-2)^(i-1) for i from 1 to 8?
Answer: A
Why: S_8 = 6(1 - 256)/(1 - (-2)) = 6(-255)/3 = -510.
Real world
A single-elimination tournament starts with 128 teams. Each round halves the number of teams still playing.
Discussion prompt
How many teams play in the sixth round, and how many games are played in the whole tournament?
Hint: Games in a round is half the teams in that round.
Answer:
\[ \text{teams: } a_n = 128\!\left(\tfrac{1}{2}\right)^{\,n-1} \;\Longrightarrow\; a_6 = 128\!\left(\tfrac{1}{32}\right) = 4 \]
\[ \text{games: } 64+32+16+8+4+2+1 = 64\!\left(\frac{1-(1/2)^7}{1-1/2}\right) = 127 \]
Four teams play in the sixth round, and 127 games are played in total.
The 127 is worth staring at: with 128 teams and one loser eliminated per game, exactly 127 teams must lose before a champion remains, so 127 games is forced by counting alone. The geometric series and the counting argument agree, which is the best kind of check — two completely different routes landing on the same number. That near-miss of 128 is characteristic of a ratio of one half: the total always falls one term short of double the first.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can the geometric sum formula total the series 7 + 7 + 7 + 7 + 7?
Correct: No — the ratio is 1, so the formula divides by zero; the sum is n times a_1, or 35.
\[ r = 1 \;\Longrightarrow\; S_n = n\,a_1 \]
Why: The series IS geometric, with a_1 equal to 7 and r equal to 1, so the first and last options are each half right. But the derivation divided both sides by 1 minus r, and that step is illegal when r is 1, which is exactly why the boxed formula carries the condition. When r is 1 every term equals a_1 and no formula is needed: five sevens is 35. This single excluded case is the geometric formula's only gap.
Explain it
They know arithmetic sequences and have never met geometric ones.
Discussion prompt
In four sentences or fewer, explain how a geometric sequence differs from an arithmetic one.
Hint: One repeats an addition; the other repeats a multiplication.
Answer:
An arithmetic sequence adds the same number every step, so its graph is a straight row of dots. A geometric one multiplies by the same number every step instead.
That makes the gaps grow as you go, so the dots curve upward rather than lining up. To test which you have, subtract consecutive terms for arithmetic and divide them for geometric.
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 exponent, test your rule at n equal to 1. For two given terms, divide one equation by the other so a_1 cancels. For signs, read the parity of the exponent n minus 1, not of n. For percents, the ratio is 1 plus the decimal for growth and 1 minus it for decay.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a geometric sequences page. Top left: write two sequences, one geometric and one not, and show the consecutive ratios under each; add a third that is arithmetic to remind yourself both tests exist. Top right: write the rule a_n = a_1 times r to the n minus 1 in large letters, label all three symbols, and note in one sentence why the exponent is n minus 1. Middle: take a term and a ratio of your own choosing, solve for a_1, write the rule, and plot the first six terms, sketching the exponential curve through them dashed. Bottom left: reproduce the subtract-r-times-it derivation and box the sum formula, writing beside it why r cannot be 1. Bottom right: solve a percent-growth problem of your own invention, showing the ratio as 1 plus the decimal and totalling several periods with the series formula.
If your percent problem used the bare decimal as the ratio, recheck it: that mistake shrinks a growing quantity to almost nothing in a single step.
Recap
Five things, and a second family of sequences to set beside the first.
| If you see | Then |
|---|---|
| Constant consecutive ratios | The sequence is geometric; r is that ratio |
| a_1 and r | Write a_n = a_1 times r to the n minus 1 |
| A term other than the first | Substitute its position and solve for a_1 |
| Two terms | Divide the equations so a_1 cancels |
| A total wanted | Use a_1 times 1 minus r to the n, over 1 minus r |
| A percent increase of p | The ratio is 1 plus p as a decimal |
Lesson 12.4 asks what happens when a geometric series never stops, and finds that when the ratio is small enough the infinite total is still a finite number.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.3 Analyze Geometric Sequences and Series §12.3, pp. 810-815 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.