UNIT 2Exponential functionsLesson 2 of 21 Exponential Growth and Decay2 Geometric Sequences

Geometric Sequences

A1.5Algebra 1Functions20 practice questions

In a geometric sequence, you multiply by the same number to get each new term.

What you'll be able to do

Find the common ratio and the next terms.
Write the explicit rule and the recursive rule.
Find a term far down the list, or the term number for a value.
Tell arithmetic, geometric and neither apart.

The big idea

Multiply each time. 5, 15, 45, 135 multiplies by 3 each time. The number 3 is the common ratio, r.

Find r by dividing. Divide any term by the term before it: 15 ÷ 5 = 3. Don't subtract.

The explicit rule. aₙ = a₁ · rⁿ⁻¹. The power is n − 1, because the first term has not been multiplied yet.

The recursive rule. Give the first term, then how to get the next one: a₁ = 5 and aₙ = 3 · aₙ₋₁.

Signs can switch. If r is negative, the signs take turns: 1, −4, 16, −64 has r = −4.

Count the steps. From term 2 to term 5 is 3 steps. So a₅ = a₂ · r³.

It is exponential. aₙ = a₁ · rⁿ⁻¹ is an exponential function. Its inputs are only 1, 2, 3 and so on.

Words you need

Sequence: A list of numbers in order.
Term: One number in a sequence. a₁ is the 1st term, and aₙ is the nth term.
Geometric sequence: A sequence that multiplies by the same number each time.
Common ratio: The number a geometric sequence multiplies by each time. It is r.
Arithmetic sequence: A sequence that adds the same number each time.
Common difference: The number an arithmetic sequence adds each time.
Explicit rule: A rule that gives any term from its term number n, like aₙ = 5 · 3ⁿ⁻¹.
Recursive rule: A rule that gives the first term and how to get each next term from the one before.
Exponential function: A rule y = a · bˣ. It multiplies by the same number b each time x goes up by 1.
Exponent: The small raised number. In 3ⁿ⁻¹, it is n − 1.
Power: A base with an exponent, like 2⁵.
Base: The number being raised to a power. In 2⁵, the base is 2.
Factor: A number you multiply by.
Product: The answer when you multiply.
Cube root: A number you use 3 times as a factor to get the first number. ∛8 = 2.
Variable: A letter that stands for a number, like n.
Expression: Numbers and letters joined by +, −, × or ÷. It has no = sign.
Logarithm: The power a base needs to make a number. log₂(16) = 4, because 2⁴ = 16.

Arithmetic or geometric?

Sequence	What it does	Type
4, 8, 12, 16	adds 4 each time	Arithmetic
4, 8, 16, 32	multiplies by 2 each time	Geometric
4, 5, 7, 10	adds 1, then 2, then 3	Neither

Sequence

What it does

Type

4, 8, 12, 16

adds 4 each time

Arithmetic

4, 8, 16, 32

multiplies by 2 each time

Geometric

4, 5, 7, 10

adds 1, then 2, then 3

Neither

Worked examples

Example 1

Find r and the next two terms of 2, −6, 18, −54.

Divide a term by the one before it: −6 ÷ 2 = −3. Check: 18 ÷ (−6) = −3.
Multiply −54 by −3 to get 162.
Multiply 162 by −3 to get −486.

Answer: r = −3, and the next two terms are 162 and −486

Example 2

Write the explicit rule and the recursive rule for 6, 12, 24, 48. Then find the 10th term.

The first term is 6, and r = 12 ÷ 6 = 2.
Explicit rule: aₙ = 6 · 2ⁿ⁻¹. Recursive rule: a₁ = 6 and aₙ = 2 · aₙ₋₁.
The 10th term is a₁₀ = 6 · 2⁹ = 6 × 512 = 3,072.
The graph shows the terms as dots. They sit on an exponential curve.

Answer: aₙ = 6 · 2ⁿ⁻¹, a₁ = 6 and aₙ = 2 · aₙ₋₁, and a₁₀ = 3,072

Example 3

The 2nd term of a geometric sequence is 12, and the 5th term is 324. Find r and the 1st term.

From term 2 to term 5 is 3 steps. So 12 · r³ = 324.
Divide by 12: r³ = 27. So r = 3.
Go back 1 step from term 2: a₁ = 12 ÷ 3 = 4.

Answer: r = 3 and a₁ = 4

Common mistakes

Using n instead of n − 1. For 6, 12, 24, 48, the 4th term is 6 · 2³ = 48, not 6 · 2⁴ = 96.

Subtracting to find r. For 4, 6, 9, r = 6 ÷ 4 = 1.5. The difference 2 is not the ratio.

Missing a negative ratio. If r² = 25, r can be 5 or −5. Check both.

Counting the steps wrong. From term 3 to term 7 is 4 steps, not 5.

Practice: 20 questions

13 of 20 are word problems. For a word problem, sketch it first and mark what you know.

Tier 1 Foundations Use the rule directly

Sequence A is 160, −80, 40, −20, ... and Sequence B is 81, 54, 36, 24, ... Find the common ratio r and the next two terms for each geometric sequence. Write any term that is not a whole number as a fraction in lowest terms.
Hugo sets up a volunteer phone tree. In Round 1, 4 volunteers are called. In each new round, every volunteer who was just called phones 3 more volunteers, so the number of people called in Rounds 1, 2, 3, and 4 is 4, 12, 36, and 108. Write both the recursive rule and the explicit rule for the number of people aₙ called in Round n. Word problem
Farida tracks a bacterial colony that starts with 5 cells in Hour 1 and doubles each hour, forming the sequence 5, 10, 20, 40 for Hours 1, 2, 3, and 4. She writes the explicit rule aₙ = 5(2)ⁿ and predicts 5(2)⁸ = 1,280 cells in Hour 8. Find her mistake, write the correct explicit rule, and find the actual number of cells in Hour 8. Word problem
Three sequences are given. Sequence A is 4, 12, 36, 108. Sequence B is 4, 12, 20, 28. Sequence C is 4, 12, 24, 40. Classify each sequence as arithmetic, geometric, or neither, and explain why. Then find the 5th term of each sequence and state which sequence has the greatest 5th term.
Jiro shrinks a poster on a copy machine. Copy 1 is 64 centimeters tall. Each new copy is 3/4 as tall as the copy before it, forming a geometric sequence where Copy 1 is the first term. How tall is Copy 5 in centimeters? Give your answer as a decimal. Word problem

Tier 2 Intermediate One more step to think through

Amina drops a rubber ball onto a gym floor. On Bounce 1, the ball reaches a peak height of 256 centimeters. On Bounce 2, it reaches 192 centimeters. On Bounce 3, it reaches 144 centimeters. The peak heights form a geometric sequence. Find the common ratio r, write the explicit rule for the height aₙ on Bounce n, and find the peak height on Bounce 5 in centimeters. Word problem
A geometric sequence has the explicit rule aₙ = 18(1/3)ⁿ⁻¹. Rewrite this rule in the exponential function form y = a · bⁿ, where the exponent is n instead of n − 1. Then rewrite the exponential function y = 10(2.5)ⁿ as an explicit geometric sequence rule aₙ = a₁ · rⁿ⁻¹ for n = 1, 2, 3, ... and state its first term a₁.
A frog population in a marsh is modeled by a geometric sequence with 16 frogs in Year 1 and 24 frogs in Year 2. Laszlo subtracts 24 − 16 = 8, says the common ratio is 8, and predicts 24 + 8 = 32 frogs in Year 3 and 32 + 8 = 40 frogs in Year 4. Find his mistake, find the true common ratio r, and find the actual predicted number of frogs in Year 4. Word problem
Thabo compares two weekly swimming plans. Plan A is an arithmetic sequence that starts at 10 laps in Week 1 and adds 6 laps each week. Plan B is a geometric sequence that starts at 2 laps in Week 1 and doubles the laps each week. In which week n does Plan B first have more laps than Plan A, and how many laps does each plan have in that week? Word problem
In the geometric sequence 3, 6, 12, 24, ..., which term number n has a value of 1,536?

Tier 3 Advanced Several steps - find the hidden structure

In a geometric sequence, the 1st term is 7 and the 4th term is 189. Find the common ratio r and the two missing terms between 7 and 189.
Renata grows bioluminescent algae in a tank. The cell count at the end of each day forms a geometric sequence. On Day 3 there are 120,000 cells, and on Day 6 there are 960,000 cells. How many cells were in the tank on Day 1, and how many will be in the tank on Day 9? Word problem
Evgeni is asked to find every possible value for the missing middle term in the geometric sequence 9, __, 144. He writes r² = 144 ÷ 9 = 16, takes r = 4, and says 36 is the only possible middle term. Explain what he missed and give the other possible middle term.
Kalpana compares two water filtration systems over repeated passes. Filter A leaves aₙ = 960(1/2)ⁿ⁻¹ milligrams of sediment per liter on Pass n. Filter B leaves bₙ = 405(2/3)ⁿ⁻¹ milligrams of sediment per liter on Pass n. On Pass 1, Filter B leaves less sediment. On which pass n do both filters leave the exact same amount of sediment, and on which pass does Filter A first leave less sediment than Filter B? Give the milligrams per liter for both filters on that pass. Word problem
Teodoro tests a vintage clock whose daily time loss forms a geometric sequence. On Day 1 the clock loses 3 seconds, and on Day 6 it loses 96 seconds. Write the explicit rule for the seconds lost aₙ on Day n, and find on which day the clock loses 768 seconds. Word problem

Tier 4 Challenge Stretch problems, some above grade level

Naledi records the number of new bamboo shoots in a greenhouse each week, which forms a geometric sequence a₁, a₂, a₃, a₄. The total of Week 1 and Week 2 is 15 shoots, and the total of Week 2 and Week 3 is 60 shoots. Find the common ratio r, the Week 1 count a₁, and the Week 4 count a₄. Word problem
Bastian tracks a message chain where 3 people receive a message in Stage 1, 9 people in Stage 2, and 27 people in Stage 3, modeled by aₙ = 3(3)ⁿ⁻¹. He wants to rewrite this rule in the exponential function form y = a · bⁿ for n = 1, 2, 3, ... He writes y = 3(3)ⁿ because the starting term is 3. Explain his mistake and write the correct exponential function y = a · bⁿ. Word problem
In Geometric Sequence X, every term is positive, and the 7th term is 9 times the 5th term. In Geometric Sequence Y, the terms alternate in sign, and the 7th term is also 9 times the 5th term. Find the common ratio r of each sequence, and compare the two sequences by finding the value of a₁₀ ÷ a₇ for Sequence X and for Sequence Y.
Sorin builds a 4-level display stand of wooden cubes where the number of cubes on each level from top to bottom forms a geometric sequence with common ratio r = 3. Level 1 is the top and Level 4 is the bottom. Level 4 has 72 more cubes than Level 2. How many cubes are on Level 1, and how many cubes are in the entire stand? Word problem
A logarithm tells you a power. For example, log₂(16) = 4 because 2⁴ = 16. Yael tracks a solar farm whose yearly energy output, in megawatt-hours, forms a geometric sequence with a₁ = 6 in Year 1 and common ratio r = 5, so aₙ = 6 · 5ⁿ⁻¹. Write an exact expression with a base-5 logarithm for the value of n when the formula reaches 300 megawatt-hours, and state between which two consecutive whole years that output is reached. Word problem Above grade level

Answers

Tier 1 Foundations

Sequence A has r = −1/2 and next two terms 10, −5; Sequence B has r = 2/3 and next two terms 16, 32/3. For Sequence A, divide the second term by the first term: r = −80 ÷ 160 = −1/2. Multiply by −1/2 twice to get the next two terms: −20 × (−1/2) = 10 and 10 × (−1/2) = −5. For Sequence B, divide the second term by the first term: r = 54 ÷ 81 = 2/3. Multiply by 2/3 twice to get the next two terms: 24 × (2/3) = 16 and 16 × (2/3) = 32/3.
Recursive rule: a₁ = 4, aₙ = 3 · aₙ₋₁; explicit rule: aₙ = 4(3)ⁿ⁻¹. The first term is a₁ = 4, and each term is 3 times the term before it, so the common ratio is r = 3. The recursive rule states the first term and how to get the next term: a₁ = 4 and aₙ = 3 · aₙ₋₁. The explicit rule uses aₙ = a₁ · rⁿ⁻¹, which gives aₙ = 4(3)ⁿ⁻¹.
Farida used the exponent n instead of n − 1, which counted one extra doubling; the correct explicit rule is aₙ = 5(2)ⁿ⁻¹, and there are 640 cells in Hour 8. Because Hour 1 is the starting term a₁ = 5, reaching Hour n takes n − 1 doublings, not n doublings. The correct explicit rule is aₙ = 5(2)ⁿ⁻¹. For Hour 8, substitute n = 8: a₈ = 5(2)⁷ = 5 × 128 = 640 cells.
Sequence A is geometric because it multiplies by 3 each step, with 5th term 324; Sequence B is arithmetic because it adds 8 each step, with 5th term 36; Sequence C is neither because neither the differences nor the ratios are constant, with 5th term 60; Sequence A has the greatest 5th term. Sequence A has a constant ratio of 12 ÷ 4 = 3, so it is geometric and its 5th term is 108 × 3 = 324. Sequence B has a constant difference of 12 − 4 = 8, so it is arithmetic and its 5th term is 28 + 8 = 36. Sequence C adds 8, then 12, then 16, so it is neither; adding the next step of 20 gives a 5th term of 40 + 20 = 60. Comparing 324, 36, and 60 shows Sequence A is greatest.
20.25 centimeters. Use the explicit rule aₙ = a₁ · rⁿ⁻¹ with a₁ = 64 and r = 3/4 = 0.75. For Copy 5, substitute n = 5: a₅ = 64 · (3/4)⁴. Evaluate: 64 × (81 / 256) = 81 / 4 = 20.25 centimeters.

Tier 2 Intermediate

r = 0.75; aₙ = 256(0.75)ⁿ⁻¹; 81 centimeters on Bounce 5. Divide Bounce 2 by Bounce 1 to find the common ratio: r = 192 ÷ 256 = 0.75. With first term a₁ = 256 and r = 0.75, the explicit rule is aₙ = 256(0.75)ⁿ⁻¹. For Bounce 5, substitute n = 5: a₅ = 256(0.75)⁴ = 256 × (81 / 256) = 81 centimeters.
y = 54(1/3)ⁿ; aₙ = 25(2.5)ⁿ⁻¹ with a₁ = 25. Use exponent rules to split (1/3)ⁿ⁻¹ as (1/3)ⁿ · (1/3)⁻¹ = 3 · (1/3)ⁿ. Multiply 18 × 3 = 54 to get y = 54(1/3)ⁿ. For y = 10(2.5)ⁿ, find the first term at n = 1: a₁ = 10(2.5)¹ = 25. With a₁ = 25 and common ratio r = 2.5, the explicit sequence rule is aₙ = 25(2.5)ⁿ⁻¹.
Laszlo found a common difference by subtracting as if the sequence were arithmetic; the true common ratio is r = 1.5, and there are 54 frogs in Year 4. In a geometric sequence, find the common ratio r by dividing consecutive terms, not subtracting: r = 24 ÷ 16 = 1.5. Multiply by 1.5 to find Year 3: 24 × 1.5 = 36 frogs. Multiply by 1.5 again to find Year 4: 36 × 1.5 = 54 frogs.
Week 6, when Plan B has 64 laps and Plan A has 40 laps. List the weekly laps for Plan A by adding 6 each week: 10, 16, 22, 28, 34, 40. List the weekly laps for Plan B by doubling each week: 2, 4, 8, 16, 32, 64. In Week 5, Plan A is still ahead with 34 laps against 32; in Week 6, Plan B passes Plan A with 64 laps against 40.
n = 10, the 10th term. Write the explicit rule with a₁ = 3 and r = 2: aₙ = 3 · 2ⁿ⁻¹. Set 3 · 2ⁿ⁻¹ = 1,536 and divide both sides by 3: 2ⁿ⁻¹ = 512. Since 2⁹ = 512, n − 1 = 9, which gives n = 10.

Tier 3 Advanced

r = 3; the two missing terms are 21 and 63. From the 1st term to the 4th term is 3 multiplication steps, so 7 · r³ = 189. Divide by 7 to get r³ = 27, and take the cube root to find r = 3. Multiply by 3 twice to find the 2nd and 3rd terms: 7 × 3 = 21 and 21 × 3 = 63.
30,000 cells on Day 1; 7,680,000 cells on Day 9. From Day 3 to Day 6 is 3 steps, so r³ = 960,000 ÷ 120,000 = 8, which gives r = 2. Work backward 2 steps from Day 3 to Day 1 by dividing by 2²: a₁ = 120,000 ÷ 4 = 30,000 cells. Move forward 3 steps from Day 6 to Day 9 by multiplying by 2³: a₉ = 960,000 × 8 = 7,680,000 cells.
Because r² = 16 has two real solutions, r = 4 and r = −4, the common ratio could also be −4, which gives −36 as the other possible middle term. From the 1st term to the 3rd term is 2 steps, so 9 · r² = 144 and r² = 16. Taking the square root gives two solutions: r = 4 or r = −4. When r = −4, the middle term is 9 × (−4) = −36, and the third term is still (−36) × (−4) = 144.
Both filters leave 120 milligrams per liter on Pass 4; Filter A first leaves less sediment on Pass 5, with 60 milligrams per liter compared to 80 milligrams per liter for Filter B. Calculate the first five passes for Filter A by multiplying by 1/2: 960, 480, 240, 120, 60. Calculate the first five passes for Filter B by multiplying by 2/3: 405, 270, 180, 120, 80. The two sequences are equal on Pass 4 at 120 milligrams per liter, and Filter A first drops below Filter B on Pass 5 with 60 milligrams per liter against 80.
aₙ = 3(2)ⁿ⁻¹; Day 9. From Day 1 to Day 6 is 5 steps, so 3 · r⁵ = 96, which gives r⁵ = 32 and r = 2. The explicit rule is aₙ = 3(2)ⁿ⁻¹. Set 3(2)ⁿ⁻¹ = 768 and divide by 3 to get 2ⁿ⁻¹ = 256 = 2⁸, so n − 1 = 8 and n = 9.

Tier 4 Challenge

r = 4, a₁ = 3 shoots, and a₄ = 192 shoots. Since a₂ = r · a₁ and a₃ = r · a₂, factor out r from the second sum: a₂ + a₃ = r(a₁ + a₂). Substitute the given sums: 60 = r(15), which gives r = 4. Since a₁ + a₂ = a₁ + 4a₁ = 5a₁ = 15, we get a₁ = 3 shoots, and Week 4 is a₄ = 3 × 4³ = 192 shoots.
In y = a · bⁿ, the number a is the value at n = 0 rather than n = 1, so writing y = 3(3)ⁿ multiplies every stage by an extra 3; the correct exponential function is y = 1(3)ⁿ, or y = 3ⁿ. In the sequence aₙ = 3(3)ⁿ⁻¹, substituting n = 1 gives 3(3)⁰ = 3, whereas Bastian's rule y = 3(3)ⁿ gives 3(3)¹ = 9 at n = 1. Rewrite 3(3)ⁿ⁻¹ using exponent laws: 3 · 3ⁿ · 3⁻¹ = (3 ÷ 3) · 3ⁿ = 1 · 3ⁿ. Therefore, a = 1 and b = 3, giving y = 1(3)ⁿ, or y = 3ⁿ.
Sequence X has r = 3 and a₁₀ ÷ a₇ = 27; Sequence Y has r = −3 and a₁₀ ÷ a₇ = −27. For both sequences, a₇ ÷ a₅ = r² = 9. Since Sequence X has all positive terms, r = 3, and a₁₀ ÷ a₇ = r³ = 3³ = 27. Since Sequence Y alternates in sign, r = −3, and a₁₀ ÷ a₇ = r³ = (−3)³ = −27.
3 cubes on Level 1; 120 cubes in the entire 4-level stand. Write Level 2 and Level 4 in terms of the top level a₁: a₂ = 3a₁ and a₄ = 3³a₁ = 27a₁. Use the difference of 72 cubes: 27a₁ − 3a₁ = 72, so 24a₁ = 72 and a₁ = 3 cubes. The four levels have 3, 9, 27, and 81 cubes, giving a total of 3 + 9 + 27 + 81 = 120 cubes.
n = log₅(50) + 1; between Year 3 and Year 4. Set 6 · 5ⁿ⁻¹ = 300 and divide both sides by 6: 5ⁿ⁻¹ = 50. By the definition of a logarithm, n − 1 = log₅(50), so n = log₅(50) + 1. Since 5² = 25 and 5³ = 125, log₅(50) is between 2 and 3, so n = log₅(50) + 1 is between 3 and 4.

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