Lesson 2: Geometric Sequences and Series
Learning Objectives
- Understand geometric sequences and identify the common ratio
- Write explicit and recursive formulas for geometric sequences
- Find specific terms in a geometric sequence
- Calculate the sum of a finite geometric series
- Determine when an infinite geometric series converges
- Calculate the sum of a convergent infinite geometric series
- Apply geometric sequences to real-world problems (population growth, compound interest, physics)
- Distinguish between geometric and arithmetic sequences
Part 1: Geometric Sequences
Geometric Sequence: A sequence in which each term after the first is obtained by multiplying the previous term by a fixed constant called the common ratio, denoted by r.
General form: a₁, a₁r, a₁r², a₁r³, ...
Finding the Common Ratio
To find the common ratio r, divide any term by the previous term:
Example 1: Identifying a Geometric Sequence
Determine if the sequence is geometric. If so, find the common ratio:
3, 6, 12, 24, 48, ...
Solution:
Check if the ratio between consecutive terms is constant:
6/3 = 2
12/6 = 2
24/12 = 2
48/24 = 2
Answer: Yes, this is a geometric sequence with common ratio r = 2.
Example 2: Common Ratio with Fractions
Find the common ratio: 80, 40, 20, 10, 5, ...
Solution:
Divide consecutive terms:
r = 40/80 = 1/2
Check: 20/40 = 1/2
10/20 = 1/2
Answer: r = 1/2 (each term is half the previous term)
Example 3: Common Ratio with Negative Values
Find the common ratio: 5, -10, 20, -40, 80, ...
Solution:
Divide consecutive terms:
r = -10/5 = -2
Check: 20/(-10) = -2
-40/20 = -2
Answer: r = -2 (signs alternate because r is negative)
Explicit Formula for Geometric Sequences
where a₁ is the first term, r is the common ratio, and n is the term number
Example 4: Finding a Specific Term
Find the 8th term of the geometric sequence: 2, 6, 18, 54, ...
Solution:
Step 1: Identify a₁ and r
a₁ = 2, r = 6/2 = 3
Step 2: Use the explicit formula
a₈ = a₁ · r⁸⁻¹
a₈ = 2 · 3⁷
a₈ = 2 · 2,187
a₈ = 4,374
Answer: The 8th term is 4,374
Example 5: Finding the First Term
In a geometric sequence with r = 4, the 5th term is 1,024. Find the first term.
Solution:
Use aₙ = a₁ · rⁿ⁻¹ and solve for a₁:
1,024 = a₁ · 4⁵⁻¹
1,024 = a₁ · 4⁴
1,024 = a₁ · 256
a₁ = 1,024/256
a₁ = 4
Answer: The first term is 4
Example 6: Finding the Common Ratio
In a geometric sequence, a₁ = 3 and a₆ = 96. Find the common ratio r.
Solution:
Use aₙ = a₁ · rⁿ⁻¹:
96 = 3 · r⁶⁻¹
96 = 3 · r⁵
32 = r⁵
r = ⁵√32 = 2
Verification: Sequence is 3, 6, 12, 24, 48, 96
Answer: r = 2
Recursive Formula for Geometric Sequences
aₙ = aₙ₋₁ · r for n ≥ 2
Example 7: Using Recursive Formula
Write the recursive formula for the sequence: 5, 15, 45, 135, ...
Solution:
Step 1: Identify a₁ and r
a₁ = 5, r = 15/5 = 3
Step 2: Write the recursive formula
a₁ = 5
aₙ = 3 · aₙ₋₁ for n ≥ 2
Answer: a₁ = 5, aₙ = 3aₙ₋₁
Example 8: Geometric Mean
Find the geometric mean between 4 and 64.
Solution:
The geometric mean of two numbers a and b is √(ab):
Geometric mean = √(4 · 64)
= √256
= 16
Verification: The sequence 4, 16, 64 is geometric with r = 4
Answer: The geometric mean is 16
Check Your Understanding: Geometric Sequences
1. Is the sequence 2, 4, 8, 14, 22 geometric? Explain.
Answer: No, this is not a geometric sequence.
Check ratios: 4/2 = 2, but 8/4 = 2, but 14/8 = 1.75, and 22/14 ≈ 1.57
The ratio is not constant, so it's not geometric. (This is actually an arithmetic sequence with differences that change.)
2. Find the 10th term of the geometric sequence: 1, -3, 9, -27, ...
Answer: -19,683
a₁ = 1, r = -3/1 = -3
a₁₀ = 1 · (-3)⁹ = 1 · (-19,683) = -19,683
3. In a geometric sequence, a₃ = 20 and a₆ = 160. Find a₁.
Answer: a₁ = 5
First find r: a₆/a₃ = r³, so 160/20 = r³, thus 8 = r³, r = 2
Then: 20 = a₁ · 2², so 20 = 4a₁, thus a₁ = 5
Part 2: Finite Geometric Series
Geometric Series: The sum of the terms of a geometric sequence.
Example: 2 + 6 + 18 + 54 + 162 is a geometric series
Sum of a Finite Geometric Series
Alternative form: Sₙ = a₁(rⁿ - 1)/(r - 1)
Which Formula to Use?
Both formulas are equivalent, but one might be easier depending on whether r is greater than or less than 1:
If |r| < 1: Use Sₙ = a₁(1 - rⁿ)/(1 - r)
If |r| > 1: Use Sₙ = a₁(rⁿ - 1)/(r - 1)
Example 9: Sum of First n Terms
Find the sum of the first 6 terms: 3, 6, 12, 24, ...
Solution:
Step 1: Identify a₁, r, and n
a₁ = 3, r = 2, n = 6
Step 2: Apply the formula (using r > 1 form)
S₆ = 3(2⁶ - 1)/(2 - 1)
S₆ = 3(64 - 1)/1
S₆ = 3(63)
S₆ = 189
Answer: The sum is 189
Example 10: Sum with Fractional Common Ratio
Find the sum: 16 + 8 + 4 + 2 + 1
Solution:
Step 1: Identify values
a₁ = 16, r = 1/2, n = 5
Step 2: Apply the formula (using r < 1 form)
S₅ = 16(1 - (1/2)⁵)/(1 - 1/2)
S₅ = 16(1 - 1/32)/(1/2)
S₅ = 16(31/32)/(1/2)
S₅ = 16(31/32) · 2
S₅ = 31
Answer: The sum is 31
Example 11: Finding Number of Terms
How many terms of the sequence 5, 10, 20, 40, ... are needed to get a sum of 315?
Solution:
Step 1: Identify known values
a₁ = 5, r = 2, Sₙ = 315
Step 2: Set up equation and solve
315 = 5(2ⁿ - 1)/(2 - 1)
315 = 5(2ⁿ - 1)
63 = 2ⁿ - 1
64 = 2ⁿ
2⁶ = 2ⁿ
n = 6
Answer: 6 terms are needed
Example 12: Sum Using Sigma Notation
Evaluate: Σ(k=1 to 8) 3(2)^(k-1)
Solution:
This represents a geometric series with a₁ = 3, r = 2, n = 8
S₈ = 3(2⁸ - 1)/(2 - 1)
S₈ = 3(256 - 1)/1
S₈ = 3(255)
S₈ = 765
Answer: 765
Check Your Understanding: Finite Geometric Series
4. Find the sum of the first 5 terms: 1, 3, 9, 27, ...
Answer: 121
a₁ = 1, r = 3, n = 5
S₅ = 1(3⁵ - 1)/(3 - 1) = (243 - 1)/2 = 242/2 = 121
5. Find the sum: 64 + 32 + 16 + 8 + 4 + 2 + 1
Answer: 127
a₁ = 64, r = 1/2, n = 7
S₇ = 64(1 - (1/2)⁷)/(1 - 1/2) = 64(1 - 1/128)/(1/2) = 64(127/128) · 2 = 127
Part 3: Infinite Geometric Series
Infinite Geometric Series: The sum of all terms in an infinite geometric sequence.
Example: 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
Convergence vs. Divergence
When Does an Infinite Series Converge?
An infinite geometric series converges (has a finite sum) if and only if |r| < 1
If |r| ≥ 1: The series diverges (sum approaches infinity or does not exist)
Sum of an Infinite Geometric Series
Important: This formula ONLY works when |r| < 1. If |r| ≥ 1, the series does not have a sum (it diverges).
Example 13: Sum of a Convergent Infinite Series
Find the sum: 1 + 1/3 + 1/9 + 1/27 + ...
Solution:
Step 1: Identify a₁ and r
a₁ = 1, r = 1/3
Step 2: Check if series converges
|r| = |1/3| = 1/3 < 1 Series converges
Step 3: Apply the formula
S = 1/(1 - 1/3)
S = 1/(2/3)
S = 3/2 = 1.5
Answer: The sum is 3/2 or 1.5
Example 14: Divergent Series
Does the series 2 + 4 + 8 + 16 + ... have a sum?
Solution:
Identify r: r = 4/2 = 2
|r| = |2| = 2 > 1
Answer: No, this series diverges. It does not have a finite sum because |r| ≥ 1.
Example 15: Repeating Decimals as Geometric Series
Express 0.777... as a fraction using geometric series.
Solution:
Step 1: Write as a series
0.777... = 0.7 + 0.07 + 0.007 + 0.0007 + ...
Step 2: Identify a₁ and r
a₁ = 0.7 = 7/10
r = 0.07/0.7 = 1/10
Step 3: Apply infinite sum formula
S = (7/10)/(1 - 1/10)
S = (7/10)/(9/10)
S = (7/10) · (10/9)
S = 7/9
Answer: 0.777... = 7/9
Example 16: More Complex Repeating Decimal
Express 0.242424... as a fraction.
Solution:
Step 1: Write as a series
0.242424... = 0.24 + 0.0024 + 0.000024 + ...
Step 2: Identify values
a₁ = 0.24 = 24/100
r = 0.0024/0.24 = 1/100
Step 3: Apply formula
S = (24/100)/(1 - 1/100)
S = (24/100)/(99/100)
S = 24/99 = 8/33
Answer: 0.242424... = 8/33
Example 17: Series with Negative Common Ratio
Find the sum: 8 - 4 + 2 - 1 + 1/2 - 1/4 + ...
Solution:
Step 1: Identify a₁ and r
a₁ = 8, r = -4/8 = -1/2
Step 2: Check convergence
|r| = |-1/2| = 1/2 < 1 Converges
Step 3: Find sum
S = 8/(1 - (-1/2))
S = 8/(1 + 1/2)
S = 8/(3/2)
S = 8 · (2/3) = 16/3
Answer: The sum is 16/3 or approximately 5.33
Check Your Understanding: Infinite Geometric Series
6. Does the series 1 + 2 + 4 + 8 + 16 + ... converge? If so, find its sum.
Answer: No, this series diverges.
r = 2, and |r| = 2 > 1, so the series does not converge.
7. Find the sum: 6 + 2 + 2/3 + 2/9 + ...
Answer: 9
a₁ = 6, r = 2/6 = 1/3
|r| = 1/3 < 1, so it converges
S = 6/(1 - 1/3) = 6/(2/3) = 6 · (3/2) = 9
8. Express 0.999... as a fraction using geometric series.
Answer: 1
0.999... = 0.9 + 0.09 + 0.009 + ...
a₁ = 9/10, r = 1/10
S = (9/10)/(1 - 1/10) = (9/10)/(9/10) = 1
This proves that 0.999... = 1!
Part 4: Real-World Applications
Example 18: Population Growth
A bacterial culture starts with 500 bacteria and doubles every hour. How many bacteria will there be after 6 hours?
Solution:
This is a geometric sequence with a₁ = 500, r = 2, n = 7 (including the initial amount)
a₇ = 500 · 2⁶
a₇ = 500 · 64
a₇ = 32,000
Answer: After 6 hours, there will be 32,000 bacteria
Example 19: Compound Interest
You invest $1,000 at 5% annual interest compounded yearly. What will your investment be worth after 10 years?
Solution:
Each year, the amount is multiplied by 1.05 (100% + 5%)
a₁ = 1000, r = 1.05, n = 11 (to get year 10 value)
a₁₁ = 1000 · (1.05)¹⁰
a₁₁ = 1000 · 1.6289
a₁₁ ≈ $1,628.90
Answer: The investment will be worth approximately $1,628.90
Example 20: Bouncing Ball
A ball is dropped from a height of 20 feet. Each bounce reaches 60% of its previous height. What is the total vertical distance traveled by the ball?
Solution:
The ball travels down 20 ft initially, then up and down for each bounce:
Downward distances: 20 + 12 + 7.2 + 4.32 + ...
This is geometric with a₁ = 20, r = 0.6
S_down = 20/(1 - 0.6) = 20/0.4 = 50 ft
Upward distances: 12 + 7.2 + 4.32 + ...
This is geometric with a₁ = 12, r = 0.6
S_up = 12/(1 - 0.6) = 12/0.4 = 30 ft
Total distance = 50 + 30 = 80 ft
Answer: The ball travels a total of 80 feet
Example 21: Medication in Bloodstream
A patient takes 100 mg of medication. Each hour, 30% of the drug is eliminated from the bloodstream. How much medication remains after 5 hours?
Solution:
Each hour, 70% (100% - 30%) of the medication remains
a₁ = 100, r = 0.7, n = 6 (to get 5 hours later)
a₆ = 100 · (0.7)⁵
a₆ = 100 · 0.16807
a₆ ≈ 16.81 mg
Answer: Approximately 16.81 mg remains after 5 hours
Example 22: Salary with Annual Raise
Your starting salary is $40,000 and you receive a 3% raise each year. What will your salary be in year 8? What is your total earnings over the first 10 years?
Solution:
Part 1: Salary in year 8
a₁ = 40,000, r = 1.03, n = 8
a₈ = 40,000 · (1.03)⁷
a₈ = 40,000 · 1.2299
a₈ ≈ $49,196
Part 2: Total earnings over 10 years
S₁₀ = 40,000(1.03¹⁰ - 1)/(1.03 - 1)
S₁₀ = 40,000(1.3439 - 1)/0.03
S₁₀ = 40,000(0.3439)/0.03
S₁₀ ≈ $458,533
Answer: Year 8 salary: ~$49,196; Total 10 years: ~$458,533
Example 23: Depreciation
A car worth $25,000 depreciates 15% per year. What is its value after 6 years?
Solution:
Each year, the car retains 85% (100% - 15%) of its value
a₁ = 25,000, r = 0.85, n = 7 (to get year 6 value)
a₇ = 25,000 · (0.85)⁶
a₇ = 25,000 · 0.3771
a₇ ≈ $9,428
Answer: The car is worth approximately $9,428 after 6 years
Example 24: Fractal Perimeter
A fractal pattern starts with a square of side length 4 cm. At each stage, the perimeter increases by a factor of 4/3. What is the limiting perimeter as the process continues infinitely?
Solution:
Initial perimeter: 4 · 4 = 16 cm
Perimeters form geometric sequence: 16, 16(4/3), 16(4/3)², ...
Since r = 4/3 > 1, the series diverges
Answer: The perimeter grows without bound (approaches infinity)
This is the famous property of the Koch snowflake: finite area but infinite perimeter!
Check Your Understanding: Applications
9. A radioactive substance decays so that 20% is lost each day. If you start with 500 grams, how much remains after 1 week?
Answer: Approximately 104.86 grams
Each day, 80% remains: r = 0.8
a₈ = 500 · (0.8)⁷ = 500 · 0.2097 ≈ 104.86 grams
10. An investment of $5,000 earns 4% annual interest compounded yearly. What is the total value after 15 years?
Answer: Approximately $9,006
a₁₆ = 5000 · (1.04)¹⁵ = 5000 · 1.8009 ≈ $9,004.50
Key Takeaways
- A geometric sequence has a constant ratio r between consecutive terms
- Explicit formula: aₙ = a₁ · rⁿ⁻¹
- Recursive formula: a₁ = (first term), aₙ = r · aₙ₋₁
- Sum of finite geometric series: Sₙ = a₁(1 - rⁿ)/(1 - r) when r ≠ 1
- An infinite geometric series converges only when |r| < 1
- Sum of infinite geometric series: S = a₁/(1 - r) when |r| < 1
- If |r| ≥ 1, an infinite geometric series diverges (no finite sum)
- Geometric sequences model exponential growth (r > 1) and decay (0 < r < 1)
- Common applications: compound interest, population growth, depreciation, medication decay
- Repeating decimals can be expressed as fractions using infinite geometric series
- When r is negative, the terms alternate in sign
- Always check if |r| < 1 before applying the infinite sum formula