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Lesson 4: Applications of Sequences and Series

Previous Lesson Module 6, Lesson 4 of 4 Practice Problems

Learning Objectives

Apply arithmetic and geometric sequences to financial problems including compound interest, annuities, and loans
Model population growth and decay using geometric sequences
Use sequences to solve physics problems involving motion, pendulums, and sound waves
Apply sequences and series to combinatorics and counting problems
Translate real-world word problems into sequence and series formulas
Choose the appropriate formula based on the problem context
Interpret solutions in the context of real-world applications
Develop strategies for multi-step problem solving

Introduction: Sequences and Series in the Real World

Sequences and series are powerful mathematical tools used across many fields including finance, biology, physics, computer science, and engineering. In this lesson, we'll explore how to apply the formulas and concepts from previous lessons to solve real-world problems.

        Problem-Solving Strategy
        Read carefully: Understand what the problem is asking
Identify the type: Is it arithmetic or geometric? Does it involve a sum?
Define variables: What are you looking for?
Extract given information: What values are provided?
Choose the formula: Which sequence or series formula applies?
Solve: Substitute values and calculate
Interpret: Answer the question in context and check reasonableness

      

Key Formulas Reference
Arithmetic: a_n = a₁ + (n - 1)d, S_n = n(a₁ + a_n)/2
Geometric: a_n = a₁ · r^n-1, S_n = a₁(1 - rⁿ)/(1 - r)
Infinite Geometric Sum: S = a₁/(1 - r) when |r| < 1

Part 1: Financial Applications

Compound Interest and Savings

Compound Interest: When interest is calculated on the initial principal and also on the accumulated interest from previous periods. This creates a geometric sequence.

Formula: A_n = P(1 + r)ⁿ

Where P = principal, r = interest rate per period, n = number of periods

Example 1: Simple Compound Interest

You deposit $5,000 into a savings account that earns 6% annual interest compounded annually. How much will you have after 10 years?

Solution:

Step 1: Identify the type.

This is a geometric sequence because each year the balance is multiplied by (1 + r).

Step 2: Extract given information.

P = $5,000 (principal)
r = 6% = 0.06 (annual rate)
n = 10 (years)

Step 3: Use the compound interest formula.

A₁₀ = 5000(1 + 0.06)¹⁰
A₁₀ = 5000(1.06)¹⁰
A₁₀ = 5000(1.7908)
A₁₀ = $8,954.24

Answer: After 10 years, you will have approximately $8,954.24.

Example 2: Finding Time to Reach a Goal

How many years will it take for $3,000 to grow to $8,000 at 8% annual interest compounded annually?

Solution:

Step 1: Set up the equation.

A_n = P(1 + r)ⁿ
8000 = 3000(1.08)ⁿ

Step 2: Solve for n.

8000/3000 = (1.08)ⁿ
2.667 = (1.08)ⁿ

Step 3: Use logarithms.

log(2.667) = n · log(1.08)
0.4260 = n · 0.0334
n = 0.4260/0.0334
n = 12.75 years

Answer: It will take approximately 12.75 years, or about 13 years.

Example 3: Monthly Compound Interest

You invest $10,000 at 4.8% annual interest compounded monthly. What is the value after 5 years?

Solution:

Step 1: Adjust for monthly compounding.

Monthly rate r = 4.8%/12 = 0.4% = 0.004
Number of periods n = 5 years × 12 months = 60 months

Step 2: Apply the formula.

A₆₀ = 10,000(1 + 0.004)⁶⁰
A₆₀ = 10,000(1.004)⁶⁰
A₆₀ = 10,000(1.2707)
A₆₀ = $12,707.42

Answer: After 5 years, the investment is worth $12,707.42.

Annuities and Regular Deposits

Annuity: A series of equal payments made at regular intervals. The future value of an annuity is the sum of a geometric series.

Future Value of Annuity Formula: FV = PMT × [(1 + r)ⁿ - 1]/r

Where PMT = regular payment, r = interest rate per period, n = number of periods

Example 4: Retirement Savings

You deposit $200 at the end of each month into a retirement account that earns 6% annual interest compounded monthly. How much will you have after 30 years?

Solution:

Step 1: Identify given values.

PMT = $200 (monthly payment)
r = 6%/12 = 0.5% = 0.005 (monthly rate)
n = 30 × 12 = 360 (total months)

Step 2: Apply the annuity formula.

FV = 200 × [(1.005)³⁶⁰ - 1]/0.005
FV = 200 × [6.0226 - 1]/0.005
FV = 200 × 5.0226/0.005
FV = 200 × 1004.52
FV = $200,904.15

Answer: After 30 years of $200 monthly deposits, you will have approximately $200,904.15.

Example 5: Finding Required Payment

You want to save $50,000 for a down payment in 8 years. If your account earns 5% annual interest compounded monthly, how much should you deposit each month?

Solution:

Step 1: Identify known values.

FV = $50,000 (desired future value)
r = 5%/12 = 0.004167 (monthly rate)
n = 8 × 12 = 96 (months)

Step 2: Solve for PMT.

50,000 = PMT × [(1.004167)⁹⁶ - 1]/0.004167
50,000 = PMT × [1.4898 - 1]/0.004167
50,000 = PMT × 117.51
PMT = 50,000/117.51
PMT = $425.38

Answer: You need to deposit $425.38 each month.

Example 6: Loan Payments

You borrow $20,000 for a car at 4.5% annual interest. If you make equal monthly payments for 5 years, what is your monthly payment?

Solution:

Step 1: Use the loan payment formula.

PMT = [P · r(1 + r)ⁿ]/[(1 + r)ⁿ - 1]
Where P = $20,000, r = 4.5%/12 = 0.00375, n = 60

Step 2: Calculate.

PMT = [20,000 · 0.00375(1.00375)⁶⁰]/[(1.00375)⁶⁰ - 1]
PMT = [20,000 · 0.00375 · 1.2516]/[1.2516 - 1]
PMT = 93.87/0.2516
PMT = $373.28

Answer: Your monthly payment will be $373.28.

Example 7: Total Interest Paid

For the car loan in Example 6, how much total interest will you pay over 5 years?

Solution:

Step 1: Calculate total amount paid.

Total payments = $373.28 × 60 months = $22,396.80

Step 2: Subtract the principal.

Total interest = $22,396.80 - $20,000 = $2,396.80

Answer: You will pay $2,396.80 in interest over 5 years.

Part 2: Population and Growth Models

Exponential Growth and Decay

Exponential Growth/Decay: Populations that grow or shrink by a constant percentage form geometric sequences.

Formula: P_n = P₀(1 + r)ⁿ

Where P₀ = initial population, r = growth rate (positive) or decay rate (negative), n = time periods

Example 8: Bacterial Growth

A bacteria culture starts with 500 bacteria and doubles every 3 hours. How many bacteria will there be after 24 hours?

Solution:

Step 1: Identify the pattern.

Doubling means r = 2 (multiply by 2)
Number of 3-hour periods in 24 hours: n = 24/3 = 8

Step 2: Apply geometric sequence formula.

a_n = a₁ · r^n-1
a₉ = 500 · 2⁸
a₉ = 500 · 256
a₉ = 128,000 bacteria

Answer: After 24 hours, there will be 128,000 bacteria.

Example 9: Population Growth

A city's population is 250,000 and grows at 2.5% per year. What will the population be in 15 years?

Solution:

Step 1: Set up the formula.

P₁₅ = 250,000(1 + 0.025)¹⁵
P₁₅ = 250,000(1.025)¹⁵

Step 2: Calculate.

P₁₅ = 250,000(1.4483)
P₁₅ = 362,075

Answer: The population will be approximately 362,075 people in 15 years.

Example 10: Radioactive Decay

A radioactive substance has a half-life of 5 years. If you start with 80 grams, how much remains after 20 years?

Solution:

Step 1: Identify the decay pattern.

Half-life means r = 0.5 (multiply by 0.5)
Number of 5-year periods: n = 20/5 = 4

Step 2: Calculate.

a₅ = 80 · (0.5)⁴
a₅ = 80 · 0.0625
a₅ = 5 grams

Answer: After 20 years, 5 grams of the substance remains.

Example 11: Wildlife Population Decline

A wildlife sanctuary has 1,200 endangered birds. Due to habitat loss, the population decreases by 8% each year. How many birds will remain after 10 years?

Solution:

Step 1: Set up decay formula.

Decrease of 8% means multiply by (1 - 0.08) = 0.92
P₁₀ = 1200(0.92)¹⁰

Step 2: Calculate.

P₁₀ = 1200(0.4344)
P₁₀ = 521 birds

Answer: Approximately 521 birds will remain after 10 years.

Example 12: Finding Growth Rate

A town's population grew from 15,000 to 20,000 in 8 years. What was the average annual growth rate?

Solution:

Step 1: Set up the equation.

20,000 = 15,000(1 + r)⁸
20,000/15,000 = (1 + r)⁸
1.3333 = (1 + r)⁸

Step 2: Solve for r using logarithms or roots.

1 + r = (1.3333)^1/8
1 + r = 1.0367
r = 0.0367 = 3.67%

Answer: The average annual growth rate was approximately 3.67%.

Part 3: Physics Applications

Motion and Falling Objects

Falling Objects: In the absence of air resistance, falling objects form arithmetic sequences for distance fallen in successive time intervals.

Distance fallen in successive seconds: 16, 48, 80, 112... (arithmetic with d = 32)

Example 13: Free Fall Distance

An object falls from rest. It falls 16 feet in the first second, 48 feet in the second second, 80 feet in the third second, and so on. How far does it fall in the 10th second?

Solution:

Step 1: Identify the sequence.

a₁ = 16, a₂ = 48, a₃ = 80
Common difference: d = 48 - 16 = 32
This is an arithmetic sequence.

Step 2: Find the 10th term.

a_n = a₁ + (n - 1)d
a₁₀ = 16 + (10 - 1)(32)
a₁₀ = 16 + 288
a₁₀ = 304 feet

Answer: The object falls 304 feet during the 10th second.

Example 14: Total Distance Fallen

For the object in Example 13, what is the total distance fallen after 10 seconds?

Solution:

Step 1: Use arithmetic series formula.

We need the sum of the first 10 terms.
S_n = n(a₁ + a_n)/2

Step 2: Calculate.

S₁₀ = 10(16 + 304)/2
S₁₀ = 10(320)/2
S₁₀ = 1,600 feet

Answer: The total distance fallen after 10 seconds is 1,600 feet.

Pendulums and Damped Motion

Example 15: Pendulum Swing

A pendulum swings 40 cm on its first swing. Due to air resistance, each subsequent swing is 95% of the previous swing. How far does it swing on the 12th swing?

Solution:

Step 1: Identify as geometric sequence.

a₁ = 40 cm
r = 0.95 (each swing is 95% of previous)
Find a₁₂

Step 2: Apply geometric formula.

a_n = a₁ · r^n-1
a₁₂ = 40 · (0.95)¹¹
a₁₂ = 40 · 0.5688
a₁₂ = 22.75 cm

Answer: The 12th swing is approximately 22.75 cm.

Example 16: Total Distance of Pendulum

For the pendulum in Example 15, what is the total distance traveled before it comes to rest?

Solution:

Step 1: Identify as infinite geometric series.

Since |r| = 0.95 < 1, the infinite sum exists.
a₁ = 40, r = 0.95

Step 2: Use infinite sum formula.

S = a₁/(1 - r)
S = 40/(1 - 0.95)
S = 40/0.05
S = 800 cm

Answer: The pendulum travels a total of 800 cm before coming to rest.

Example 17: Bouncing Ball

A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 80% of its previous height. What is the total vertical distance traveled by the ball before it stops bouncing?

Solution:

Step 1: Analyze the motion.

Initial drop: 10 m (down)
First bounce up: 10(0.8) = 8 m, then down 8 m
Second bounce up: 8(0.8) = 6.4 m, then down 6.4 m
Pattern: 10 + 2[8 + 6.4 + 5.12 + ...]

Step 2: The bounce heights form a geometric series.

Bounce heights: 8, 6.4, 5.12, ... (a₁ = 8, r = 0.8)
Sum of bounces = 8/(1 - 0.8) = 8/0.2 = 40 m

Step 3: Calculate total distance.

Total = initial drop + 2(sum of bounce heights)
Total = 10 + 2(40) = 10 + 80 = 90 m

Answer: The ball travels a total vertical distance of 90 meters.

Sound and Waves

Example 18: Sound Wave Amplitude

A sound wave has an initial amplitude of 100 decibels. Due to distance and absorption, the amplitude decreases by 12% for each meter traveled. What is the amplitude after the sound travels 8 meters?

Solution:

Step 1: Set up geometric sequence.

a₁ = 100 decibels
r = 1 - 0.12 = 0.88 (decreases by 12%)
Find a₉ (after 8 meters means 9th term)

Step 2: Calculate.

a₉ = 100 · (0.88)⁸
a₉ = 100 · 0.3596
a₉ = 35.96 decibels

Answer: The amplitude is approximately 36 decibels after 8 meters.

Part 4: Combinatorics and Counting

Patterns and Arrangements

Example 19: Seating Arrangements

A theater has 20 rows. The first row has 15 seats, the second row has 18 seats, the third row has 21 seats, and so on. How many seats are in the 20th row?

Solution:

Step 1: Identify the pattern.

a₁ = 15, a₂ = 18, a₃ = 21
d = 3 (arithmetic sequence)

Step 2: Find the 20th term.

a₂₀ = a₁ + (n - 1)d
a₂₀ = 15 + (20 - 1)(3)
a₂₀ = 15 + 57
a₂₀ = 72 seats

Answer: The 20th row has 72 seats.

Example 20: Total Theater Capacity

For the theater in Example 19, what is the total seating capacity?

Solution:

Step 1: Use arithmetic series formula.

S_n = n(a₁ + a_n)/2
S₂₀ = 20(15 + 72)/2

Step 2: Calculate.

S₂₀ = 20(87)/2
S₂₀ = 1,740/2
S₂₀ = 870 seats

Answer: The total seating capacity is 870 seats.

Example 21: Stacking Blocks

You're building a pyramid with blocks. The top level has 1 block, the second level has 4 blocks, the third level has 9 blocks, the fourth has 16 blocks. How many blocks are in the 10th level?

Solution:

Step 1: Identify the pattern.

Level 1: 1 = 1²
Level 2: 4 = 2²
Level 3: 9 = 3²
Level 4: 16 = 4²
Pattern: a_n = n²

Step 2: Find the 10th term.

a₁₀ = 10² = 100 blocks

Answer: The 10th level has 100 blocks.

Example 22: Handshakes Problem

At a party, every person shakes hands with every other person exactly once. If there are n people, the total number of handshakes follows the pattern: 0, 1, 3, 6, 10, 15, ... How many handshakes occur at a party with 20 people?

Solution:

Step 1: Identify the pattern.

These are triangular numbers: 1, 1+2=3, 1+2+3=6, 1+2+3+4=10
Formula: H_n = n(n-1)/2

Step 2: Calculate for 20 people.

H₂₀ = 20(20-1)/2
H₂₀ = 20(19)/2
H₂₀ = 380/2
H₂₀ = 190 handshakes

Answer: There are 190 handshakes at a party with 20 people.

Part 5: Multi-Step Problem Solving

Complex Real-World Problems

Example 23: Job Offer Comparison

You have two job offers. Company A offers $50,000 in the first year with a $2,000 raise each year. Company B offers $48,000 in the first year with a 5% raise each year. Which job pays more over 10 years?

Solution:

Step 1: Analyze Company A (arithmetic).

a₁ = 50,000, d = 2,000
a₁₀ = 50,000 + 9(2,000) = 68,000
S₁₀ = 10(50,000 + 68,000)/2 = 590,000

Step 2: Analyze Company B (geometric).

a₁ = 48,000, r = 1.05
S₁₀ = 48,000[(1.05)¹⁰ - 1]/(1.05 - 1)
S₁₀ = 48,000[1.6289 - 1]/0.05
S₁₀ = 48,000(12.578) = 603,744

Step 3: Compare.

Company B: $603,744
Company A: $590,000
Difference: $13,744

Answer: Company B pays $13,744 more over 10 years.

Example 24: Drug Dosage

A patient takes 100 mg of a medication daily. The body eliminates 20% of the drug each day. After many days, the amount of drug in the body reaches an equilibrium. What is the maximum amount of drug in the body?

Solution:

Step 1: Model the situation.

Day 1: 100 mg
Day 2: 100 + 100(0.8) = 100 + 80 = 180 mg
Day 3: 100 + 180(0.8) = 100 + 144 = 244 mg
Pattern: 100 + 100(0.8) + 100(0.8)² + ...

Step 2: Recognize as geometric series.

This is 100[1 + 0.8 + 0.8² + 0.8³ + ...]
Infinite geometric series with a₁ = 1, r = 0.8

Step 3: Calculate equilibrium amount.

Sum = 1/(1 - 0.8) = 1/0.2 = 5
Total = 100 × 5 = 500 mg

Answer: The equilibrium amount is 500 mg in the body.

Example 25: Chain Letter

A chain letter asks you to send it to 5 friends. Each of them sends it to 5 more friends, and so on. If everyone participates and no one receives the letter twice, how many people receive the letter in the 8th generation?

Solution:

Step 1: Identify the pattern.

Generation 1: 5 people
Generation 2: 5 × 5 = 25 people
Generation 3: 25 × 5 = 125 people
Geometric sequence with a₁ = 5, r = 5

Step 2: Find 8th generation.

a₈ = 5 · 5⁷
a₈ = 5 · 78,125
a₈ = 390,625 people

Answer: The 8th generation has 390,625 people.

Check Your Understanding

1. You invest $8,000 at 5% annual interest compounded annually. How much will you have after 12 years?

Answer: $14,365.84

A = 8000(1.05)¹² = 8000(1.7959) = $14,365.84

2. You deposit $150 per month into an account earning 6% annual interest compounded monthly. How much will you have after 20 years?

Answer: $69,385.78

r = 0.06/12 = 0.005, n = 240 months
FV = 150[(1.005)²⁴⁰ - 1]/0.005 = 150(462.572) = $69,385.78

3. A bacteria population doubles every 4 hours. Starting with 200 bacteria, how many will there be after 28 hours?

Answer: 25,600 bacteria

n = 28/4 = 7 doubling periods
a₈ = 200 · 2⁷ = 200 · 128 = 25,600

4. A city's population of 180,000 is declining at 3% per year. What will the population be in 15 years?

Answer: 113,706 people

P₁₅ = 180,000(1 - 0.03)¹⁵ = 180,000(0.6317) = 113,706

5. A pendulum swings 50 cm on its first swing. Each swing is 90% of the previous. What is the total distance traveled before it stops?

Answer: 500 cm

Infinite geometric series: S = 50/(1 - 0.9) = 50/0.1 = 500 cm

6. A ball drops from 20 feet and bounces to 75% of its previous height each time. What is the total vertical distance traveled?

Answer: 140 feet

Initial drop: 20 ft
Bounce heights: 15, 11.25, 8.44, ... (a₁ = 15, r = 0.75)
Sum of bounces = 15/(1 - 0.75) = 60
Total = 20 + 2(60) = 140 feet

7. An auditorium has 25 rows. The first row has 12 seats and each row has 3 more seats than the previous. How many seats total?

Answer: 1,200 seats

a₁ = 12, d = 3, n = 25
a₂₅ = 12 + 24(3) = 84
S₂₅ = 25(12 + 84)/2 = 25(96)/2 = 1,200

8. You want $100,000 in 15 years. If you can earn 7% compounded annually, how much should you invest now?

Answer: $36,244.60

100,000 = P(1.07)¹⁵
P = 100,000/2.7590 = $36,244.60

9. A medication has a half-life of 6 hours. If you take 400 mg, how much remains after 30 hours?

Answer: 12.5 mg

n = 30/6 = 5 half-lives
a₆ = 400 · (0.5)⁵ = 400 · 0.03125 = 12.5 mg

10. Job A starts at $45,000 with $1,500 annual raises. Job B starts at $43,000 with 4% annual raises. Which pays more over 8 years?

Answer: Job A pays more

Job A: S₈ = 8(45,000 + 55,500)/2 = $402,000
Job B: S₈ = 43,000[(1.04)⁸ - 1]/0.04 = $398,774
Job A pays $3,226 more over 8 years

Key Takeaways

Compound interest and annuities use geometric sequences; regular deposits create geometric series
Population growth and decay follow geometric sequences with r = 1 + growth rate or 1 - decay rate
Half-life problems use geometric sequences with r = 0.5 for each half-life period
Falling objects create arithmetic sequences; distance fallen in successive time periods increases linearly
Pendulums and bouncing balls with air resistance form geometric sequences that converge to zero
Total distance for bouncing objects requires careful analysis: initial + 2(sum of bounces)
Infinite geometric series with |r| < 1 can model equilibrium situations (drug dosage, pendulum total distance)
Theater seating, stacking patterns, and similar arrangements often follow arithmetic sequences
Always identify whether the problem involves arithmetic or geometric sequences before choosing a formula
Multi-step problems may require comparing two different types of sequences
Check that answers make sense in the real-world context of the problem
Financial formulas: Future Value = P(1 + r)ⁿ, Annuity = PMT[(1 + r)ⁿ - 1]/r

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