Module 6 Study Guide
Sequences and Series
1. Arithmetic Sequences
1.1 Definition
Common difference: d = 4
Each term is 4 more than the previous term.
1.2 Finding the Common Difference
Subtract any term from the term that follows it.
1.3 Explicit Formula (nth Term)
Where:
- an = the nth term
- a1 = the first term
- n = the term number
- d = common difference
a1 = 5, d = 4
a20 = 5 + (20 - 1)(4) = 5 + 76 = 81
1.4 Arithmetic Series (Sum of Terms)
This is the sum of the first n terms.
a1 = 6, d = 5
a15 = 6 + (15 - 1)(5) = 76
S15 = 15(6 + 76)/2 = 15(82)/2 = 615
1.5 Applications
- Saving money with constant deposits
- Seating arrangements with constant increases
- Linear depreciation
- Evenly spaced objects
2. Geometric Sequences
2.1 Definition
Common ratio: r = 2
Each term is twice the previous term.
2.2 Finding the Common Ratio
Divide any term by the term before it.
2.3 Explicit Formula (nth Term)
Where:
- an = the nth term
- a1 = the first term
- n = the term number
- r = common ratio
a1 = 64, r = 1/2
a8 = 64 · (1/2)7 = 64 · (1/128) = 1/2
2.4 Geometric Series (Sum of Terms)
This is the sum of the first n terms.
a1 = 2, r = 3
S6 = 2(1 - 36)/(1 - 3) = 2(-728)/(-2) = 728
2.5 Infinite Geometric Series
a1 = 12, r = 1/2
Since |r| = 1/2 < 1, the series converges.
S = 12/(1 - 1/2) = 12/(1/2) = 24
2.6 Applications
- Compound interest and investments
- Population growth (exponential)
- Radioactive decay
- Bouncing ball problems
- Depreciation (exponential)
3. Summation Notation
3.1 Sigma Notation
Components:
- k = index of summation (variable)
- Lower limit = starting value of k
- Upper limit = ending value of k
- f(k) = expression to evaluate for each k
= (2·1 + 1) + (2·2 + 1) + (2·3 + 1) + (2·4 + 1)
= 3 + 5 + 7 + 9 = 24
3.2 Properties of Summation
Σ[f(k) + g(k)] = Σ f(k) + Σ g(k) (sum of functions)
Σ[f(k) - g(k)] = Σ f(k) - Σ g(k) (difference of functions)
3.3 Common Summation Formulas
Σ k from k=1 to n = n(n + 1)/2 (sum of first n integers)
Σ k2 from k=1 to n = n(n + 1)(2n + 1)/6 (sum of squares)
Σ k3 from k=1 to n = [n(n + 1)/2]2 (sum of cubes)
Using the formula: = 10(11)(21)/6 = 2310/6 = 385
3.4 Writing Series in Summation Notation
- Identify the pattern in the terms
- Determine the general term f(k)
- Identify the starting value (lower limit)
- Identify the ending value (upper limit)
- Write using Σ notation
Pattern: Each term is 4k where k = 1, 2, 3, 4, 5
Answer: Σ(4k) from k=1 to 5
4. Comparison: Arithmetic vs. Geometric
| Feature | Arithmetic Sequence | Geometric Sequence |
|---|---|---|
| Pattern | Add constant (d) | Multiply by constant (r) |
| Common Value | d = an - an-1 | r = an/an-1 |
| nth Term | an = a1 + (n - 1)d | an = a1 · rn-1 |
| Sum (Finite) | Sn = n(a1 + an)/2 | Sn = a1(1 - rn)/(1 - r) |
| Sum (Infinite) | Does not exist | S = a1/(1 - r) when |r| < 1 |
| Growth Type | Linear | Exponential |
| Example | 2, 5, 8, 11, 14, ... | 2, 6, 18, 54, 162, ... |
5. Problem-Solving Strategies
5.1 Identifying Sequence Type
- Calculate differences between consecutive terms (for arithmetic)
- Calculate ratios of consecutive terms (for geometric)
- If differences are constant, it is arithmetic
- If ratios are constant, it is geometric
- If neither is constant, it may be another type of sequence
5.2 Finding Missing Terms
- Identify the sequence type first
- Find the common difference or common ratio
- Use the appropriate formula
- Check your answer by verifying the pattern
5.3 Word Problem Steps
- Read carefully and identify what is being asked
- Determine if the situation is arithmetic or geometric
- Identify a1, d or r, and n if given
- Choose the appropriate formula
- Solve and check if the answer makes sense
5.4 Common Mistakes to Avoid
• Confusing arithmetic and geometric sequences
• Using the wrong formula for sums
• Forgetting that n is the term number, not always the value
• In word problems, carefully determine if you need the nth term or the sum
• For infinite geometric series, check that |r| < 1 before using the formula
6. Practice Problems with Solutions
Problem 1: Arithmetic Sequence
Solution:
a1 = 10, d = 4
a25 = 10 + (25 - 1)(4) = 10 + 96 = 106
Problem 2: Geometric Sequence
Solution:
a1 = 5, r = 3
a6 = 5 · 35 = 5 · 243 = 1,215
Problem 3: Arithmetic Series
Solution:
a1 = 7, d = 4
a30 = 7 + (30 - 1)(4) = 123
S30 = 30(7 + 123)/2 = 30(130)/2 = 1,950
Problem 4: Geometric Series
Solution:
a1 = 4, r = 2
S7 = 4(1 - 27)/(1 - 2) = 4(-127)/(-1) = 508
Problem 5: Infinite Geometric Series
Solution:
a1 = 8, r = 1/2
Since |r| = 1/2 < 1, the series converges.
S = 8/(1 - 1/2) = 8/(1/2) = 16
Problem 6: Summation
Solution:
= (3·1 - 1) + (3·2 - 1) + (3·3 - 1) + (3·4 - 1) + (3·5 - 1)
= 2 + 5 + 8 + 11 + 14 = 40
Problem 7: Application - Savings
Solution:
This is arithmetic: a1 = 100, d = 25
a12 = 100 + (12 - 1)(25) = 375
S12 = 12(100 + 375)/2 = 2,850
She will have saved $2,850.
Problem 8: Application - Population Growth
Solution:
This is geometric: a1 = 300, r = 2
After 10 hours = 5 doubling periods
a6 = 300 · 25 = 300 · 32 = 9,600 bacteria
7. Test-Taking Tips
- Always identify the sequence type first (arithmetic or geometric)
- Write down known values: a1, d or r, n
- Choose the correct formula based on what you are finding
- For infinite geometric series, check that |r| < 1 before applying the formula
- In word problems, determine if you need a single term or the sum
- Double-check calculations, especially with exponents
- Verify your answer makes sense in context
- For summation notation, expand a few terms to check your work
- Show your work for partial credit