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Module 6 Study Guide

Sequences and Series

1. Arithmetic Sequences

1.1 Definition

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant value called the common difference (d) to the previous term.
Example: 3, 7, 11, 15, 19, ...
Common difference: d = 4
Each term is 4 more than the previous term.

1.2 Finding the Common Difference

d = an - an-1

Subtract any term from the term that follows it.

1.3 Explicit Formula (nth Term)

an = a1 + (n - 1)d

Where:

  • an = the nth term
  • a1 = the first term
  • n = the term number
  • d = common difference
Example: Find the 20th term of 5, 9, 13, 17, ...
a1 = 5, d = 4
a20 = 5 + (20 - 1)(4) = 5 + 76 = 81

1.4 Arithmetic Series (Sum of Terms)

Sn = n(a1 + an)/2

This is the sum of the first n terms.

Alternative formula: Sn = n[2a1 + (n - 1)d]/2
Example: Find the sum of the first 15 terms of 6, 11, 16, 21, ...
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

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a constant value called the common ratio (r).
Example: 3, 6, 12, 24, 48, ...
Common ratio: r = 2
Each term is twice the previous term.

2.2 Finding the Common Ratio

r = an / an-1

Divide any term by the term before it.

2.3 Explicit Formula (nth Term)

an = a1 · rn-1

Where:

  • an = the nth term
  • a1 = the first term
  • n = the term number
  • r = common ratio
Example: Find the 8th term of 64, 32, 16, 8, ...
a1 = 64, r = 1/2
a8 = 64 · (1/2)7 = 64 · (1/128) = 1/2

2.4 Geometric Series (Sum of Terms)

Sn = a1(1 - rn)/(1 - r) when r ≠ 1

This is the sum of the first n terms.

Alternative formula: Sn = a1(rn - 1)/(r - 1) when r ≠ 1
Example: Find the sum of the first 6 terms of 2, 6, 18, 54, ...
a1 = 2, r = 3
S6 = 2(1 - 36)/(1 - 3) = 2(-728)/(-2) = 728

2.5 Infinite Geometric Series

An infinite geometric series converges (has a finite sum) only when |r| < 1.
S = a1/(1 - r) when |r| < 1
Example: Find the sum of 12 + 6 + 3 + 1.5 + ...
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

The Greek letter Σ (sigma) is used to denote summation. It provides a compact way to write the sum of many terms.
Σ f(k) from k=1 to n = f(1) + f(2) + f(3) + ... + f(n)

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
Example: Evaluate Σ(2k + 1) from k=1 to 4
= (2·1 + 1) + (2·2 + 1) + (2·3 + 1) + (2·4 + 1)
= 3 + 5 + 7 + 9 = 24

3.2 Properties of Summation

Σ(c · f(k)) = c · Σ f(k) (constant multiple)

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 c from k=1 to n = nc (sum of constants)

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)
Example: Evaluate Σ k2 from k=1 to 10
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
Example: Write 4 + 8 + 12 + 16 + 20 in summation 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

Watch out for:
• 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

Find the 25th term of: 10, 14, 18, 22, ...

Solution:
a1 = 10, d = 4
a25 = 10 + (25 - 1)(4) = 10 + 96 = 106

Problem 2: Geometric Sequence

Find the 6th term of: 5, 15, 45, 135, ...

Solution:
a1 = 5, r = 3
a6 = 5 · 35 = 5 · 243 = 1,215

Problem 3: Arithmetic Series

Find the sum of the first 30 terms of: 7, 11, 15, 19, ...

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

Find the sum of the first 7 terms of: 4, 8, 16, 32, ...

Solution:
a1 = 4, r = 2
S7 = 4(1 - 27)/(1 - 2) = 4(-127)/(-1) = 508

Problem 5: Infinite Geometric Series

Find the sum: 8 + 4 + 2 + 1 + ...

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

Evaluate: Σ(3k - 1) from k=1 to 5

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

Maria saves $100 in month 1, $125 in month 2, $150 in month 3, and so on. How much will she save in total after 12 months?

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

A bacteria culture starts with 300 bacteria and doubles every 2 hours. How many bacteria after 10 hours?

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