Lesson 1: Arithmetic Sequences and Series
Learning Objectives
- Understand sequences and sequence notation, including subscript and function notation
- Identify arithmetic sequences and find the common difference
- Use the explicit formula to find any term of an arithmetic sequence
- Use the recursive formula to generate terms of an arithmetic sequence
- Find the sum of a finite arithmetic series using the sum formula
- Distinguish between finite and infinite arithmetic series
- Apply arithmetic sequences to real-world problems involving patterns
- Solve application problems involving seat arrangements, savings plans, and linear growth
Introduction: What is a Sequence?
Sequence: An ordered list of numbers. Each number in the sequence is called a term.
For example: 3, 7, 11, 15, 19, ...
Sequences can be finite (having a specific number of terms) or infinite (continuing forever). We use special notation to represent sequences and their terms.
Sequence Notation
We denote a sequence using subscript notation. The general form is:
a₁, a₂, a₃, a₄, ..., aₙ, ...
Where:
- aₙ represents the nth term of the sequence
- a₁ is the first term
- a₂ is the second term
- n is the term number (position in the sequence)
We can also write a sequence using function notation: a(n) or f(n), where n represents the position of the term.
Example 1: Finding Terms of a Sequence
Find the first five terms of the sequence defined by aₙ = 2n + 1.
Solution:
We substitute n = 1, 2, 3, 4, 5 into the formula:
a₁: a₁ = 2(1) + 1 = 2 + 1 = 3
a₂: a₂ = 2(2) + 1 = 4 + 1 = 5
a₃: a₃ = 2(3) + 1 = 6 + 1 = 7
a₄: a₄ = 2(4) + 1 = 8 + 1 = 9
a₅: a₅ = 2(5) + 1 = 10 + 1 = 11
Answer: The first five terms are 3, 5, 7, 9, 11.
Example 2: Finding a Specific Term
Given the sequence aₙ = n² - 4, find a₁₀.
Solution:
Substitute n = 10 into the formula:
a₁₀ = (10)² - 4
a₁₀ = 100 - 4
a₁₀ = 96
Answer: The 10th term is 96.
Example 3: Writing Terms from a Pattern
Write the first six terms of the sequence: 5, 8, 11, 14, ...
Solution:
We observe the pattern: each term increases by 3.
a₁ = 5
a₂ = 8
a₃ = 11
a₄ = 14
a₅ = 17 (adding 3 to previous term)
a₆ = 20 (adding 3 to previous term)
Answer: The first six terms are 5, 8, 11, 14, 17, 20.
Section 1: Arithmetic Sequences
Arithmetic Sequence: A sequence in which the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by d.
For consecutive terms: d = aₙ₊₁ - aₙ
In an arithmetic sequence, you can find any term by adding the common difference to the previous term.
Example 4: Identifying the Common Difference
Determine whether the sequence is arithmetic. If so, find the common difference: 4, 7, 10, 13, 16, ...
Solution:
Check the difference between consecutive terms:
7 - 4 = 3
10 - 7 = 3
13 - 10 = 3
16 - 13 = 3
Since the difference is constant, this is an arithmetic sequence.
Answer: Yes, it is arithmetic with common difference d = 3.
Example 5: Non-Arithmetic Sequence
Determine whether the sequence is arithmetic: 2, 4, 8, 16, 32, ...
Solution:
Check the difference between consecutive terms:
4 - 2 = 2
8 - 4 = 4
16 - 8 = 8
The differences are not constant (2, 4, 8, ...).
Answer: No, this is not an arithmetic sequence. (Note: This is a geometric sequence.)
Example 6: Arithmetic Sequence with Negative Common Difference
Find the common difference: 20, 15, 10, 5, 0, -5, ...
Solution:
Check the difference between consecutive terms:
15 - 20 = -5
10 - 15 = -5
5 - 10 = -5
0 - 5 = -5
Answer: The common difference is d = -5. (The sequence is decreasing.)
The Explicit Formula for Arithmetic Sequences
Instead of finding each term one by one, we can use a formula to find any term directly.
Explicit Formula:
aₙ = a₁ + (n - 1)d
where a₁ is the first term, d is the common difference, and n is the term number
Example 7: Finding the nth Term Using the Explicit Formula
Find the 20th term of the arithmetic sequence: 5, 9, 13, 17, ...
Solution:
Step 1: Identify a₁ and d.
a₁ = 5
d = 9 - 5 = 4
Step 2: Use the formula aₙ = a₁ + (n - 1)d with n = 20:
a₂₀ = 5 + (20 - 1)(4)
a₂₀ = 5 + (19)(4)
a₂₀ = 5 + 76
a₂₀ = 81
Answer: The 20th term is 81.
Example 8: Finding the Explicit Formula
Write the explicit formula for the sequence: 12, 7, 2, -3, -8, ...
Solution:
Step 1: Identify a₁ and d.
a₁ = 12
d = 7 - 12 = -5
Step 2: Substitute into the formula aₙ = a₁ + (n - 1)d:
aₙ = 12 + (n - 1)(-5)
aₙ = 12 - 5(n - 1)
aₙ = 12 - 5n + 5
aₙ = 17 - 5n
Answer: The explicit formula is aₙ = 17 - 5n or aₙ = 12 + (n - 1)(-5).
Example 9: Finding Which Term Has a Given Value
In the arithmetic sequence 3, 8, 13, 18, ..., which term has the value 78?
Solution:
Step 1: Identify a₁ and d.
a₁ = 3
d = 8 - 3 = 5
Step 2: Set up the formula with aₙ = 78:
78 = 3 + (n - 1)(5)
Step 3: Solve for n:
78 = 3 + 5n - 5
78 = -2 + 5n
80 = 5n
n = 16
Answer: 78 is the 16th term of the sequence.
Example 10: Finding the First Term Given Another Term
An arithmetic sequence has a common difference of 4 and a₁₂ = 50. Find a₁.
Solution:
Use the formula aₙ = a₁ + (n - 1)d with n = 12, aₙ = 50, and d = 4:
50 = a₁ + (12 - 1)(4)
50 = a₁ + (11)(4)
50 = a₁ + 44
a₁ = 6
Answer: The first term is a₁ = 6.
The Recursive Formula for Arithmetic Sequences
A recursive formula defines each term based on the previous term(s).
Recursive Formula:
a₁ = (first term value)
aₙ = aₙ₋₁ + d for n ≥ 2
Example 11: Using the Recursive Formula
Given the recursive formula a₁ = 7 and aₙ = aₙ₋₁ + 3, find the first five terms.
Solution:
Start with a₁ and add 3 to find each subsequent term:
a₁ = 7
a₂ = a₁ + 3 = 7 + 3 = 10
a₃ = a₂ + 3 = 10 + 3 = 13
a₄ = a₃ + 3 = 13 + 3 = 16
a₅ = a₄ + 3 = 16 + 3 = 19
Answer: The first five terms are 7, 10, 13, 16, 19.
Example 12: Writing a Recursive Formula
Write the recursive formula for the sequence: 25, 21, 17, 13, ...
Solution:
Step 1: Identify a₁ and d.
a₁ = 25
d = 21 - 25 = -4
Step 2: Write the recursive formula:
a₁ = 25
aₙ = aₙ₋₁ - 4 for n ≥ 2
Answer: a₁ = 25, aₙ = aₙ₋₁ - 4 for n ≥ 2.
Section 2: Arithmetic Series
Arithmetic Series: The sum of the terms of an arithmetic sequence.
For example, the series 2 + 5 + 8 + 11 + 14 is the sum of the first five terms of the arithmetic sequence 2, 5, 8, 11, 14, ...
We use the notation Sₙ to represent the sum of the first n terms of a sequence.
The Sum Formula for Arithmetic Series
Instead of adding all terms individually, we can use formulas to find the sum efficiently.
Sum Formula (using first and last term):
Sₙ = n/2 (a₁ + aₙ)
where n is the number of terms, a₁ is the first term, and aₙ is the last term
Alternative Sum Formula (using common difference):
Sₙ = n/2 [2a₁ + (n - 1)d]
where n is the number of terms, a₁ is the first term, and d is the common difference
Example 13: Finding the Sum of an Arithmetic Series
Find the sum of the first 10 terms of the sequence: 3, 7, 11, 15, ...
Solution:
Step 1: Identify a₁, d, and n.
a₁ = 3
d = 7 - 3 = 4
n = 10
Step 2: Find a₁₀ using the explicit formula:
a₁₀ = 3 + (10 - 1)(4) = 3 + 36 = 39
Step 3: Use the sum formula Sₙ = n/2 (a₁ + aₙ):
S₁₀ = 10/2 (3 + 39)
S₁₀ = 5(42)
S₁₀ = 210
Answer: The sum of the first 10 terms is 210.
Example 14: Using the Alternative Sum Formula
Find the sum of the first 20 terms of the sequence with a₁ = 5 and d = 3.
Solution:
Use the formula Sₙ = n/2 [2a₁ + (n - 1)d] with n = 20:
S₂₀ = 20/2 [2(5) + (20 - 1)(3)]
S₂₀ = 10[10 + (19)(3)]
S₂₀ = 10[10 + 57]
S₂₀ = 10(67)
S₂₀ = 670
Answer: The sum of the first 20 terms is 670.
Example 15: Sum of a Decreasing Sequence
Find the sum: 50 + 45 + 40 + 35 + ... + 5
Solution:
Step 1: Identify a₁, aₙ, and d.
a₁ = 50
aₙ = 5
d = 45 - 50 = -5
Step 2: Find n using aₙ = a₁ + (n - 1)d:
5 = 50 + (n - 1)(-5)
5 = 50 - 5n + 5
5 = 55 - 5n
5n = 50
n = 10
Step 3: Use the sum formula:
S₁₀ = 10/2 (50 + 5)
S₁₀ = 5(55)
S₁₀ = 275
Answer: The sum is 275.
Example 16: Sum with Given First and Last Terms
An arithmetic series has 25 terms. The first term is 8 and the last term is 104. Find the sum.
Solution:
Use Sₙ = n/2 (a₁ + aₙ) with n = 25, a₁ = 8, and a₂₅ = 104:
S₂₅ = 25/2 (8 + 104)
S₂₅ = 25/2 (112)
S₂₅ = 25(56)
S₂₅ = 1400
Answer: The sum is 1400.
Example 17: Finding the Number of Terms in a Series
The sum of an arithmetic series is 400. The first term is 5 and the last term is 75. How many terms are in the series?
Solution:
Use Sₙ = n/2 (a₁ + aₙ) with Sₙ = 400, a₁ = 5, and aₙ = 75:
400 = n/2 (5 + 75)
400 = n/2 (80)
400 = 40n
n = 10
Answer: There are 10 terms in the series.
Finite vs. Infinite Arithmetic Series
Finite Arithmetic Series: Has a specific number of terms and a finite sum. We can calculate the sum using the formulas above.
Infinite Arithmetic Series: Continues forever. For arithmetic series, an infinite series does NOT have a finite sum (it diverges to ±∞) unless all terms are zero.
Important Note
Unlike geometric series (which we'll study next), arithmetic series with a non-zero common difference do NOT converge to a finite sum. The sum formulas only apply to finite arithmetic series.
Section 3: Applications of Arithmetic Sequences and Series
Arithmetic sequences and series appear in many real-world situations involving constant change or linear growth.
Example 18: Theater Seating
A theater has 20 rows of seats. The first row has 15 seats, and each subsequent row has 2 more seats than the previous row. How many seats are in the theater?
Solution:
Step 1: Identify the sequence parameters.
a₁ = 15 (first row)
d = 2 (each row adds 2 seats)
n = 20 (number of rows)
Step 2: We need the total number of seats, which is the sum S₂₀.
Step 3: Use the sum formula Sₙ = n/2 [2a₁ + (n - 1)d]:
S₂₀ = 20/2 [2(15) + (20 - 1)(2)]
S₂₀ = 10[30 + 19(2)]
S₂₀ = 10[30 + 38]
S₂₀ = 10(68)
S₂₀ = 680
Answer: The theater has 680 seats total.
Example 19: Savings Plan
Maria starts a savings plan where she saves $50 in January, $75 in February, $100 in March, and continues increasing by $25 each month. How much will she have saved after 12 months?
Solution:
Step 1: Identify the sequence: 50, 75, 100, ...
a₁ = 50
d = 25
n = 12
Step 2: Find the total saved using the sum formula:
S₁₂ = 12/2 [2(50) + (12 - 1)(25)]
S₁₂ = 6[100 + 11(25)]
S₁₂ = 6[100 + 275]
S₁₂ = 6(375)
S₁₂ = 2250
Answer: Maria will have saved $2,250 after 12 months.
Example 20: Stacking Boxes
A warehouse stacks boxes in rows. The bottom row has 35 boxes, the next row has 32 boxes, and each row above has 3 fewer boxes than the row below. If the top row has 5 boxes, how many boxes are stacked in total?
Solution:
Step 1: Identify parameters (note: we're working from bottom to top).
a₁ = 35 (bottom row)
d = -3 (decreasing)
aₙ = 5 (top row)
Step 2: Find n using aₙ = a₁ + (n - 1)d:
5 = 35 + (n - 1)(-3)
5 = 35 - 3n + 3
5 = 38 - 3n
3n = 33
n = 11
Step 3: Find the total using the sum formula:
S₁₁ = 11/2 (35 + 5)
S₁₁ = 11/2 (40)
S₁₁ = 11(20)
S₁₁ = 220
Answer: There are 220 boxes stacked in total.
Example 21: Construction Project Timeline
A construction company completes 12 units in the first week, 16 units in the second week, 20 units in the third week, and so on. How many weeks will it take to complete 300 units total?
Solution:
Step 1: Identify the sequence: 12, 16, 20, ...
a₁ = 12
d = 4
Sₙ = 300 (total units needed)
Step 2: Use the sum formula Sₙ = n/2 [2a₁ + (n - 1)d] and solve for n:
300 = n/2 [2(12) + (n - 1)(4)]
300 = n/2 [24 + 4n - 4]
300 = n/2 [20 + 4n]
600 = n(20 + 4n)
600 = 20n + 4n²
4n² + 20n - 600 = 0
n² + 5n - 150 = 0
Step 3: Use the quadratic formula:
n = [-5 ± √(25 + 600)] / 2
n = [-5 ± √625] / 2
n = [-5 ± 25] / 2
Taking the positive solution:
n = 20/2 = 10
Answer: It will take 10 weeks to complete 300 units.
Example 22: Odometer Reading
A delivery driver's odometer reads 45,000 miles at the start of the year. Each week, the driver adds approximately 350 miles. What will the odometer read after 52 weeks (one year)?
Solution:
Step 1: This is an arithmetic sequence where each week adds 350 miles.
Starting mileage = 45,000
Weekly addition = 350 miles
Number of weeks = 52
Step 2: The total miles added is an arithmetic series:
First addition: a₁ = 350
Common difference: d = 0 (same amount each week)
Number of terms: n = 52
Wait, this is actually simpler: 350 × 52 = 18,200 miles added.
Step 3: Add to starting mileage:
Final reading = 45,000 + 18,200 = 63,200 miles
Answer: The odometer will read 63,200 miles after one year.
Example 23: Salary with Annual Raises
An employee starts with a salary of $45,000 and receives a $2,000 raise each year. What will be the employee's total earnings over 10 years?
Solution:
Step 1: The yearly salaries form an arithmetic sequence:
Year 1: $45,000
Year 2: $47,000
Year 3: $49,000
...
a₁ = 45,000
d = 2,000
n = 10
Step 2: Find total earnings using the sum formula:
S₁₀ = 10/2 [2(45,000) + (10 - 1)(2,000)]
S₁₀ = 5[90,000 + 9(2,000)]
S₁₀ = 5[90,000 + 18,000]
S₁₀ = 5(108,000)
S₁₀ = 540,000
Answer: The employee will earn a total of $540,000 over 10 years.
Example 24: Pattern Recognition
Find the sum of all odd numbers from 1 to 99.
Solution:
Step 1: The odd numbers form an arithmetic sequence: 1, 3, 5, 7, ..., 99
a₁ = 1
d = 2
aₙ = 99
Step 2: Find n using aₙ = a₁ + (n - 1)d:
99 = 1 + (n - 1)(2)
99 = 1 + 2n - 2
99 = 2n - 1
100 = 2n
n = 50
Step 3: Find the sum:
S₅₀ = 50/2 (1 + 99)
S₅₀ = 25(100)
S₅₀ = 2,500
Answer: The sum of all odd numbers from 1 to 99 is 2,500.
Example 25: Pyramid Structure
A pyramid is built with blocks arranged in layers. The top layer has 1 block, the second layer has 3 blocks, the third has 5 blocks, and so on, with each layer having 2 more blocks than the one above it. If there are 15 layers, how many blocks are in the pyramid?
Solution:
Step 1: The layers form an arithmetic sequence: 1, 3, 5, 7, ...
a₁ = 1
d = 2
n = 15
Step 2: Find the total using the sum formula:
S₁₅ = 15/2 [2(1) + (15 - 1)(2)]
S₁₅ = 15/2 [2 + 14(2)]
S₁₅ = 15/2 [2 + 28]
S₁₅ = 15/2 (30)
S₁₅ = 15(15)
S₁₅ = 225
Answer: The pyramid contains 225 blocks.
Check Your Understanding
1. Find the 15th term of the arithmetic sequence: 7, 11, 15, 19, ...
Answer: 63
a₁ = 7, d = 4
a₁₅ = 7 + (15 - 1)(4) = 7 + 56 = 63
2. Is the sequence 2, 6, 18, 54, ... arithmetic? Why or why not?
Answer: No, it is not arithmetic.
The differences between consecutive terms are not constant: 6 - 2 = 4, but 18 - 6 = 12. This is a geometric sequence (each term is multiplied by 3).
3. Write the explicit formula for the sequence: 100, 95, 90, 85, ...
Answer: aₙ = 100 + (n - 1)(-5) or aₙ = 105 - 5n
a₁ = 100, d = -5
aₙ = 100 + (n - 1)(-5) = 100 - 5n + 5 = 105 - 5n
4. Find the sum of the first 25 terms of the sequence with a₁ = 4 and d = 3.
Answer: 1,000
S₂₅ = 25/2 [2(4) + (25 - 1)(3)]
S₂₅ = 12.5[8 + 72] = 12.5(80) = 1,000
5. In the sequence 5, 12, 19, 26, ..., which term equals 96?
Answer: The 14th term
a₁ = 5, d = 7
96 = 5 + (n - 1)(7)
96 = 5 + 7n - 7
96 = 7n - 2
98 = 7n
n = 14
6. Find the sum: 8 + 13 + 18 + 23 + ... + 78
Answer: 645
a₁ = 8, aₙ = 78, d = 5
First find n: 78 = 8 + (n - 1)(5) → 70 = 5n - 5 → n = 15
S₁₅ = 15/2 (8 + 78) = 15/2 (86) = 645
7. Write the recursive formula for the sequence: 30, 24, 18, 12, ...
Answer: a₁ = 30, aₙ = aₙ₋₁ - 6 for n ≥ 2
The first term is 30 and the common difference is -6.
8. An auditorium has 30 rows. The first row has 20 seats, and each row has 3 more seats than the previous row. How many seats total?
Answer: 1,905 seats
a₁ = 20, d = 3, n = 30
S₃₀ = 30/2 [2(20) + (30 - 1)(3)]
S₃₀ = 15[40 + 87] = 15(127) = 1,905
9. If a₅ = 23 and a₉ = 39 in an arithmetic sequence, find a₁.
Answer: a₁ = 7
Find d first: a₉ - a₅ = 4d → 39 - 23 = 4d → d = 4
Then use a₅ = a₁ + 4d: 23 = a₁ + 4(4) → a₁ = 7
10. Find the sum of all even integers from 2 to 100.
Answer: 2,550
Sequence: 2, 4, 6, ..., 100
a₁ = 2, aₙ = 100, d = 2
Find n: 100 = 2 + (n - 1)(2) → n = 50
S₅₀ = 50/2 (2 + 100) = 25(102) = 2,550
Key Takeaways
- A sequence is an ordered list of numbers; an arithmetic sequence has a constant difference between consecutive terms
- The common difference d = aₙ₊₁ - aₙ can be positive, negative, or zero
- The explicit formula aₙ = a₁ + (n - 1)d allows you to find any term directly
- The recursive formula defines each term based on the previous term: aₙ = aₙ₋₁ + d
- An arithmetic series is the sum of terms in an arithmetic sequence
- Two sum formulas: Sₙ = n/2 (a₁ + aₙ) or Sₙ = n/2 [2a₁ + (n - 1)d]
- Choose the first sum formula when you know the first and last terms; use the second when you know the common difference
- Arithmetic sequences model situations with constant change: savings plans, seating arrangements, linear growth
- Infinite arithmetic series (with d ≠ 0) do not have a finite sum
- Always identify a₁, d, and n before applying formulas