Pre-Calculus 11: Arithmetic Sequences
Introduction to Arithmetic Sequences
An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the "common difference." Arithmetic sequences appear throughout mathematics and have numerous applications in real-world situations, from calculating financial growth to analyzing patterns in nature.
Note: An arithmetic sequence is also sometimes called an arithmetic progression. These terms can be used interchangeably.
Identifying Arithmetic Sequences
To determine if a sequence is arithmetic, calculate the differences between consecutive terms. If all differences are equal, the sequence is arithmetic.
Example: Is the sequence 5, 11, 17, 23, 29, ... arithmetic?
Calculate the differences:
11 - 5 = 6
17 - 11 = 6
23 - 17 = 6
29 - 23 = 6
Since all differences are equal to 6, this is an arithmetic sequence with a common difference d = 6.
Example: Is the sequence 2, 4, 8, 16, 32, ... arithmetic?
Calculate the differences:
4 - 2 = 2
8 - 4 = 4
16 - 8 = 8
32 - 16 = 16
Since the differences are not equal (2 4 8 16), this is not an arithmetic sequence. (In fact, this is a geometric sequence.)
Understanding the Common Difference
The common difference (d) in an arithmetic sequence is the constant amount by which each term differs from the preceding term. You can find the common difference by subtracting any term from the term that follows it:
d = a - a = a - a = a - a = ...
Example: Consider the sequence: 3, 7, 11, 15, 19, ...
The common difference is: 7 - 3 = 4, 11 - 7 = 4, and so on.
Therefore, d = 4 for this arithmetic sequence.
The common difference can be positive, negative, or zero:
- When d > 0, the sequence is increasing
- When d < 0, the sequence is decreasing
- When d = 0, all terms are the same (a constant sequence)
Examples with different common differences:
- Increasing: 2, 5, 8, 11, 14, ... (d = 3 > 0)
- Decreasing: 14, 11, 8, 5, 2, ... (d = -3 < 0)
- Constant: 7, 7, 7, 7, 7, ... (d = 0)
Terms of an Arithmetic Sequence
The terms in an arithmetic sequence are typically denoted as a, a, a, ..., a, where:
- a is the first term
- a is the second term
- a is the third term
- a is the nth term
Example: In the sequence 5, 8, 11, 14, 17, ...
- a = 5
- a = 8
- a = 11
- a = 14
- a = 17
The common difference is d = 3.
Recursive Formula
A recursive formula defines each term of a sequence by relating it to the preceding term. For an arithmetic sequence, the recursive formula is:
a = first term; a = a + d (for n > 1)
Example: Write the recursive formula for the sequence 4, 7, 10, 13, 16, ...
The first term is a = 4, and the common difference is d = 3.
The recursive formula is: a = 4; a = a + 3 (for n > 1)
Note: To find a specific term using a recursive formula, you need to know all previous terms. For example, to find a, you first need to find a, then a, then a, and finally a.
Explicit Formula
Unlike recursive formulas, which rely on previous terms, explicit formulas allow you to find any term directly without knowing the preceding terms. For an arithmetic sequence, the explicit formula is:
a = a + (n-1)d
Where:
- a is the nth term
- a is the first term
- n is the term number
- d is the common difference
Example: Find the 10th term of the sequence 2, 8, 14, 20, 26, ...
First, identify the known values: a = 2, d = 6, n = 10
Apply the explicit formula: a = 2 + (10-1)6 = 2 + 9(6) = 2 + 54 = 56
The 10th term is 56.
Properties of Arithmetic Sequences
Understanding the properties of arithmetic sequences can help simplify problems and deepen your understanding:
- Three Consecutive Terms: The middle term of any three consecutive terms is the average of the other two: (a + a)/2 = a
- Sum of Terms at Equal Distance: For any arithmetic sequence, the sum of terms at equal distance from the beginning and end is constant: a + a = a + a = a + a = ...
- Linear Growth: Arithmetic sequences grow linearly, which means they form a straight line when plotted on a graph.
Example: In the arithmetic sequence 3, 7, 11, 15, 19, ...
For the first property: (7 + 15)/2 = 22/2 = 11, which is indeed the middle term.
For the second property: 3 + 19 = 22, 7 + 15 = 22, 11 + 11 = 22, confirming the property.
Determining the Sum of an Arithmetic Sequence
The sum (S) of the first n terms of an arithmetic sequence can be found using:
S = n(a + a) = n[2a + (n-1)d]
Example: Find the sum of the first 15 terms of the sequence: 3, 7, 11, 15, ...
First, identify the known values: a = 3, d = 4, n = 15
Find the 15th term using the explicit formula: a = 3 + (15-1)4 = 3 + 14(4) = 3 + 56 = 59
Now find the sum: S = 15(3 + 59) = 15(62) = 930
The sum of the first 15 terms is 930.
Special Case: Sum of First n Positive Integers
The sequence of positive integers 1, 2, 3, 4, 5, ... is an arithmetic sequence where a = 1 and d = 1.
The sum of the first n positive integers is:
S = n(n+1)/2
Example: Find the sum of the first 100 positive integers.
S = 100(100+1)/2 = 100(101)/2 = 50(101) = 5,050
Solving Applied Problems
Arithmetic sequences can model various real-world situations. Let's look at some applications.
Example 1: A theater has 20 seats in the first row, 24 seats in the second row, 28 seats in the third row, and so on. How many seats are in the 15th row?
Solution: This is an arithmetic sequence where a = 20 and d = 4.
Using the explicit formula: a = 20 + (15-1)4 = 20 + 14(4) = 20 + 56 = 76
The 15th row has 76 seats.
Example 2: Maria saves $50 in the first month, $55 in the second month, $60 in the third month, and so on. How much will she save in the 12th month? What is her total savings after 12 months?
Solution: This is an arithmetic sequence where a = 50 and d = 5.
Monthly savings (12th month): a = 50 + (12-1)5 = 50 + 11(5) = 50 + 55 = $105
Total savings after 12 months: S = 12(50 + 105) = 12(155) = $1,860
Practice Problems
Problem 1: Find the 25th term of the arithmetic sequence: 5, 10, 15, 20, ...
Solution: a = 5, d = 5, n = 25
a = 5 + (25-1)5 = 5 + 24(5) = 5 + 120 = 125
The 25th term is 125.
Problem 2: Find the sum of the first 20 terms of the arithmetic sequence: 2, 6, 10, 14, ...
Solution: a = 2, d = 4, n = 20
First, find the 20th term: a = 2 + (20-1)4 = 2 + 19(4) = 2 + 76 = 78
Now find the sum: S = 20(2 + 78) = 20(80) = 1,600
The sum of the first 20 terms is 1,600.
Problem 3: In an arithmetic sequence, a = 3 and a = 30. Find the common difference and the explicit formula.
Solution: Using the explicit formula a = a + (n-1)d:
a = 3 + (10-1)d = 3 + 9d = 30
9d = 27, so d = 3
The explicit formula is: a = 3 + (n-1)3 = 3 + 3n - 3 = 3n
Problem 4: Find the common difference if the 5th term of an arithmetic sequence is 21 and the 9th term is 33.
Solution: Using the explicit formula for both terms:
a = a + 4d = 21
a = a + 8d = 33
Subtracting the first equation from the second: 4d = 12
Therefore, d = 3
Conclusion
Arithmetic sequences are fundamental building blocks in Pre-Calculus 11. By mastering the concepts of common difference, recursive and explicit formulas, and sum formulas, you'll be well-equipped to tackle more complex mathematical topics. The ability to recognize arithmetic patterns in data and apply these formulas to solve problems is a valuable mathematical skill with applications across numerous fields.
As you continue your study of sequences and series, remember that arithmetic sequences are just one type of pattern you'll encounter. Geometric sequences, harmonic sequences, and Fibonacci sequences will expand your understanding of mathematical patterns even further. The skills you develop with arithmetic sequences will serve as a solid foundation for exploring these more advanced topics.
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