What Is An Arithmetic Sequence?
An arithmetic sequence is an infinite sequence of numbers in which the difference between each pair of consecutive numbers is always the same. For example, in the sequence 1, 3, 5, 7, 9 . . . the difference between one number and the next is always 2.
What is the constant difference (d) between any two consecutive numbers in the following sequences?
- -5, -3, -1, 1, 3, 5 . . .
- .5, 1, 1.5, 2 . . .
- 10, 6, 2, -2 . . .
In the first sequence, d = 2 because you can add 2 to any number in the sequence to get the next number. For example, -3 + 2 = -1 and 1 +2 = 3. In the second sequence, d = .5. In the third sequence, each number is 4 less than the previous number, so d = -4.
The Recursive Formula For An Arithmetic Sequence
One way of finding a number within a sequence is to use the recursive formula. To write the formula, we use the following notation:
a is a term in the sequence.
n is the number of terms in the sequence.
d is the constant difference between terms.
Thus, an = an-1 + d
In other words, to find the 5th number in a sequence with a constant difference of 6, we need to know the 4th number (an-1) and add 6 to it. If we are given the sequence 5, 11, 17, 23, and we need to find the next number, we can easily apply this formula by adding 6 to 23 and getting 29. In other words, if d = 6 and if an – 1 = 23, then
an = 23 + 6 = 29
The Explicit Formula For An Arithmetic Sequence
If we only have the first number in a sequence (a1), however, the explicit formula can be a more useful way to find another number in the sequence. To understand how the explicit formula is derived, let’s start with the following sequence where d = -7:
100, 93, 86, 79 . . .
To get the first number, we start with 100 and add -7 zero times. So a1 = 100 + (-7 x 0). To get the second number, we subtract 7 one time. So a2 = 100 + (-7 x 1). The next number in the series is a3 = 100 + (-7 x 2), and so on. Each time we are adding -7 exactly one less time than the number of terms in the sequence. Therefore, we can write a general formula to express this pattern as follows:
an = a1 + (n-1) x d
If we want to find, for example, the 17th number in a series that begins with 3 and has a constant difference of .5, we can plug that information into the formula like this:
a17 = 3 + (17-1) x .5 = 11
Practice
Problem 1: What is the constant difference (d) in the following sequence?
24, 32, 40, 48, 56 . . .
Solution: In this sequence d = 8 because we can add 8 to each number to get the next number.
Problem 2: What is the next number in the sequence above?
Solution: Using the recursive formula, we know that the 6th number (a6) is equal to the 5th number (a6-1) plus the constant difference (d). Since 56 + 8 = 64, the next number in the series is 64.
Problem 3: Write an explicit formula for the sequence in Problem 1 and use that formula to find the 11th number in the sequence.
Solution: Since an = 24 + (n-1) x 8, and n = 11, then a11 = 24 + (11-1) x 8 = 104.
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