Arithmetic vs. Geometric Sequences

Published: Aug 7th, 2021

Learning Objectives:
By the end of this lesson, you should be able to:
1. Identity and classify sequences
2. Create functions for classifying sequences

In the last lesson, we learned what sequences are and how to write functions to represent them. Today, we're going to learn about two key types of sequences: arithmetic and geometric. Let's get started!

An arithmetic sequence is a sequence in which the next term is generated by adding a fixed number the previous term. We call that fixed number the common difference of the sequence.

For example, consider the sequence, \(5, 8, 11, 14,...\) Notice that each term is simply the previous term plus 3. This means that this is an arithmetic sequence with a common difference of 3.

Now that you know what an arithmetic sequence is, let's talk about its function and general term forms. The general term of an arithmetic sequence is \(t_n = a + (n - 1)d, n ∈ N\), but you can also represent them using function notation: \(f(n) = a + (n - 1)d, n ∈ N\). In both cases, \(a\) refers to the first term of the sequence and \(d\) refers to its common difference. You can represent arithmetic sequences in either form, as they're identical aside from \(f(n)\) changing to \(t_n\) in the general term, but the general term is preferable due to its benefits when working with recursive sequences.

Makes sense? I hope so! Let's try an example!

For the sequence, \(3, 7, 11, 15,...\), find the general term and \(t_{10}\)

First, let's find the common difference between the terms. Notice that \(7 - 3 = 11 - 7 = 15 - 11 = 4\), meaning that the common difference is 4.

As the first term is 3 and the common difference is 4, the general term is \(t_n = 3 + (n - 1)4\).

To find \(t_{10}\), we plug 10 into our general term as \(n\) like so:

\(t_{10} = 3 + (10 - 1)4\)

\(t_{10} = 3 + (9)4\)

\(t_{10} = 3 + 36\)

\(t_{10} = 39\)

And there we have it! Now that we've understood how to identify and write functions for arithmetic sequences, let's do the same for geometric ones.

A geometric sequence is a sequence in which the next term is the previous one multiplied by a fixed number, which we call the common ratio of that sequence.

For instance, take a look at the sequence, \(1, 2, 4, 8,...\) Each term is the previous term multiplied by 2. Therefore, this is a geometric sequence with a common ratio of 2.

Just as we did with arithmetic sequences, we can write geometric sequences in both function and general term form. Geometric sequences have a general term of \(t_n = ar^{n - 1}, n ∈ N, r ∈ R\) and a function form of \(f(n) = ar^{n - 1}, n ∈ N, r ∈ R\). In both forms, \(a\) refers to the first term of the sequence and \(r\) refers to its common ratio. The statement, \(r ∈ R\), means that \(r\) can be any real number, or any number on the number line. As with arithmetic sequences, both function and general term form are valid, but the latter is preferred.

With that, let's try another example!

For the sequence, \(3, 6, 12, 24,...\), find the general term and \(t_6\)

First of all, we need to find the common ratio. Notice that \(\frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2\). This shows us that the common ratio of this sequence is 2.

Since the first term is 3, our general term becomes \(t_n = 3(2)^{n - 1}\)

Using this equation, we find that \(t_6 = 3(2)^{6 - 1}\)

\(t_6 = 3(2)^5\)

\(t_6 = 3(32)\)

\(t_6 = 96\)

I hope that made sense! We're almost done, but before we finish, let's try one more problem to make sure we know what we're doing.

For the sequence, \(128, 64, 32, 16,...\), identify the sequence type and find its general term

How do we know if this is an arithmetic or geometric sequence (or neither)? Let's start by looking at the first differences between the terms.

We find that 128 - 64 = 64, but \(64 - 32 = 32.\) Since \(64 ≠ 32\), this is not an arithmetic sequence because the difference between each pair of terms isn't always the same. Could it still be a geometric sequence? Let's look at the ratios of each term to the one before it to find out if it is.

Since \(\frac{64}{128} = \frac{32}{64} = \frac{16}{32} = \frac{1}{2}\), this sequence has a common ratio of \(\frac{1}{2}\), meaning that it is a geometric sequence.

As its first term is 128, the general term of this sequence is \(t_n = 128(\frac{1}{2})^{n - 1}\)

Well, that's about it for arithmetic and geometric sequences. Be sure to try out the practice problems to test your knowledge.


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