Explicit and Recursive Formulas for Geometric Sequences

(Video) A Formula for the $n^{\text{th}}$ Term

by PatrickJMT

This video by PatrickJMT covers formulas to find the $n$th term of a sequence.

$\large{\text{Summary}}$

The general formula for the $n$th term of a geometric sequence is: $$a_n=a_1 \cdot r^{n-1}$$ where $a_1=\text{first term}$ and $r=\text{common ratio}$.

How?

To show we got this formula, let's look at a sequence: $$3,6,12,24,48,\cdots$$ First off, we know it's a geometric sequence because each term is multiplied by $2$ to get to the next term. Since each term is just being multiplied by $2$ again, we can rewrite our terms as: $$\begin{align} a_1&=3\\ a_2&=3\cdot2\\ a_3&=3\cdot2\cdot2\\ a_4&=3\cdot2\cdot2\cdot2\\ a_5&=3\cdot2\cdot2\cdot2\cdot2\\ &\cdots \end{align}$$ We can even combine like terms and simplify more. $$\begin{align} a_1&=3\\ a_2&=3\cdot2\\ a_3&=3\cdot2^2\\ a_4&=3\cdot2^3\\ a_5&=3\cdot2^4\\ &\cdots \end{align}$$ So the $n$th term in this sequence would be, following this pattern: $$a_n=3\cdot2^{n-1}$$ Since we know that our first term is $3$ and our common ratio is $2$, we can write our general formula from above. $$a_n=a_1\cdot r^{n-1}$$

Using the Formula

Suppose we now want to find the tenth term in the sequence: $$3,6,12,24,48,\cdots$$ We can use the formula we found and plug in $n=10$. $$\begin{align} a_n&=3\cdot2^{n-1}\\ a_{10}&=3\cdot2^{10-1}\\ a_{10}&=3\cdot2^9\\ a_{10}&=3\cdot512\\ a_{10}&=1536 \end{align}$$

A Harder Problem

If $a_2=32$ and $a_5=-4$, find $a_9$.

First, we want to find how many terms are between the two we're given. We have the $5$th term and the $2$nd term. So there's, $$5-2=3,$$ spaces between them.

We know the $5$th term will equal the $2$nd term multiplied by $3$ powers of whatever the common ratio is. In other words, $$a_5=a_2\cdot r^3$$

Since we know the values of $a_5$ and $a_2$, we can solve for the common ratio, $r$. $$\begin{align} a_5&=a_2\cdot r^3\\ -4&=32\cdot r^3\\ -\frac18&=r^3\\ \sqrt[3]{-\frac18}&=r\\ \frac{1}{\sqrt[3]{-8}}&=r\\ -\frac12&=r \end{align}$$

Next, we want to find $a_9$. We're going to use the same method. Take either of the terms given and see how many spaces are between it and $a_9$. I'm going to choose $a_5$. $$9-5=4$$ There's $4$ spaces which means you multiply $4$ times by the common ratio to get from $a_5$ to $a_9$. We can set up another equation with these two and solve for $a_9$. $$\begin{align} a_9&=a_5\cdot r^4\\ a_9&=-4\cdot\left(-\frac12\right)^4\\ a_9&=-4\cdot\frac{1}{(-2)^4}\\ a_9&=-4\cdot\frac{1}{16}\\ a_9&=-\frac14 \end{align}$$

Two Ways to Represent Geometric Sequences $\\$

Writing out the sequence takes up a lot of space and requires people to figure out the pattern. There are better, more concise ways to represent a geometric sequences.

Explicit Formulas

In explicit formulas, if you want to figure out what the $n^{th}$ term of a geometric sequence is, you only need the initial term and the common ratio.

Recursive Formulas

In recursive formulas, if you want to figure out what the $n^{th}$ term of a geometric sequence is, you need the initial term, the common ratio and the term before the desired term.

Here is another graphic outlining what we have learned!

Try it! To get a better sense of both formulas work, apply it to the following sequence. $$2,4,6,8,10...$$ This sequence starts with $2$ and has a common ratio of $2$.

What is the $9$th term of this sequence?

Try to answer this question by:

• Using the explicit formula
• Using the recursive formula
• Writing of the sequence and check what the $9$th term is

Maximum Depth Exceeded: Fun with Sequences and Python, Part 5

To see the previous part of this story, go here.

An example of a recursive formula for a geometric sequence is

$$b_n = 3 \times b_{n - 1}$$

because $b_n$ is written in terms of an earlier element in the sequence, in this case $b_{n - 1}.$ We often want to find an explicit formula for $b_n,$ which is a formula for which $b_{n - 1}, b_{n - 2}, \dots, b_1, b_0$ don't appear. To do this, it's easiest to plug our recursive formula into a Python program:

Now we find $b(n)$ for the values $n = 0, 1, 2..., 20$ (we could keep going if we wanted but there's no need):

To get to $b(n),$ we multiply $b(0)$ by $3$ a total of $n$ times, which produces a factor of $3^n.$ This means our explicit formula is

$$b(n) = 11 \times 3^n.$$

This pattern is much more clear because we listed out a bunch of terms first.