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# 8.5: Geometric Sequences

Created by: CK-12

## Learning Objectives

• Identify a geometric sequence
• Graph a geometric sequence.
• Solve real-world problems involving geometric sequences.

## Introduction

Consider the following question:

Which would you prefer, being given one million dollars, or one penny the first day, double that penny the next day, and then double the previous day's pennies and so on for a month?

At first glance it’s easy to say "Give me the million!" But why don’t we do a few calculations to see how the other choice stacks up?

You start with a penny the first day and keep doubling each day. Doubling means that we keep multiplying by 2 each day for one month (30 days).

On the first day, you get 1 penny, or $2^0$ pennies.

On the second day, you get 2 pennies, or $2^1$ pennies.

On the third day, you get 4 pennies, or $2^2$ pennies. Do you see the pattern yet?

On the fourth day, you get 8 pennies, or $2^3$ pennies. Each day, the exponent is one less than the number of that day.

So on the thirtieth day, you get $2^{29}$ pennies, which is 536,870,912 pennies, or \$5,368,709.12. That’s a lot more than a million dollars, even just counting the amount you get on that one day!

This problem is an example of a geometric sequence. In this section, we’ll find out what a geometric sequence is and how to solve problems involving geometric sequences.

## Identify a Geometric Sequence

A geometric sequence is a sequence of numbers in which each number in the sequence is found by multiplying the previous number by a fixed amount called the common ratio. In other words, the ratio between any term and the term before it is always the same. In the previous example the common ratio was 2, as the number of pennies doubled each day.

The common ratio, $r$, in any geometric sequence can be found by dividing any term by the preceding term.

Here are some examples of geometric sequences and their common ratios.

$& 4, 16, 64, 256, \ldots \qquad \qquad \ r = 4 \quad \ \ \ \ (\text{divide 16 by 4 to get 4})\\& 15, 30, 60, 120, \ldots \qquad \qquad r = 2 \quad \ \ \ \ (\text{divide 30 by 15 to get 2})\\& 11, \frac{11}{2}, \frac{11}{4}, \frac{11}{8}, \frac{11}{16}, \ldots \qquad \ r = \frac{1}{2} \quad \ \ \left (\text{divide} \ \frac{11}{2} \ \text{by 11 to get} \ \frac{1}{2}\right )\\& 25, -5, 1, -\frac{1}{5}, \frac{1}{25}, \ldots \qquad \ \ r = - \frac{1}{5} \quad \left (\text{divide 1 by -5 to get} - \frac{1}{5} \right )$

If we know the common ratio $r$, we can find the next term in the sequence just by multiplying the last term by $r$. Also, if there are any terms missing in the sequence, we can find them by multiplying the term before each missing term by the common ratio.

Example 1

Fill is the missing terms in each geometric sequence.

a) 1, ___, 25, 125, ___

b) 20, ___, 5, ___, 1.25

Solution

a) First we can find the common ratio by dividing 125 by 25 to obtain $r = 5$.

To find the first missing term, we multiply 1 by the common ratio: $1 \cdot 5 = 5$

To find the second missing term, we multiply 125 by the common ratio: $125 \cdot 5 = 625$

Sequence (a) becomes: 1, 5, 25, 125, 625,...

b) We need to find the common ratio first, but how do we do that when we have no terms next to each other that we can divide?

Well, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice: once to get to the second term in the sequence, and again to get to five. So we can say $20 \cdot r \cdot r = 5$, or $20 \cdot r^2 = 5$.

Dividing both sides by 20, we get $r^2 = \frac{5}{20} = \frac{1}{4}$, or $r = \frac{1}{2}$ (because $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$).

To get the first missing term, we multiply 20 by $\frac{1}{2}$ and get 10.

To get the second missing term, we multiply 5 by $\frac{1}{2}$ and get 2.5.

Sequence (b) becomes: 20, 10, 5, 2.5, 1.25,...

You can see that if we keep multiplying by the common ratio, we can find any term in the sequence that we want—the tenth term, the fiftieth term, the thousandth term.... However, it would be awfully tedious to keep multiplying over and over again in order to find a term that is a long way from the start. What could we do instead of just multiplying repeatedly?

Let’s look at a geometric sequence that starts with the number 7 and has common ratio of 2.

$& \text{The} \ 1^{st} \ \text{term is:} && 7 && \text{or} \qquad 7 \cdot 2^0\\& \text{We obtain the} \ 2^{nd} \ \text{term by multiplying by 2}: && 7 \cdot 2 && \text{or} \qquad 7 \cdot 2^1\\& \text{We obtain the} \ 3^{rd} \ \text{term by multiplying by 2 again:} && 7 \cdot 2 \cdot 2 && \text{or} \qquad 7 \cdot 2^2\\& \text{We obtain the} \ 4^{th} \ \text{term by multiplying by 2 again:} && 7 \cdot 2 \cdot 2 \cdot 2 && \text{or} \qquad 7 \cdot 2^3\\& \text{We obtain the} \ 5^{th} \ \text{term by multiplying by 2 again:} && 7 \cdot 2 \cdot 2 \cdot 2 \cdot 2 && \text{or} \qquad 7 \cdot 2^4\\& \text{The nth term would be:} && 7 \cdot 2^{n - 1}$

The nth term is $7 \cdot 2^{n - 1}$ because the 7 is multiplied by 2 once for the $2^{nd}$ term, twice for the third term, and so on—for each term, one less time than the term’s place in the sequence. In general, we write a geometric sequence with $n$ terms like this:

$a, ar, ar^2, ar^3, \ldots, ar^{n-1}$

The formula for finding a specific term in a geometric sequence is:

$n^{th}$ term in a geometric sequence: $a_n = a_1 r^{n-1}$

($a_1 =$ first term, $r =$ common ratio)

Example 2

For each of these geometric sequences, find the eighth term in the sequence.

a) 1, 2, 4,...

b) 16, -8, 4, -2, 1,...

Solution

a) First we need to find the common ratio: $r = \frac{2}{1} = 2$.

The eighth term is given by the formula $a_8 = a_1r^7 = 1 \cdot 2^7 = 128$

In other words, to get the eighth term we start with the first term, which is 1, and then multiply by 2 seven times.

b) The common ratio is $r = \frac{-8}{16} = \frac{-1}{2}$

The eighth term in the sequence is $a_8 = a_1 r^7 = 16 \cdot \left ( \frac{-1}{2} \right )^7 = 16 \cdot \frac{(-1)^7}{2^7} = 16 \cdot \frac{-1}{2^7} = \frac{-16}{128} = - \frac{1}{8}$

Let’s take another look at the terms in that second sequence. Notice that they alternate positive, negative, positive, negative all the way down the list. When you see this pattern, you know the common ratio is negative; multiplying by a negative number each time means that the sign of each term is opposite the sign of the previous term.

## Solve Real-World Problems Involving Geometric Sequences

Let’s solve two application problems involving geometric sequences.

Example 3

A courtier presented the Indian king with a beautiful, hand-made chessboard. The king asked what he would like in return for his gift and the courtier surprised the king by asking for one grain of rice on the first square, two grains on the second, four grains on the third, etc. The king readily agreed and asked for the rice to be brought. (From Meadows et al. 1972, via Porritt 2005) How many grains of rice does the king have to put on the last square?

Solution

A chessboard is an $8 \times 8$ square grid, so it contains a total of 64 squares.

The courtier asked for one grain of rice on the first square, 2 grains of rice on the second square, 4 grains of rice on the third square and so on. We can write this as a geometric sequence:

1, 2, 4,...

The numbers double each time, so the common ratio is $r = 2$.

The problem asks how many grains of rice the king needs to put on the last square, so we need to find the $64^{th}$ term in the sequence. Let’s use our formula:

$a_n = a_1 r^{n-1}$, where $a_n$ is the nth term, $a_1$ is the first term and $r$ is the common ratio.

$a_{64} = 1 \cdot 2^{63} = 9,223,372,036,854,775,808$ grains of rice.

The problem we just solved has real applications in business and technology. In technology strategy, the Second Half of the Chessboard is a phrase, coined by a man named Ray Kurzweil, in reference to the point where an exponentially growing factor begins to have a significant economic impact on an organization's overall business strategy.

The total number of grains of rice on the first half of the chessboard is $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 \ldots + 2,147,483,648$, for a total of exactly 4,294,967,295 grains of rice, or about 100,000 kg of rice (the mass of one grain of rice being roughly 25 mg). This total amount is about $\frac{1}{1,000,000^{th}}$ of total rice production in India in the year 2005 and is an amount the king could surely have afforded.

The total number of grains of rice on the second half of the chessboard is $2^{32} + 2^{33} + 2^{34} \ldots + 2^{63}$, for a total of 18, 446, 744, 069, 414, 584, 320 grains of rice. This is about 460 billion tons, or 6 times the entire weight of all living matter on Earth. The king didn’t realize what he was agreeing to—perhaps he should have studied algebra! [Wikipedia; GNU-FDL]

Example 4

A super-ball has a 75% rebound ratio—that is, when it bounces repeatedly, each bounce is 75% as high as the previous bounce. When you drop it from a height of 20 feet:

a) how high does the ball bounce after it strikes the ground for the third time?

b) how high does the ball bounce after it strikes the ground for the seventeenth time?

Solution

We can write a geometric sequence that gives the height of each bounce with the common ratio of $r = \frac{3}{4}$:

$20, 20 \cdot \frac{3}{4}, 20 \cdot \left ( \frac{3}{4} \right )^2, 20 \cdot \left ( \frac{3}{4} \right )^3 \ldots$

a) The ball starts at a height of 20 feet; after the first bounce it reaches a height of $20 \cdot \frac{3}{4} = 15 \ feet$.

After the second bounce it reaches a height of $20 \cdot \left ( \frac{3}{4} \right )^2 = 11.25 \ feet$.

After the third bounce it reaches a height of $20 \cdot \left ( \frac{3}{4} \right )^3 = 8.44 \ feet$.

b) Notice that the height after the first bounce corresponds to the second term in the sequence, the height after the second bounce corresponds to the third term in the sequence and so on.

This means that the height after the seventeenth bounce corresponds to the $18^{th}$ term in the sequence. You can find the height by using the formula for the $18^{th}$ term:

$a_{18} = 20 \cdot \left ( \frac{3}{4} \right )^{17} = 0.15 \ feet.$

Here is a graph that represents this information. (The heights at points other than the top of each bounce are just approximations.)

For more practice finding the terms in geometric sequences, try the browser game at http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/sequences_and_series/geometric_sequences.html.

## Review Questions

Determine the first five terms of each geometric sequence.

1. $a_1 = 2, r = 3$
2. $a_1 = 90, r = -\frac{1}{3}$
3. $a_1 = 6, r = -2$
4. $a_1 = 1, r = 5$
5. $a_1 = 5, r = 5$
6. $a_1 = 25, r = 5$
7. What do you notice about the last three sequences?

Find the missing terms in each geometric sequence.

1. 3, __ , 48, 192, __
2. 81, __ , __ , __ , 1
3. $\frac{9}{4} \ , \underline{\;\;\;\;} \ , \underline{\;\;\;\;} \ , \frac{2}{3} \ , \underline{\;\;\;\;}$
4. 2, __ , __ , -54, 162

Find the indicated term of each geometric sequence.

1. $a_1 = 4, r = 2$; find $a_6$
2. $a_1 = -7, r = - \frac{3}{4}$; find $a_4$
3. $a_1 = -10, r = -3$; find $a_{10}$
4. In a geometric sequence, $a_3 = 28$ and $a_5 = 112$; find $r$ and $a_1$.
5. In a geometric sequence, $a_2 = 28$ and $a_5 = 112$; find $r$ and $a_1$.
6. As you can see from the previous two questions, the same terms can show up in sequences with different ratios.
1. Write a geometric sequence that has 1 and 9 as two of the terms (not necessarily the first two).
2. Write a different geometric sequence that also has 1 and 9 as two of the terms.
3. Write a geometric sequence that has 6 and 24 as two of the terms.
4. Write a different geometric sequence that also has 6 and 24 as two of the terms.
5. What is the common ratio of the sequence whose first three terms are 2, 6, 18?
6. What is the common ratio of the sequence whose first three terms are 18, 6, 2?
7. What is the relationship between those ratios?
7. Anne goes bungee jumping off a bridge above water. On the initial jump, the bungee cord stretches by 120 feet. On the next bounce the stretch is 60% of the original jump and each additional bounce the rope stretches by 60% of the previous stretch.
1. What will the rope stretch be on the third bounce?
2. What will be the rope stretch be on the $12^{th}$ bounce?

Feb 23, 2012

Jul 08, 2014