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# 8.7: Geometric Sequences and Exponential Functions

Difficulty Level: At Grade 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 please!”

However, let’s do a few calculations before we decide in order to see how the pennies add 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 1st day we getOn the 2nd day we getOn the 3rd day we getOn the 4th day we getOn the 30th day we get1 penny2 pennies4 pennies8 pennies=20 pennies=21 pennies=22 pennies=23 pennies=229 penniesLook at the exponent on the 2.Can you see the pattern?The exponent increases by 1 each day.So we calculate that\ldots\begin{align*}\text{On the}\ 1^{st}\ \text{day we get} && 1\ \text{penny} & =2^0\ \text{pennies} && \text{Look at the exponent on the} \ 2.\\ \text{On the}\ 2^{nd}\ \text{day we get} && 2\ \text{pennies} & =2^1\ \text{pennies} && \text{Can you see the pattern?}\\ \text{On the}\ 3^{rd}\ \text{day we get} && 4\ \text{pennies} & =2^2\ \text{pennies} && \text{The exponent increases by}\ 1\ \text{each day.}\\ \text{On the}\ 4^{th}\ \text{day we get} && 8\ \text{pennies} & =2^3\ \text{pennies} && \text{So we calculate that\ldots}\\ \text{On the}\ 30^{th}\ \text{day we get} &&& =2^{29}\ \text{pennies} \end{align*}

229=536,870,912\begin{align*}2^29= 536,870,912\end{align*} pennies or \$5,368,709 which is well over 5 times greater than one million dollars.

So even just considering the pennies given on the final day, the pennies win!

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

## Identify a Geometric Sequence

The problem above is an example of 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 a term and the previous term 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\begin{align*}r\end{align*}, 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.

\begin{align*}& 4, 16, 64, 256, \ldots && r = 4 && (\text{divide} \ 16 \div 4 \ \text{to get} \ 4) \\ & 15, 30, 60, 120,\ldots && r = 2 && (\text{divide} \ 30 \div \ 15 \ \text{to get} \ 2) \\ & 11, \frac{11} {2}, \frac{11} {4}, \frac{11} {8}, \frac{11} {16},\ldots && r = \frac{1} {2} && \left (\text{divide} \ \frac{1} {2} \div 11 \ \text{to get} \ \frac{1} {2}\right ) \\ & 25, -5, 1, -\frac{1} {5}, - \frac{1} {25},\ldots && r = - \frac{1} {5} && \left (\text{divide} \ 1 \div (-5) \ \text{to get} - \frac{1} {5}\right )\end{align*}

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

Example 1

Fill is the missing terms in the geometric sequences.

a) \begin{align*}1,\underline{\;\;\;\;\;\;\;\;}, 25, 125,\underline{\;\;\;\;\;\;\;\;}\end{align*}

b) \begin{align*}20,\underline{\;\;\;\;\;\;\;\;}, 5,\underline{\;\;\;\;\;\;\;\;}, 1.25\end{align*}

Solution

a) First we can find the common ratio by dividing 125 by 25 to obtain \begin{align*}r=5\end{align*}.

To find the \begin{align*}1^{st}\end{align*} missing term we multiply 1 by the common ratio \begin{align*}1 \cdot 5 = 5\end{align*}

To find the \begin{align*}2^{nd}\end{align*} missing term we multiply 125 by the common ratio \begin{align*}125 \cdot 5 = 625\end{align*}

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

b) We first need to find the common ratio, but we run into difficulty because we have no terms next to each other that we can divide.

However, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice. We multiply it once to get to the second term in the sequence and again to get to five. So we can say

\begin{align*}20 \cdot r\cdot r=5\end{align*} or \begin{align*}20 \cdot r^2=5\end{align*}

Divide both sides by 20 and find \begin{align*} r^2 = \frac{5} {20} = \frac{1} {4}\end{align*} or \begin{align*} r = \frac{1} {2}\end{align*} (because \begin{align*} \frac{1} {2} \cdot \frac{1} {2} = \frac{1} {4}\end{align*}).

To get the \begin{align*}1^{st}\end{align*} missing term we multiply 20 by \begin{align*} \frac{1} {2}\end{align*} and get \begin{align*} 20 \cdot \frac{1} {2} = 10\end{align*}.

To get the \begin{align*}2^{nd}\end{align*} missing term we multiply 5 by \begin{align*} \frac{1} {2}\end{align*} and get: \begin{align*} 5 \cdot \frac{1} {2} = 2.5\end{align*}.

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

You see that we can find any term in a geometric sequence simply by multiplying the last term by the common ratio. Then, we keep multiplying by the common ratio until we get to a term in the sequence that we want. However, if we want to find a term that is a long way from the start it becomes tedious to keep multiplying over and over again. There must be a better way to do this.

Because we keep multiplying by the same number, we can use exponents to simplify the calculation. For example, let’s take a geometric sequence that starts with the number 7 and has common ratio of 2.

\begin{align*}& \text{The} \ 1^{st} \ \text{term is:} && 7 \\ & \text{We obtain the}\ 2^{nd} \ \text{term by multiplying by} \ 2. && 7 \cdot 2 \\ & \text{We obtain the}\ 3^{rd} \ \text{term by multiplying by}\ 2\ \text{again.} && 7 \cdot 2 \cdot 2 \\ & \text{We obtain the}\ 4^{th} \ \text{term by multiplying by}\ 2\ \text{again.} && 7 \cdot 2 \cdot 2 \cdot 2 \\ & \text{The} \ n^{th} \ \text{term would be} && 7\cdot 2^{n-1}\end{align*}

The \begin{align*}n^{th}\end{align*} term is \begin{align*}7 \cdot 2^{n-1}\end{align*} because the 7 is multiplied by one factor of two for the \begin{align*}2^{nd}\end{align*} term, two factors of 2 for the third term and always by one less factor of 2 than the term’s place in the sequence. In general terms we write geometric sequence with n terms like this

\begin{align*}a, ar,ar^2, ar^3, \ldots , ar^{n-1}\end{align*}

The general formula for finding specific terms in a geometric sequence is

\begin{align*}n^{th}\end{align*} term in a geometric sequence \begin{align*}a_{n}=ar_{1} r^{n-1}(a_{1}=\text{first\ term},\ r= \text{common\ ratio})\end{align*}

Example 2

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

a) \begin{align*}1, 2, 4,\ldots\end{align*}

b) \begin{align*}16, -8, 4, -2, 1,\ldots \end{align*}

Solution

a) First we need to find the common ratio \begin{align*} r = \frac{2} {1} = 2.\end{align*}

The eighth term is given by the formula \begin{align*}2 = 1 \cdot 2^7 =128\end{align*}.

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

b) The common ratio is \begin{align*} r = \frac{-8} {16} = \frac{-1} {2}\end{align*}

The eighth term in the sequence is: \begin{align*} a_g = 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}\end{align*}

Look again at the terms in b).

When a common ratio is negative the terms in the sequence alternate positive, negative, positive, negative all the way down the list. When you see this, you know the common ratio is negative.

## Graph a Geometric Sequence

Geometric sequences and exponential functions are very closely related. You just learned that to get to the next term in a geometric sequence you multiply the last term in the sequence by the common ratio. In Sections 8.5 and 8.6, you learned that an exponential function is multiplied by the same factor every time the independent value is increased by one unit. As a result, geometric sequences and exponential functions look very similar.

The fundamental difference between the two concepts is that a geometric sequence is discrete while an exponential function is continuous.

Discrete means that the sequence has values only at distinct points (the \begin{align*}1^{st}\end{align*} term, \begin{align*}2^{nd}\end{align*} term, etc)

Continuous means that the function has values for all possible values of \begin{align*}x\end{align*}. The integers are included, but also all the numbers in between.

As a result of this difference, we use a geometric series to describe quantities that have values at discrete points, and we use exponential functions to describe quantities that have values that change continuously.

Here are two examples one discrete and one continuous.

Example 3 Discrete sequence

An ant walks past several stacks of Lego blocks. There is one block in the first stack, 3 blocks in the \begin{align*}2^{nd}\end{align*} stack and 9 blocks in the \begin{align*}3^{rd}\end{align*} stack. In fact, in each successive stack there are triple the number of blocks than there were in the previous stack.

In this example, each stack has a distinct number of blocks and the next stack is made by adding a certain number of whole pieces all at once. More importantly, however, there are no values of the sequence between the stacks. You cannot ask how high the stack is between the \begin{align*}2^{nd}\end{align*} and \begin{align*}3^{rd}\end{align*} stack, as no stack exists at that position!

Example 4 Continuous Function

A population of bacteria in a Petri dish increases by a factor of three every 24 hours. The starting population is 1 million bacteria. This means that on the first day the population increases to 3 million on the second day to 9 million and so on.

In this example, the population of bacteria is continuous. Even though we only measured the population every 24 hours we know that it does not get from 1 million to 3 million all at once but that the population changes bit by bit over the 24 hours. In other words, the bacteria are always there, and you can, if you so wish, find out what the population is at any time during a 24 hours period.

When we graph an exponential function, we draw the graph with a solid curve to signify that the function has values at any time during the day. On the other hand, when we graph a geometric sequence, we draw discrete points to signify that the sequence only has value at those points but not in between.

Here are graphs for the two examples we gave before:

## Solve Real-World Problems Involving Geometric Sequences

Let’s solve two application problems involving geometric sequences.

Example 5 Grains of rice on a chessboard

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, p.29 via Porritt 2005) How many grains of rice does the king have to put on the last square?

[Wikipedia; GNU-FDL]

Solution

A chessboard is an \begin{align*}8 \times 8\end{align*} 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.

\begin{align*}1, 2, 4,\ldots \end{align*}

The numbers double each time, so the common ratio is \begin{align*}r=2\end{align*}.

The problem asks how many grains of rice the kind needs to put on the last square. What we need to find is the \begin{align*}64^{th}\end{align*} term in the sequence.

This means multiplying the starting term, 1, by the common ratio 64 times in a row. Instead of doing this, let’s use the formula we found earlier.

\begin{align*}a_n = a_1 r^{n-1}\end{align*}, where \begin{align*}a_n\end{align*} is the \begin{align*}n^{th}\end{align*} term, \begin{align*}a_1\end{align*} is the first term and \begin{align*}r\end{align*} is the common ratio.

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

Second Half of the Chessboard

The problem we just solved has real applications in business and technology. In technology the strategy, 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 \begin{align*} 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 +\ldots + 2,147,483,648\end{align*}, for a total of exactly 4,294,967,295 grains of rice, or about 100,000 kg of rice, with the mass of one grain of rice being roughly 25 mg. This total amount is about \begin{align*}\frac{1}{1,000,000^{th}}\end{align*} of total rice production in India in year 2005 and was considered economically viable to the emperor of India.

The total number of grains of rice on the second half of the chessboard is \begin{align*}2^{32} +2^{33}+2^{34}+\ldots +2^{63}\end{align*}, 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 did not realize what he was agreeing. Next time maybe he should read the fine print! [Wikipedia; GNU-FDL]

Example 6 Bouncing Ball

A super-ball has a 75% rebound ratio. When you drop it from a height of 20 feet, it bounces and bounces and bounces ...

(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

75% rebound ratio means that after the ball bounces on the ground, it reaches a maximum height that is 75% or \begin{align*}\left ( \frac{3}{4} \right )\end{align*} of its previous maximum height. We can write a geometric sequence that gives the maximum heights of the ball after each bounce with the common ratio of \begin{align*}r = \frac{3}{4}\end{align*}.

\begin{align*} 20, 20 \cdot \frac{3} {4} , 20 \cdot \left (\frac{3} {4}\right )^2, 20 \cdot \left (\frac{3} {4}\right )^3 \ldots\end{align*}

a) The ball starts at a height of 20 feet, after the first bounce it reaches a height of \begin{align*}20 \cdot \frac{3} {4} = 15 \ feet\end{align*}

After the second bounce, it reaches a height of \begin{align*} 20 \cdot \left (\frac{3} {4}\right )^2 = 11.25 \ feet\end{align*}

After the third bounce, it reaches a height of \begin{align*} 20 \cdot \left (\frac{3} {4}\right )^3 = 8.44 \ feet\end{align*}

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.

b) This means that the height after the seventeenth bounce corresponds to the \begin{align*}18^{th}\end{align*} term in the sequence. You can find the height by using the formula for the \begin{align*}18^{th}\end{align*} term:

\begin{align*} a_{18} = 20 \cdot \left (\frac{3} {4}\right )^{17} = 0.15\ feet\end{align*}

Here is a graph that represents this information.

## Review Questions

Determine the first five terms of each geometric sequence.

1. \begin{align*}a_1 =2, r=3\end{align*}
2. \begin{align*} a_1 = 90, r = - \frac{1} {3}\end{align*}
3. \begin{align*}a_1 = 6, r=-2\end{align*}

Find the missing terms in each geometric series:

1. \begin{align*}3,\underline{\;\;\;\;\;\;\;}, 48, 192,\underline{\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}81,\underline{\;\;\;\;\;\;\;} , \underline{\;\;\;\;\;\;\;} , \underline{\;\;\;\;\;\;\;} , 1\end{align*}
3. \begin{align*}\frac{9} {4}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;},\frac{2} {3},\underline{\;\;\;\;\;\;\;}\end{align*}

Find the indicated term of each geometric series.

1. \begin{align*}a_1=4, r=2\end{align*} Find \begin{align*}a_6\end{align*}.
2. \begin{align*} a_1 = -7, r = - \frac{3} {4}\end{align*} Find \begin{align*} a_4\end{align*}.
3. \begin{align*}a_1 =-10, r=-3\end{align*} Find \begin{align*}a_{10}\end{align*}.
4. 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 stretches the rope 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 \begin{align*}12^{th}\end{align*} bounce?

1. \begin{align*}2, 6, 18, 54, 162\end{align*}
2. \begin{align*}90, -30, 10, -\frac{10}{3}, \frac{10}{9}\end{align*}
3. \begin{align*}6, -12, 24, -48, 96\end{align*}
4. \begin{align*}3, 12, 48, 192, 768 \end{align*}
5. \begin{align*}81, 27, 9, 3, 1 \end{align*}
6. \begin{align*}\frac{9}{4}, \frac{3}{2}, 1, \frac{2}{3}, \frac{4}{9} \end{align*}
7. \begin{align*}a_6 =128 \end{align*}
8. \begin{align*}a_4 =2.95\end{align*}
9. \begin{align*}a_{10} = 196830 \end{align*}
1. 43.2 feet
2. 0.44 feet

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