# 8.7: Geometric Sequences and Exponential Functions

Difficulty Level: At Grade Created by: CK-12

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 you would think that a million dollars would be more. However, let’s use geometric sequences before we decide to see how the pennies add up. You start with a penny the first day and keep doubling each day until the end of the month (30 days).

1st Day2nd Day3rd Day4th Day30th Day1 penny2 pennies4 pennies8 pennies=20=21=22=23=229\begin{align*}1^{st} \ \text{Day} && 1 \ \text{penny} & = 2^0\\ 2^{nd} \ \text{Day} && 2 \ \text{pennies} & = 2^1\\ 3^{rd} \ \text{Day} && 4 \ \text{pennies} & = 2^2\\ 4^{th} \ \text{Day} && 8 \ \text{pennies} & = 2^3\\ 30^{th} \ \text{Day} && & = 2^{29}\end{align*}

229=536,870,912\begin{align*}2^{29} =536,870,912\end{align*} pennies or $5,368,709, which is well over five times greater than$1,000,000.00.

Geometric Sequence: 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.

nth\begin{align*}n^{th}\end{align*} term in a geometric sequence an=a1rn1\begin{align*}a_n= a_1 r^{n-1}\end{align*} (a1=\begin{align*}a_1 =\end{align*} first term, r=\begin{align*}r =\end{align*} common ratio)

## The Common Ratio of a Geometric Sequence

The common ratio, r\begin{align*}r\end{align*}, in any geometric sequence can be found by dividing any term by the preceding term. If we know the common ratio in a sequence then we can find any term in the sequence.

Example 1: Find the eighth term in the geometric sequence.

1, 2, 4,...

Solution First we need to find the common ratio r=21=2\begin{align*}r=\frac{2}{1}=2\end{align*}.

The eighth term is given by the formula 2=127=128\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 two seven times.

## Graphing Geometric Sequences

Exponential graphs and geometric sequence graphs look very much alike. Exponential graphs are continuous, however, and the sequence graphs are discrete with distinct points (1st\begin{align*}1^{st}\end{align*} term and 2nd\begin{align*}2^{nd}\end{align*} term, etc).

Example: 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.

Solution: The population of bacteria is continuous. Even though we measured the population only every 24 hours, we know that it does not get from 1 million to 3 million all at once. Rather, 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-hour period.

When we graph an exponential function, we draw the graph with a solid curve to show 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 show the sequence has value only at those points but not in between.

Here are graphs for the two examples previously given:

## Solving Real-World Problems Involving Geometric Sequences

Example: 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?

Solution: A chessboard is an 8×8\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.

1, 2, 4,...

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

The problem asks how many grains of rice the king needs to put on the last square. What we need to find is the 64th\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. Let’s use the formula.

an=a1rn1\begin{align*}a_n=a_1 r^{n-1}\end{align*}, where an\begin{align*}a_n\end{align*} is the nth\begin{align*}n^{th}\end{align*} term, a1\begin{align*}a_1\end{align*} Is the first term and r\begin{align*}r\end{align*} is the common ratio.

a64=1263=9,223,372,036,854,775,808\begin{align*}a_{64}=1 \cdot 2^{63}=9, 223, 372, 036, 854, 775, 808\end{align*} grains of rice.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. Define a geometric sequence.
2. Using the chessboard example, how many grains of rice should be placed on the:
1. Fourth square?
2. Tenth square?
3. Twenty-fifth square?
4. 60th\begin{align*}60^{th}\end{align*} square?
3. Is the following an example of a geometric sequence? Explain your answer. –1, 1, –1, 1, –1, ...
4. If you are given one penny the first day, double that penny the next day, and then double the previous day's pennies and so on for 20 days, how much money would you have?

Determine the constant ratio, r\begin{align*}r\end{align*}, for each geometric sequence.

1. 8, 6,...
2. 2, 4, 8, 16,...
3. 9,3,13,19,...\begin{align*}9,3,\frac{1}{3},\frac{1}{9}, ...\end{align*}
4. 2, –8, 32, –128

Determine the first five terms of each geometric sequence.

1. a1=2,r=3\begin{align*}a_1=2, r=3\end{align*}
2. a1=90,r=13\begin{align*}a_1=90, r=-\frac{1}{3}\end{align*}
3. a1=6,r=2\begin{align*}a_1=6, r=-2\end{align*}

Find the missing terms in each geometric series:

1. 3, ____, 48, 192, ____
2. 81, ____, ____, ____, 1
3. 94\begin{align*}\frac{9}{4}\end{align*}, ____, ____, 23\begin{align*}\frac{2}{3}\end{align*}, ____

Find the indicated term of each geometric series.

1. a1=4,r=2\begin{align*}a_1=4, r=2\end{align*} Find a6\begin{align*}a_6\end{align*}.
2. a1=7,r=34\begin{align*}a_1=-7, r=-\frac{3}{4}\end{align*} Find a4\begin{align*}a_4\end{align*}.
3. a1=10,r=3\begin{align*}a_1=-10, r=-3\end{align*} Find a10\begin{align*}a_{10}\end{align*}.
1. A ball is tossed from a height of four feet. Each bounce is 80% as high as the previous bounce.
1. Write an equation to represent the situation.
2. How high is the ball after the fifth bounce?
2. An ant walks past several stacks of Lego blocks. There is one block in the first stack, three blocks in the second stack, and nine blocks in the third stack. In fact, in each successive stack there are triple the number of blocks there were in the previous stack.
1. How many blocks are in the eighth stack?
2. When is the stack 343 blocks high?
3. A super-ball has a 75% rebound ratio. When you drop it from a height of 20 feet, it bounces and bounces and bounces...
1. How high does the ball bounce after it strikes the ground for the third time?
2. How high does the ball bounce after it strikes the ground for the seventeenth time?
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 the rope stretch be on the 12th\begin{align*}12^{th}\end{align*} bounce?
5. A virus population doubles every 30 minutes. It begins with a population of 30.
1. How many viral cells will be present after 5 hours?
2. When will it reach 1,000,000 cells? Round to the nearest half-hour.
6. The half-life of the prescription medication Amiodarone is 25 days. Suppose a patient has a single dose of 12 mg of this drug in her system.
1. How much Amiodarone will be in the patient’s system after four half-life periods?
2. When will she have less than 3 mg of the drug in her system?
3. What is the growth factor of this situation?

Mixed Review

1. Translate into an algebraic sentence: A number squared is less than 15 more than twice that number.
2. Give the slope and y\begin{align*}y-\end{align*}intercept of y=23x7\begin{align*}y=\frac{2}{3} x-7\end{align*}.
3. Evaluate 10!
4. Convert 6 miles to yards.
5. Simplify 5y23y24y11\begin{align*}\frac{5y^2-3y^2}{4y^{11}}\end{align*}.
6. Simplify 3x2x6+4x3x5\begin{align*}3x^2 \cdot x^6+4x^3 x^5\end{align*}.
7. Evaluate \begin{align*}\left(\frac{27}{64}\right)^{-\frac{1}{3}}\end{align*}.

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