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There are many situations where a quantity increases or decreases by a constant factor, like interest rates, population growth, etc. These situations can be modeled with a special type of sequence called a geometric sequence. This lesson will introduce geometric sequences for specific everyday life applications and will show how they can be described using explicit rules and recursive rules.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Ratio
Pattern
Sequences
Exponential Function

Explore

How Does the Thickness of a Sheet of Paper Increase as It Is Folded in Half?

A regular sheet of paper has a thickness of about $0.1$ millimeter. Every time the paper is folded in half, its thickness doubles. Try the applet below to explore how the thickness increases as the sheet keeps getting folded in half.

Interactive applet showing how the thickness of a sheet of paper increases by continuously getting folded in half

Challenge

Predicting the Height of a Paper Sheet Folded Multiple Times

Using the same applet, now try to predict values according to the pattern.

If a regular sheet of paper with a thickness of $0.1$ millimeters could be folded in half $15$ times, how tall would it be? Would it be taller than an average person? For reference, the height of an average person is $1.7$ meters.
If it could be folded in half $20$ times, how tall would it be? Would it be taller than a $10$ story building? For reference, a $10$ story building is about $45$ meters tall.

Discussion

Introducing Geometric Sequences

As can be seen in the previous applet, each time the sheet of paper was folded in half, its thickness doubled. The values for the paper thickness after each fold can be represented by terms of a specific type of sequence called a geometric sequence.

Concept

Geometric Sequence

A geometric sequence is a sequence in which the ratio

r

between consecutive terms is a nonzero constant. This ratio is called the common ratio. The following is an example geometric sequence with first term

3

and common ratio

2 .

Geometric Sequence: 3, 6, 12, 24, 48, ... with a common ratio of 2

The behavior of geometric sequences generally depends on the values of the first term

a_{1}

and the common ratio

r .

The following table shows the effects of these parameters.

	$a_{1} > 0$	$a_{1} < 0$
$r > 1$	Increasing $3 \to \times 2 6 \to \times 2 12 \to \times 2 24 \to \times 2 48 \dots$	Decreasing $- 3 \to \times 2 - 6 \to \times 2 - 12 \to \times 2 - 24 \to \times 2 - 48 \dots$
$r = 1$	Constant $3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \dots$	Constant $- 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \dots$
$0 < r < 1$	Decreasing $48 \to \times \frac{1}{2} 24 \to \times \frac{1}{2} 12 \to \times \frac{1}{2} 6 \to \times \frac{1}{2} 3 \dots$	Increasing $- 48 \to \times \frac{1}{2} - 24 \to \times \frac{1}{2} - 12 \to \times \frac{1}{2} - 6 \to \times \frac{1}{2} - 3 \dots$
$r < 0$	Alternating $3 \to \times (- 2) - 6 \to \times (- 2) 12 \to \times (- 2) - 24 \to \times (- 2) 48 \dots$	Alternating $- 3 \to \times (- 2) 6 \to \times (- 2) - 12 \to \times (- 2) 24 \to \times (- 2) - 48 \dots$

Like for any other sequence, the first term of a geometric sequence is denoted by $a_{1},$ the second by $a_{2},$ and so on. Since geometric sequences have a common ratio $r,$ once one term is known, the next term can always be found by multiplying the known term by $r .$

General geometric sequence a_1, a_2, a_3, a_4, a_5, ... with a common ratio of r.

In fact, the sequence can be found using only $a_{1}$ and $r,$ since all the subsequent terms can be found by multiplying $a_{1}$ by $r$ a specific number of times. Because of this, geometric sequences have the following general form.

$a_{1}, a_{1} r, a_{1} r^{2}, a_{1} r^{3}, a_{1} r^{4}, \dots$

Pop Quiz

Identifying Geometric Sequences

The following applet shows the first five terms of an infinite sequence. Analyze them carefully and determine whether or not the sequence is geometric.

Interactive applet showing different sequences

Discussion

Describing Geometric Sequences Using an Explicit Rule

Geometric sequences can be described by using a formula that uses the positions of the terms to calculate their values. This formula is called an explicit rule of the geometric sequence.

Rule

Explicit Rule of Geometric Sequences

Every geometric sequence can be described by a function known as the explicit rule, whose input is the position of a term $n$ and whose output is the term's value $a_{n} .$ An explicit rule for a geometric sequence has the following general form.

$a_{n} = a_{1} \cdot r^{n - 1}$

Here, $a_{1}$ is the first term of the sequence and $r$ is the common ratio.

Proof

Proof by Induction

Recall that every geometric sequence has a common ratio

r .

Therefore, it is possible to find every term of the sequence by multiplying the first term

a_{1}

by this common ratio a particular number of times. Therefore, knowing

a_{1}

and

r

is enough to describe the whole geometric sequence.

It is easier to identify a pattern that can be used to write a general expression for the explicit rule by making a table. Note that by the Zero Exponent Property,

r^{0}

is equal to

1 .

Furthermore,

r

can be written as

r^{1} .

$n$	$a_{n}$	Using $a_{1}$ and $r$
$1$	$a_{1}$	$a_{1} \cdot r^{0}$
$2$	$a_{2}$	$a_{1} \cdot r^{1}$
$3$	$a_{3}$	$a_{1} \cdot r^{2}$
$4$	$a_{4}$	$a_{1} \cdot r^{3}$

It can be seen that the exponent of the common ratio is always $1$ less than the value of the position $n .$ With this pattern, it is possible to write the explicit rule in the same form as the formula given at the beginning.

$a_{n} = a_{1} \cdot r^{n - 1}$

Example

Modeling Multiplication of Bacteria Using Geometric Sequences

Jordan is studying her biology notes. She finds out that a bacterium can divide into two bacteria in a period of time of about $20$ minutes. These two bacteria can then divide into two bacteria each, and so on.

Having read her notes, she is now ready for the lab practice. In a glass slide she has prepared a sample with $7$ isolated bacteria.

Microscope and sample holder with 7 bacteria

For her practice, Jordan had to check every $20$ minutes and count the number of bacteria. The results can be written as a sequence.

Representing the situation using a sequence

a Show that this situation can be modeled by using a geometric sequence.

b Find the next three terms of the sequence.

Answer

a Demonstration: See solution.

b Terms:

a_{5} = 112, a_{6} = 224, a_{7} = 448

Hint

a Recall that a sequence is geometric if it has a common ratio.

b In a geometric sequence, any term can be found by multiplying the preceding term by the common ratio.

Solution

a Recall that a sequence is geometric if it has a common ratio. Since the the number of bacteria is doubling after every

20

minutes, the common ratio for this sequence is

r = 2 .

Since this sequence has a common ratio, it is, by definition, a geometric sequence.

b Once one term is known, the next term can be found by multiplying the known term by the common ratio

r .

For example, knowing that

a_{4} = 56,

the next term

a_{5}

can be obtained by multiplying

56

by the value of

r .

a_{5} = a_{4} r

▼

Substitute values and evaluate

SubstituteII

$a_{4} = 56$ , $r = 2$

a_{5} = (56) (2)

Multiply

a_{5} = 112

Therefore, the next term of the sequence is

112 .

The process can be repeated to find the next two terms. Just keep multiplying each found term by the common ratio

r = 2 .

Example

An Ecology Project that Follows a Geometric Sequence

Ramsha's ecology teacher asks each of the $30$ students in her class to plant a seed. Then, they explain that if each student asked $3$ people to do the same by tomorrow, and these $3$ people did the same by the next day, the amount of planted seeds could modeled by a geometric sequence with the explicit rule $a_{n} = 30 \cdot 3^{n - 1} .$

Teacher holding a plant next to the blackboard

a The variable

n

represents the day number and

a_{n}

the number of seeds planted on the

n^{th}

day. The teacher told the students that this number could be bigger than they would think in just a few days. Help Ramsha to find the number of seeds that would be planted on the

5^{th}

day.

b On the

1 0^{th}

day, how many seeds would be planted?

Hint

a The explicit rule must be evaluated for

n = 5

since the number of seeds planted on the

5^{th}

day is to be calculated.

b The explicit rule must be evaluated for

n = 10

since the number of seeds planted on the

1 0^{th}

day is to be calculated.

Solution

a To find the number of seeds planted on a specific day, the day number should be used as the

n -

value in the explicit rule. Therefore, to find the number of seeds planted on the

5^{th}

day, the rule will be evaluated for

n = 5 .

a_{n} = 30 \cdot 3^{n - 1}

▼

Substitute

5

for

n

and evaluate

Substitute

$n = 5$

a_{5} = 30 \cdot 3^{5 - 1}

SubTerm

Subtract term

a_{5} = 30 \cdot 3^{4}

CalcPow

Calculate power

a_{5} = 30 \cdot 81

Multiply

a_{5} = 2430

b Just as in the previous part, the explicit rule will be evaluated for the

n -

value that represents the desired day — in this case,

n = 10 .

a_{n} = 30 \cdot 3^{n - 1}

▼

Substitute

10

for

n

and evaluate

Substitute

$n = 10$

a_{10} = 30 \cdot 3^{10 - 1}

SubTerm

Subtract term

a_{10} = 30 \cdot 3^{9}

CalcPow

Calculate power

a_{10} = 30 \cdot 19683

Multiply

a_{10} = 590490

The number of seeds planted on just the

1 0^{th}

day would already be over half a million!

Example

The Geometric Sequence Behind the Story of Chess

There is a famous story about the invention of chess. When the game was presented to the king, he was so happy about it that he told the inventor to choose any payment. The inventor asked the king to put a single grain of rice on the first square, two grains on the second, four on the third and so on. The amount on the final square was the desired payment.

Chess board with grains of rice following the patter described and the first three terms of the corresponding sequence and common ratio

The king was surprised and believed that this was such a bad decision for the inventor, as the king thought this debt could be paid with no more than a bag of rice. However, when he ordered his treasurer to pay the agreed amount, it turned out that this wealthy king was not rich enough as to pay the debt. In fact, it is impossible for anyone to pay it!

a To explore this in detail, first find an explicit rule to model this situation.

b Find the how many of grains of rice are needed to pay this debt if a board of chess has

64

squares. Give the answer, rounded to

3

significant figures.

c A single grain of rice weighs about

4.4 \times 1 0^{- 6}

kilograms. The amount of rice produced in the entire world annually is about

480

million metric tons, or

4.8 \times 1 0^{11}

kilograms. If the global annual rice production were used, how many years would be needed to pay this debt? Round the answer to the nearest year.

Hint

a The general form for the explicit rule of a geometric sequence is

a_{n} = a_{1} \cdot r^{n - 1} .

b Evaluate the rule found in Part A for

n = 64 .

c First calculate the weight of the total amount of grains of the debt, then find the years needed to produce that amount.

Solution

a This situation can be modeled by a sequence with the first term

a_{1} = 1,

since a single grain is used on the first square, and a common ratio

r = 2,

since the terms double every time. First, the general form for the explicit rule of a geometric sequence will be recalled.

a_{n} = a_{1} \cdot r^{n - 1}

Now, the values

a_{1} = 1

and

r = 2

will be substituted to find the explicit rule of the presented sequence.

a_{n} = a_{1} \cdot r^{n - 1} ⇓ a_{n} = 1 \cdot 2^{n - 1}

b To find the total number of grains of rice of the king's debt, the explicit formula will be evaluated for

n = 64,

since the chessboard has

64

squares.

a_{n} = 1 \cdot 2^{n - 1}

▼

Substitute

64

for

n

and evaluate

Substitute

$n = 64$

a_{64} = 1 \cdot 2^{64 - 1}

SubTerm

Subtract term

a_{64} = 1 \cdot 2^{63}

CalcPow

Calculate power

a_{64} = 1 \cdot 9.223372 \dots \times 1 0^{18}

OneMult

$1 \cdot a = a$

a_{64} = 9.223372 \dots \times 1 0^{18}

RoundSigDig

Round to $3$ significant digit(s)

a_{64} = 9.22 \times 1 0^{18}

c First the total weight of the debt will be found. To do this, the number of grains will be multiplied by the weight of each individual grain:

4.4 \times 1 0^{- 6}

kilograms.

(9.22 \times 1 0^{18}) (4.4 \times 1 0^{- 6})

▼

Simplify

CommutativePropMult

Commutative Property of Multiplication

(9.22) (4.4) (1 0^{18}) (1 0^{- 6})

Multiply

40.568 (1 0^{18}) (1 0^{- 6})

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

40.568 (1 0^{12})

Write in scientific notation

4.0568 \times 1 0^{13}

That is

4.0568 \times 1 0^{13}

kilograms of rice. Recall that the whole Earth's production of rice per year is about

4.8 \times 1 0^{11}

kilograms. Therefore, the number of years needed for the whole planet to produce the rice of the debt can be obtained by dividing the total weight of the debt by the amount produced each year.

\frac{4 . 0 5 6 8 \times 1 0 ^{13}}{4 . 8 \times 1 0 ^{11}}

▼

Simplify

WriteProdFrac

Write as a product of fractions

\frac{4 . 0 5 6 8}{4 . 8} \cdot \frac{1 0 ^{13}}{1 0 ^{11}}

DivPow

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

\frac{4 . 0 5 6 8}{4 . 8} \cdot 1 0^{2}

CalcPow

Calculate power

\frac{4 . 0 5 6 8}{4 . 8} \cdot 100

CalcQuot

Calculate quotient

0.845166 \dots \cdot 100

Multiply

84.516666 \dots

RoundInt

Round to nearest integer

85

It would take all the rice produced in the world over

85

years to pay the king's debt! This is definitely much more than what the king thought at first.

Example

Investigating a Police Case Using Geometric Sequences

A criminal mastermind started a big scam. They emailed some number of people and convinced them to send money by providing suspicious information for an investment plan. The criminal told each victim that all they needed to do was to contact $5$ people and ask them to send the same amount of money, and the mastermind's company would take care of the rest.

Scammer using their computer to gather money from their victims

The scammer gave each victim one week to find $5$ people. Then, they gave one week to the new people to repeat the process. However, because of a blunder, the police caught the criminal.

a Which of the following options represents an explicit rule that can be used to find the number of new victims of this pyramid scheme at a given week?

b The criminal refuses to talk, but the police have gathered enough evidence to conclude that the operation has been going on for

4

weeks, and that on the fourth week, the criminal received

$ 200000

from the victims. Moreover, from some victim statements, the police also know that each victim was asked to send

$ 80 .

What was the starting number of victims?

Hint

a The general form for the explicit rule of a geometric sequence is

a_{n} = a_{1} \cdot r^{n - 1} .

b Since the total amount of money received together with the amount that each victim sent is known, the number of victims at the fourth week can be found. Use this information and the result from Part A to calculate the initial number of victims.

Solution

a The formula of an explicit rule for a geometric sequence with common ratio

r

and initial term

a_{1}

has the following form.

a_{n} = a_{1} \cdot r^{n - 1}

Here,

n

is the position of the

n^{th}

term and

a_{n}

is the

n^{th}

term's value. Since each victim had to contact

5

people, who would then contact

5

more people, and so on, it can be concluded that each week the number of victims increased

5

times. Thus, the common ratio is

r = 5 .

This value will be substituted into the above formula.

a_{n} = a_{1} \cdot r^{n - 1} ⇓ a_{n} = a_{1} \cdot 5^{n - 1}

In this explicit rule,

a_{n}

represents the total number of victims on the

n^{th}

week. So far, the initial number of victims

a_{1}

is unknown.

b It is known that on the

4^{th}

week, the scammer received a total amount of

$ 200000

from the victims. Because

a_{4}

represents the number of victims on the

4^{th}

week, and since each victim sent

$ 80,

80 a_{4}

should equal the total amount received.

80 a_{4} = 200000 \Rightarrow a_{4} = 2500

Now that

a_{4}

is known, the value

n = 4

will be used in the explicit rule from Part A to determine the starting number of victims

a_{1} .

a_{n} = a_{1} \cdot 5^{n - 1}

Substitute

$n = 4$

a_{4} = a_{1} \cdot 5^{4 - 1}

Substitute

$a_{4} = 2500$

2500 = a_{1} \cdot 5^{4 - 1}

▼

Solve for

a_{1}

SubTerm

Subtract term

2500 = a_{1} \cdot 5^{3}

CalcPow

Calculate power

2500 = a_{1} \cdot 125

DivEqn

$LHS / 125 = RHS / 125$

20 = a_{1}

RearrangeEqn

Rearrange equation

a_{1} = 20

Therefore, the criminal started by scamming

20

victims.

Discussion

Defining Geometric Sequences Recursively

It has been shown how an explicit rule can describe a geometric sequence with a function that receives the term position as input and returns the term's value as output. However, a geometric sequence can also be described with a recursive relation.

Concept

Recursive Rule of a Geometric Sequence

A recursive rule of a geometric sequence is a pair made of a recursive equation telling how the term $a_{n}$ is related to its preceding term $a_{n - 1},$ and the first term of the sequence $a_{1} .$

$a_{1}, a_{n} = a_{n - 1} \cdot r$

In the equation above,

r

represents the common ratio. The following applet gives an example recursive rule for a geometric sequence and shows how it can be used to determine the first five terms of the sequence.

Interactive applet showing the use of the recursive rule for a geometric sequence

Note that if the first term is not given, the recursive equation by itself describes all different geometric sequences with the same common ratio.

The recursive equation, a_n =(a_n-1)(3), describes both the geometric sequence 2,6,18,54,162,... as well as the geometric sequence -5,-15,-45,-135,-405,...

This is why the first term should be specified in the recursive rule to uniquely define the specific geometric sequence.

Now it will be explained, step by step, how to write the recursive rule for a geometric sequence.

Method

Writing a Recursive Rule for a Geometric Sequence

The recursive rule of a geometric sequence includes the first term of the sequence and a recursive equation.

$a_{1}, a_{n} = a_{n - 1} \cdot r$

Consider the following example geometric sequence.

a_{1} 2, a_{2} 6, a_{3} 18, a_{4} 54, \dots

There are 3 steps to follow to write the recursive rule for this geometric sequence.

Find the Common Ratio

To find the common ratio

r,

identify any two

consecutive

terms

and divide the latter term by the former one.

a_{1} 2, a_{2} 6, a_{3} 18, a_{4} 54, \dots \frac{1 8}{6} = 3 \Rightarrow r = 3 (l a l

What to Do When No Consecutive Terms Are Known

The common ratio can still be found even if the sequence is known to be geometric but no consecutive terms are known. In general, it is enough to know any two terms of a geometric sequence and their positions. For example, suppose that it is known that

r

is positive and only

a_{2}

and

a_{4}

are known.

a_{1} ?, a_{2} 6, a_{3} ?, a_{4} 54, \dots

Recall the general form of a geometric sequence.

Since each term increases by factor of

r,

a_{2}

is equal to

a_{1} r

and

a_{4}

is equal to

a_{1} r^{3} .

a_{4} = a_{1} r^{3} a_{2} = a_{1} r^{3} \Rightarrow 54 = a_{1} r^{3} \Rightarrow 46 = a_{1} r

A single equation in terms of only

r

can be written by dividing the corresponding sides of the resulting equations. This equation can then be solved.

\frac{5 4}{6} = \frac{a _{1} r ^{3}}{a _{1} r}

▼

Simplify

CalcQuot

Calculate quotient

9 = \frac{a _{1} r ^{3}}{a _{1} r}

CancelCommonFac

Cancel out common factors

9 = \frac{a _{1} r ^{3}}{a _{1} r}

SimpQuot

Simplify quotient

9 = \frac{r ^{3}}{r}

$\frac{a ^{m}}{a} = a^{m - 1}$

9 = r^{2}

▼

Solve for

r

SqrtEqn

$LHS = RHS$

3 = r

RearrangeEqn

Rearrange equation

r = 3

Note that when solving for

r,

only the principal root was considered because it was known from the beginning that

r

is positive.

Identify the First Term of the Sequence

To completely define the geometric sequence, the first term should also be determined. Without it, the recursive equation would describe any geometric sequence with common ratio

r = 3 .

From the sequence, it can be seen that

a_{1} = 2 .

a_{1} 2, a_{2} 6, a_{3} 18, a_{4} 54, \dots

What to Do If $a_{1}$ Is Not Known

Reconsider the case where only

a_{2}

and

a_{4}

are known. Two equations were obtained.

5446 = a_{1} r^{3} = a_{1} r

Since it was determined that

r = 3,

this value can be substituted into either equation to find

a_{1} .

This will be illustrated using the latter equation.

6 = a_{1} 3 \Rightarrow a_{1} = 2

Write the Equation Rule

Finally, the recursive rule for the sequence can be written by putting together the previous results and substituting the common ratio and the value of the first term into the general formula.

a_{1}, a_{n} = a_{n - 1} \cdot r ⇓ a_{1} = 2, a_{n} = a_{n - 1} \cdot 3

Example

Using a Recursive Rule for Carbon Dating a Sample

The study of the elements present in the atmosphere makes carbon dating possible. In particular, the ratio

R

carbon - 14

carbon - 12

is of main importance.

\frac{carbon - 1 4}{carbon - 1 2} = R

Since

carbon - 14

is radioactive,

R

reduces by half after a specific period of time called a half life. Because organisms take carbon from the atmosphere while breathing, the same carbon ratio of the atmosphere is present in their bodies while they are alive. However, when they die, they stop inhaling carbon from the atmosphere and the ratio starts to decrease as the

carbon - 14

decays over time.

Geometric sequence: 50, 25,12.5, ... and image of bones buried in the ground

The diagram above shows a sequence representing the percentage of the original ratio of carbon $R$ present in the organism. The term's value $a_{n}$ is the percentage left of the original ratio $R$ in the sample after $n$ half life periods.

a Prove that this sequence is geometric and find a recursive rule for it.

b Use the recursive rule to find the next two terms of the sequence.

c An examined sample has only

6.25 %

of the original carbon ratio from the atmosphere. Use the terms found in the Part B to estimate the age of the sample, considering that the half life period of

carbon - 14

is about

5730

years.

Answer

a Proof: See solution.
Recursive Rule:

a_{1} = 50, a_{n} = a_{n - 1} \cdot \frac{1}{2}

b Terms:

a_{4} = 6.25, a_{5} = 3.125

c Estimated Age:

22920

years

Hint

a The general form for the recursive rule of a geometric sequence is

a_{1}, a_{n} = a_{n - 1} \cdot r

b Use the recursive rule found in Part A. To find a specific term, the previous term needs to be known.

c Note that the positions of the terms represent the number of half life periods that have passed.

Solution

a It is known that the percentage decreases by half from one term to the next. Therefore, the sequence has a common ratio of

r = \frac{1}{2} .

Because it has a common ratio, it is, by definition, a geometric sequence. Recall the form of a general recursive rule for a geometric sequence.

a_{1}, a_{n} = a_{n - 1} \cdot r

Since the common ratio is already known, the first term

a_{1}

must be identified. This can be found in the sequence.

Geometric sequence with first term and common ratio highlighted in different colors

Now the values of the first term and the common ratio will be substituted in the general form to obtain the desired recursive rule.

a_{1}, a_{n} = a_{n - 1} \cdot r ⇓ a_{1} = 50, a_{n} = a_{n - 1} \cdot \frac{1}{2}

b The next two terms can be found by using the recursive formula consecutively. First

a_{4}

will be found by substituting

n = 4

and

a_{3} = 12.5 .

a_{n} = a_{n - 1} \cdot \frac{1}{2}

▼

Substitute values and evaluate

Substitute

$n = 4$

a_{4} = a_{4 - 1} \cdot \frac{1}{2}

SubTerm

Subtract term

a_{4} = a_{3} \cdot \frac{1}{2}

Substitute

$a_{3} = 12.5$

a_{4} = 12.5 \cdot \frac{1}{2}

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

a_{4} = \frac{1 2 . 5}{2}

CalcQuot

Calculate quotient

a_{4} = 6.25

The same process needs to be repeated to find

a_{5} .

The following table shows a summary of these calculations.

$a_{1} = 50, (a_{n} = a_{n - 1} \cdot \frac{1}{2}$
$n$	Substitution	$a_{n}$
$n = 4$	$a_{4} = a_{3} \cdot \frac{1}{2} ⇓ a_{4} = 12.5 \cdot \frac{1}{2}$	$a_{4} = 6.25$
$n = 5$	$a_{5} = a_{4} \cdot \frac{1}{2} ⇓ a_{5} = 6.25 \cdot \frac{1}{2}$	$a_{5} = 3.125$

Now, the next two terms can be added to the sequence.

c The first thing to do is to identify the position of the term

6.25

in the sequence.

As can be seen, the position of the corresponding term is

n = 4 .

This means that in order for the sample to have only

6.25 %

of the original ratio

R

left,

4

half life periods must have passed. The age of the sample will be found by multiplying

4

and the length of the half life period, which is given to be

5730

years.

Age of the Sample 4 \times 5730 = 22920

The estimated age of the sample is

22920

years.

Example

Modeling a Bouncing Ball Using Geometric Sequences

Ali is curious about what would happen if he dropped a ball from the top of the building he lives in, which is about $81$ meters tall. He has noticed that when the ball bounces, it reaches one third the height it was dropped from, so the ball would reach first $\frac{8 1}{3} = 27$ meters high on the first bounce. To calculate the height for the next bounces, he writes a recursive rule.

$Bouncing ball dropped from top of a building and recursive rule a_1=27, a_n=a_{n-1}(1/3)$

a Use the recursive rule to calculate the values of the first three terms.

b Find the corresponding explicit rule and calculate the height of the

5^{t h}

bounce. Give the answer rounded to two decimal places.

c Ali did some research online and found a bouncing ball of a special material that, when dropped from the same height of

81

meters, its bouncing height can be modeled according to the following explicit rule.

a_{n} = 72.9 \cdot 0 . 9^{n - 1}

Find the corresponding recursive rule for this new bouncing ball.

Answer

a Terms:

a_{1} = 27, a_{2} = 9, a_{3} = 3

b Explicit Rule:

a_{n} = 27 \cdot (\frac{1}{3})^{n - 1}

Height of the Fifth Bounce:

0.33

meters.

c Recursive Rule:

a_{1} = 72.9, a_{n} = a_{n - 1} \cdot 0.9

Hint

a Use the given recursive rule repeatedly. Note that to find a specific term, the previous term should be known.

b Recall that the general form for the explicit rule of a geometric sequence is

a_{n} = a_{1} \cdot r^{n - 1} .

c Use the formula for the recursive rule of a geometric sequence.

Solution

a The first three terms can be found by using the recursive rule repeatedly. It is already known that

a_{1} = 27 .

Therefore, the next term

a_{2}

will be found by substituting

n = 2

and

a_{1} = 27 .

a_{n} = a_{n - 1} \cdot \frac{1}{3}

▼

Substitute values and evaluate

Substitute

$n = 2$

a_{2} = a_{2 - 1} \cdot \frac{1}{3}

SubTerm

Subtract term

a_{2} = a_{1} \cdot \frac{1}{3}

Substitute

$a_{1} = 27$

a_{2} = 27 \cdot \frac{1}{3}

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

a_{2} = \frac{2 7}{3}

CalcQuot

Calculate quotient

a_{2} = 9

The third term

a_{3}

can be found by repeating the same process. The following table shows a summary of these calculations.

$a_{n} = a_{n - 1} \cdot \frac{1}{3} ($
$n$	Substitution	$a_{n}$
$n = 1$	$a_{1} = 27$	$a_{1} = 27$
$n = 2$	$a_{2} = a_{1} \cdot \frac{1}{3} ⇓ a_{2} = 27 \cdot \frac{1}{3}$	$a_{2} = 9$
$n = 3$	$a_{3} = a_{2} \cdot \frac{1}{3} ⇓ a_{3} = 9 \cdot \frac{1}{3}$	$a_{3} = 3$

b To find the explicit rule of the sequence, first identify the first term and the common ratio from the known recursive rule. In order to do this, start by reviewing the general form of a recursive rule.

a_{1}, a_{n} = a_{n - 1} \cdot r

Here,

a_{1}

represents the first term of the sequence and

r

is the common ratio. Their values will be identified by comparing this general form and the given recursive rule.

General Recursive Rule a_{1}, a_{n} = a_{n - 1} \cdot r Particular Recursive Rule a_{1} = 27, a_{n} = a_{n - 1} \cdot \frac{1}{3}

It can be concluded that

a_{1} = 27

and

r = \frac{1}{3} .

Now the general form of an explicit rule will be shown and these values will be substituted to find the desired explicit rule for this situation.

General Explicit Rule a_{n} = a_{1} \cdot r^{n - 1} Particular Explicit Rule a_{n} = 27 \cdot (\frac{1}{3})^{n - 1}

Now, to find the fifth term

a_{5},

the explicit rule will be evaluated at

n = 5 .

a_{n} = 27 \cdot (\frac{1}{3})^{n - 1}

▼

Substitute

5

for

n

and evaluate

Substitute

$n = 5$

a_{5} = 27 \cdot (\frac{1}{3})^{5 - 1}

SubTerm

Subtract term

a_{5} = 27 \cdot (\frac{1}{3})^{4}

CalcPow

Calculate power

a_{5} = 27 \cdot \frac{1}{8 1}

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

a_{5} = \frac{2 7}{8 1}

CalcQuot

Calculate quotient

a_{5} = 0.333333 \dots

RoundDec

Round to $2$ decimal place(s)

a_{5} = 0.33

Therefore, after

5

bounces the ball will reach a height of approximately

0.33

meters, or

33

centimeters.

c Similar to the previous part, in order to write a recursive rule, the first term and the common ratio should be identified. To do this, the general form of an explicit rule will be reviewed.

a_{n} = a_{1} \cdot r^{n - 1}

Here,

a_{1}

represents the first term of the sequence and

r

is the common ratio. Now, the general equation will be compared to the specific explicit rule provided.

General Explicit Rule a_{n} = a_{1} \cdot r^{n - 1} Particular Explicit Rule a_{n} = 72.9 \cdot 0.9^{n - 1}

It can be concluded that

a_{1} = 72.9

and

r = 0.9 .

Finally, the general form of a recursive rule will be written and these values will be substituted to find the recursive rule corresponding to the given explicit rule.

General Explicit Rule a_{1}, a_{n} = a_{n - 1} \cdot r Particular Explicit Rule a_{1} = 72.9, a_{n} = a_{n - 1} \cdot 0.9

Discussion

Geometric Sequences and Exponential Functions

Because the values of the terms of a geometric sequence increase by a constant factor, every geometric sequence shows an exponential relation, where the common ratio can be considered as the constant multiplier of the associated exponential function. Consider the following geometric sequence.

Geometric sequence: a_1 =6, a_2=10, a_3=12, a_4=14, a_5=16,...

The sequence can be represented by a table of values, where the independent variable $n$ represents the term's position and the dependent variable $a_{n}$ represents the term's value.

Table of values of the given geometric sequence

In this case, the common ratio is $2 .$ Therefore, all the terms of the sequence lie on the graph of an exponential curve with a constant multiplier of $2 .$

Graph of the sequence terms overlapping the associated exponential function

The relationship of

n

and

a_{n}

satisfies the definition of an exponential function. Therefore, geometric sequences are exponential functions themselves — their domain and range just happen to be discrete. In other words, the variables

n

and

a_{n}

only take specific values. This is why their graphs are series of isolated points and not a solid line.

Closure

Modeling the Thickness of a Sheet of Paper using Geometric sequences

Now that geometric sequences have been introduced and it has been shown how to describe them by using explicit rules, the challenge at the beginning of the lesson will be solved.

a If a regular sheet of paper, with a thickness of

0.1

millimeter, could be folded in half

15

times, how tall would it be? Write the answer in meters rounded to one decimal place. Would it be taller than an average person? For reference, the height of an average person is

1.7

meters.

b If it could be folded in half

20

times, how tall would it be? Write the answer in meters rounded to one decimal place. Would it be taller than a

10

story building? For reference, a

10

story building is about

45

meters tall.

Answer

a Height:

3.3

meters
Taller Than a Person? Yes

b Height:

104.9

meters
Taller Than a $10$ Story Building? Yes

Hint

a Find an explicit rule for the sequence and evaluate it to find the corresponding term value.

b Use the explicit rule from the previous part and evaluate it again. To use the correct

n -

value recall what the term position's represent.

Solution

a If the height of the paper is represented by the term value and the term's position represents the number of folds, the situation can be written as a sequence. Moreover, since the terms of this sequence double their value each time, this is a geometric sequence with a common ratio

r = 2 .

Geometric sequence representing the thickness of the folded sheet of paper

From the sequence, it can be seen that the first term is

a_{1} = 0.2 .

Since both the common ratio and the initial term are already known, they can be substituted into the formula of the explicit rule of a geometric sequence.

a_{n} = a_{1} \cdot r^{n - 1} ⇓ a_{n} = 0.2 \cdot 2^{n - 1}

Now, to find how tall the folded paper sheet would be after being folded

15

times, the explicit rule will be evaluated for

n = 15 .

a_{n} = 0.2 \cdot 2^{n - 1}

▼

Substitute

15

for

n

and evaluate

Substitute

$n = 15$

a_{15} = 0.2 \cdot 2^{15 - 1}

SubTerm

Subtract term

a_{15} = 0.2 \cdot 2^{14}

CalcPow

Calculate power

a_{15} = 0.2 \cdot 16384

Multiply

a_{15} = 3276.8

The folded sheet of paper would be

3276.8

millimeters tall. Since a meter has

1000

millimeters, the height in meters will be obtained by dividing it by

1000 .

\frac{3 2 7 6 . 8}{1 0 0 0} = 3.2768

This rounds to about

3.3

meters. Therefore, the folded paper would be definitely taller than an average person.

b To find the how tall the folded paper sheet would be after being folded

20

times, the explicit rule will now be evaluate for

n = 20 .

a_{n} = 0.2 \cdot 2^{n - 1}

▼

Substitute

20

for

n

and evaluate

Substitute

$n = 20$

a_{15} = 0.2 \cdot 2^{20 - 1}

SubTerm

Subtract term

a_{15} = 0.2 \cdot 2^{19}

CalcPow

Calculate power

a_{15} = 0.2 \cdot 524288

Multiply

a_{15} = 104857.6

The folded sheet of paper would be

104857.6

millimeters tall. Now this quantity will be divided by

1000

to convert it to meters.

\frac{1 0 4 8 5 7 . 6}{1 0 0 0} = 104.8576

This rounds to about

104.9

meters. Therefore, it would be definitely taller than a

10

story building.

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Catch-Up and Review

Proof

Answer

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

What to Do When No Consecutive Terms Are Known

What to Do If a1​ Is Not Known

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

What to Do If $a_{1}$ Is Not Known