Topics
6. Linear and Exponential Models
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United States of America
Grades 9-12

6.04 Connecting sequences and functions

Lesson
Practice

Introduction

So far, we have discussed how to write sequences using a recursive notation, but it is not very convenient to use that notation if we need to find a high-numbered term like the 100th term. We would have to find the previous 99 terms before knowing what the 100th term is.

In this lesson, we are going to derive a different notation, one that will make it easier to find any term in an arithmetic or geometric sequence. Then, we will explore the relationship between sequences and functions.

Explicit arithmetic sequences

Exploration

Create an arithmetic sequence and state the first term and the common difference.

  1. How many times would you need to add the common difference to find the 3rd term?
  2. How many times would you need to add the common difference to find the 10th term?
  3. How many times would you need to add the common difference to find the 100th term?
  4. What is the relationship between the number of the term and the number of times you need to add the common difference?
  5. Can you think of an equation that could help you find any term when given the first term and common difference?

When we know the first term of an arithmetic sequence, we repeatedly add the common difference to find the other terms in the sequence. If the number of the term we are looking for is n, then the number of times we need to add the common difference is \left(n-1\right). Since repeated addition is multiplication, we can use this relationship to create a general formula.

The nth term, a_n, of an arithmetic sequence is given by the explicit rule or general formula:

\displaystyle a_n=a_1 + d\left(n-1\right)
\bm{a_n}
current term
\bm{a_1}
first term
\bm{d}
common difference
\bm{n}
current term number

If we know the explicit formula, we can easily convert it to the recursive formula. The only things we need to know are the common difference and the first term, which are already given in the explicit formula.

\displaystyle a_n=a_{n-1}+d, \text{ } a_1=c
\bm{a_n}
current term
\bm{a_{n-1}}
previous term
\bm{d}
common difference
\bm{a_1}
first term
\bm{c}
value of the first term

Examples

Example 1

The points on this graph represent an arithmetic sequence:

1
2
3
4
n
-8
-6
-4
-2
2
4
a_n
a

Complete the following table of values:

n1234
a_n
Worked Solution
Create a strategy

The second row of the table represents the values of a_n, so we are looking for the output values of the points.

Apply the idea
n1234
a_n3-1-5-9
Reflect and check

These values of a_n represent the terms in the arithmetic sequence.

b

Write the explicit rule that represents the sequence.

Worked Solution
Create a strategy

The explicit rule of an arithmetic sequence is a_n=a_1+d\left(n-1\right). We need to find the common difference and the first term and then substitute them into the formula.

Apply the idea

To find the common difference, we can subtract any term from the following term:

d=-5-\left(-1\right)=-4

From the table or the graph, we see the first term of the sequence is a_1=3.

The explicit rule for finding any term in this sequence is a_n=3-4\left(n-1\right).

Reflect and check

We can simplify this formula if we want:

\displaystyle a_n\displaystyle =\displaystyle 3-4\left(n-1\right)Given equation
\displaystyle =\displaystyle 3-4n+4Distributive property
\displaystyle =\displaystyle -4n+7Evaluate the addition
c

Find the 15th term of the sequence.

Worked Solution
Create a strategy

Since we are looking for the 15th term, n=15.

Apply the idea
\displaystyle a_n\displaystyle =\displaystyle 3-4\left(n-1\right)Explicit formula
\displaystyle a_{15}\displaystyle =\displaystyle 3-4\left(15-1\right)Substitute n=15
\displaystyle =\displaystyle 3-4\left(14\right)Evaluate the subtraction
\displaystyle =\displaystyle 3-56Evaluate the multiplication
\displaystyle =\displaystyle -53Evaluate the subtraction

The 15th term is -53.

Reflect and check

Using the simplified form of the equation instead:

\displaystyle a_n\displaystyle =\displaystyle -4n+7Simplified explicit formula
\displaystyle a_{15}\displaystyle =\displaystyle -4\left(15\right)+7Substitute n=15
\displaystyle =\displaystyle -60+7Evaluate the multiplication
\displaystyle =\displaystyle -53Evaluate the addition

Example 2

An arithmetic sequence is defined by the rule T_n=T_{n-1}+1.95 where T_1=1.5.

a

Write the explicit rule of this sequence.

Worked Solution
Create a strategy

When given a recursive rule, we are given the common difference and the first term. We can take these from the recursive notation and substitute them into the explicit formula.

Apply the idea

In recursive notation, T_n=T_{n-1}+d, the common difference is the number being added. For this rule, d=1.95.

The explicit rule is T_n=1.5+1.95(n-1).

b

Find T_{10}.

Worked Solution
Create a strategy

It is much quicker to find T_{10} using the explicit rule. If we used the recursive rule, we would need to find all the terms between T_1 and T_{10}.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle 1.5+1.95\left(n-1\right)Explicit formula from part (a)
\displaystyle T_{10}\displaystyle =\displaystyle 1.5+1.95\left(10-1\right)Substitute n=10
\displaystyle T_{10}\displaystyle =\displaystyle 1.5+1.95\left(9\right)Evaluate the subtraction
\displaystyle T_{10}\displaystyle =\displaystyle 1.5+17.55Evaluate the multiplication
\displaystyle T_{10}\displaystyle =\displaystyle 19.05Evaluate the addition

Example 3

In an arithmetic sequence, a_7=43 and a_{14}=85.

a

Find the common difference.

Worked Solution
Create a strategy

One way to solve this is by substituting the known information about each term into the explicit formula. Since there are 2 unknowns, a_1 and d, this will create a system of equations that we can solve.

Apply the idea

From the 7th term, we know a_7=43, n=7. Substituting this into the explicit formula, we have 43=a_1+d\left(7-1\right).

From the 14th term, we know a_{14}=85, n=14. Substituting this into the explicit formula, we have 85=a_1+d\left(14-1\right).

After simplifying, the system of equations is:

\begin{aligned}43&=a_1+6d\\85&=a_1+13d\end{aligned}

Subtracting the second equation from the first one, we have:

\begin{aligned}43&=a_1+6d\\-(85&=a_1+13d) \\ \hline -42&=-7d \\ 6&=d \\ d&=6 \end{aligned}

Reflect and check

Another way to solve this is by determining how many times we need to add the common difference to the 7th term to get the 14th term.

We would need to add the common difference 7 times. Using this, we can set up an equation.

\displaystyle 43+d+d+d+d+d+d+d\displaystyle =\displaystyle 85
\displaystyle 43+7d\displaystyle =\displaystyle 85
\displaystyle 7d\displaystyle =\displaystyle 42
\displaystyle d\displaystyle =\displaystyle 6
b

Find the first term.

Worked Solution
Create a strategy

We can use either of the equations from the system we set up before because the answer for the first term will be the same.

Apply the idea

For this problem, we will use the equation for the 7th term and substitute d=6.

\displaystyle 43\displaystyle =\displaystyle a_1+6\left(7-1\right)Equation for 7th term
\displaystyle 43\displaystyle =\displaystyle a_1+36Evaluate the subtraction and multiplication
\displaystyle 7\displaystyle =\displaystyle a_1Subtract 36 from both sides

The first term of this sequence is a_1=7.

Reflect and check

Another method we could use is to determine how many times we would need to subtract the common difference to get from the 7th term to the 1st term.

We would need to subtract the common difference 6 times.

\displaystyle a_1\displaystyle =\displaystyle 43-d-d-d-d-d-d
\displaystyle =\displaystyle 43-6d
\displaystyle =\displaystyle 43-6\left(6\right)
\displaystyle =\displaystyle 43-36
\displaystyle =\displaystyle 7
c

Write the explicit rule for this sequence.

Worked Solution
Create a strategy

Now that we know the first term and the common difference, we can substitute them into the explicit formula.

Apply the idea

a_n=7+6\left(n-1\right)

Reflect and check

We can check the rule by finding the 7th and 14th terms.

a_7=7+6\left(7-1\right)=43

a_{14}=7+6\left(14-1\right)=85

These are the same as the given information, so our rule is correct.

Idea summary

The nth term, a_n, of an arithmetic sequence is given by the explicit formula:

\displaystyle a_n=a_1 + d\left(n-1\right)
\bm{a_n}
current term
\bm{a_1}
first term
\bm{d}
common difference
\bm{n}
current term number

Explicit geometric sequences

Exploration

Think of a geometric sequence and state the first term and the common ratio.

  1. How many times would you need to multiply by the common ratio to find the 5th term?
  2. How many times would you need to multiply by the common ratio to find the 50th term?
  3. What is the relationship between the number of the term and the number of times you need to multiply by the common ratio?
  4. Can you think of an equation that could help you find any term when given the first term and common ratio?

When we know the first term of a geometric sequence, we repeatedly multiply by the common ratio to find the other terms in the sequence. If the number of the term we are looking for is n, then the number of times we need to multiply by the common ratio is \left(n-1\right). Since repeated multiplication is denoted by exponents, we can use this relationship to create a general formula.

The nth term, a_n, of a geometric sequence is given by the explicit rule or general formula:

\displaystyle a_n=a_1 \left(r\right)^{n-1}
\bm{a_n}
current term
\bm{a_1}
first term
\bm{r}
common ratio
\bm{n}
current term number

Similar to arithmetic sequences, we can easily convert an explicit rule for a geometric sequence to its recursive rule. We simply need to know the common ratio and the first term which are already given in the explicit formula.

\displaystyle a_n=r\cdot a_{n-1}, \text{ } a_1=c
\bm{a_n}
current term
\bm{a_{n-1}}
previous term
\bm{r}
common ratio
\bm{a_1}
first term
\bm{c}
value of the first term

Examples

Example 4

Consider the following geometric sequence:

n1234
t_n-93.6-1.440.576

Write the explicit rule for this sequence.

Worked Solution
Create a strategy

From the table, we see the first term is t_1=-9. To find the common ratio, we need to divide any term by the previous term.

Apply the idea

r=\dfrac{3.6}{-9}=-0.4

Substituting t_1 and r into the explicit formula for geometric sequences, we get:

t_n=-9\left(-0.4\right)^{n-1}

Reflect and check

Let's verify the rule by using it to determine t_4.

\displaystyle t_4\displaystyle =\displaystyle -9(-0.4)^{4-1}Substitute n=4
\displaystyle =\displaystyle -9(-0.4)^3Evaluate the subtraction
\displaystyle =\displaystyle -9(-0.064)Evaluate the exponent
\displaystyle =\displaystyle 0.576Evaluate the multiplication

Example 5

A geometric sequence is defined by T_n=5\cdot T_{n-1} where T_1=-1.

a

Write the explicit rule for this geometric sequence.

Worked Solution
Create a strategy

In a recursive formula, the number being multiplied to the previous term is r, and the first term is given. Once we identify these values from the recursive rule, we can substitute them into the explicit formula.

Apply the idea

r=5, \, T_1=-1

The explicit rule for this sequence is T_n=-1(5)^{n-1}.

b

Find the 6th term.

Worked Solution
Create a strategy

We will use the explicit formula to find T_6 because it will take less work and less time than it would if we used the recursive formula.

Apply the idea
\displaystyle T_6\displaystyle =\displaystyle -1(5)^{6-1}Substitute n=6
\displaystyle =\displaystyle -1(5)^5Evaluate the subtraction
\displaystyle =\displaystyle -3125Evaluate the exponent and multiplication

Example 6

In a geometric sequence, a_2=-\dfrac{3}{5} and a_5=\dfrac{3}{625}.

a

Find the common ratio.

Worked Solution
Create a strategy

One way to solve this is by determining how many times we need to multiply by the common ratio to get from the 2nd term to the 5th term. From there, we can set up an equation and solve for r.

Apply the idea

To get from the 2nd term to the 5th term, we would need to multiply by the common ratio 3 times.

\displaystyle -\dfrac{3}{5}\cdot r \cdot r \cdot r\displaystyle =\displaystyle \dfrac{3}{625}Given equation
\displaystyle -\dfrac{3}{5} r^3\displaystyle =\displaystyle \dfrac{3}{625}Rewrite using exponents
\displaystyle r^3\displaystyle =\displaystyle -\dfrac{1}{125}Divide both sides by -\dfrac{3}{5}
\displaystyle r\displaystyle =\displaystyle -\dfrac{1}{5}Cube root property
Reflect and check

Another way to solve this is by creating a system of equations using the given information about each term.

From the second term, we know a_2=-\dfrac{3}{5}, n=2. Substituting this information into the explicit formula, we get:

-\dfrac{3}{5}=a_1(r)^{1}

From the fifth term, we know a_5=\dfrac{3}{625}, n=5. Substituting this into the explicit formula, we get:

\dfrac{3}{625}=a_1(r)^{4}

Next, we will solve the first equation for a_1.

a_1=\dfrac{-3}{5r}

Then, substitute this into the second equation and solve.

\displaystyle \dfrac{3}{625}\displaystyle =\displaystyle -\dfrac{3}{5r}\cdot r^4Substitute a_1=\dfrac{-3}{5r}
\displaystyle \dfrac{3}{625}\displaystyle =\displaystyle -\dfrac{3}{5}r^3Quotient of powers property
\displaystyle -\dfrac{1}{125}\displaystyle =\displaystyle r^3Divide both sides by -\dfrac{3}{5}
\displaystyle -\dfrac{1}{5}\displaystyle =\displaystyle rCube root property
b

Find the first term.

Worked Solution
Create a strategy

We can use the explicit formula with the information given about one of the terms to solve for the common ratio. For this problem, we will use a_2=-\dfrac{3}{5}, \, n=2, \, r=-\dfrac{1}{5}.

Apply the idea
\displaystyle a_n\displaystyle =\displaystyle a_1\left(-\dfrac{1}{5}\right)^{n-1}Explicit formula
\displaystyle -\dfrac{3}{5}\displaystyle =\displaystyle a_1\left(-\dfrac{1}{5}\right)^{2-1}Substitute a_2=-\dfrac{3}{5}, n=2, and r=-\dfrac{1}{5}
\displaystyle -\dfrac{3}{5}\displaystyle =\displaystyle -\dfrac{1}{5}a_1Evaluate the subtraction
\displaystyle 3\displaystyle =\displaystyle a_1Multiply both sides by -5

The first term is 3.

Reflect and check

Since we knew the second term and the common ratio, we could have divided the second term by the common ratio to find the first term.

\displaystyle a_1\displaystyle =\displaystyle -\dfrac{3}{5}\div r
\displaystyle =\displaystyle -\dfrac{3}{5}\div -\dfrac{1}{5}
\displaystyle =\displaystyle -\dfrac{3}{5}\times -\dfrac{5}{1}
\displaystyle =\displaystyle 3
c

Write the explicit rule for the sequence.

Worked Solution
Create a strategy

Now that we know the first term and the common ratio, we can substitute them into the explicit formula for geometric sequences.

Apply the idea

a_n=3\left(-\dfrac{1}{5}\right)^{n-1}

Reflect and check

We can verify our rule by using it to confirm the 2nd and 5th terms.

a_2=3\left(-\dfrac{1}{5}\right)^{2-1}=-\dfrac{3}{5}

\begin{aligned} a_5&=3\left(-\dfrac{1}{5}\right)^{5-1} \\ &=3\left(-\dfrac{1}{5}\right)^4\\ &=3\left(\dfrac{1}{625}\right) \\ &=\dfrac{3}{625} \end{aligned}

Idea summary

The nth term, a_n, of a geometric sequence is given by the explicit formula:

\displaystyle a_n=a_1 \left(r\right)^{n-1}
\bm{a_n}
current term
\bm{a_1}
first term
\bm{r}
common ratio
\bm{n}
current term number

Connecting sequences and functions

Sequences are functions whose domain is a subset of the integers. Arithmetic sequences have a linear relationship because the terms share a common difference.

When we represent an arithmetic sequence as a linear function whose domain is a subset of the integers, we generally use function notation and then simplify:

\displaystyle a(n)=a(1)+d(n-1)
\bm{a(n)}
value of the nth term
\bm{a(1)}
first term
\bm{n}
term number
\bm{d}
common difference (the slope of the line)

After simplifying, the equation will be in the slope-intercept form of a linear function. The common difference is the slope of the line, or the rate of change. Recall that linear functions have a domain of all real numbers, but the domain of a sequence is a subset of the integers, beginning from 0 or 1, depending on how it is defined.

-4
-3
-2
-1
1
2
3
4
x
-8
-6
-4
-2
2
4
6
8
y
a(n)=3+2(n-1)
-4
-3
-2
-1
1
2
3
4
x
-8
-6
-4
-2
2
4
6
8
y
f(x)=2x+1

Geometric sequences have an exponential relationship because the terms share a common ratio. When we represent a geometric sequence as an exponential function whose domain is a subset of the integers, we use function notation and simplify:

\displaystyle a(n)=a(1)\cdot r^{n-1}
\bm{a(n)}
value of the nth term
\bm{a(1)}
first term
\bm{n}
term number
\bm{r}
common ratio, the constant factor

Geometric sequences with a common ratio greater than 1 can model exponential growth.

1
2
3
4
5
6
x
4
8
12
16
20
24
28
32
36
y
a(n)=\left(2\right)^{n-1}
1
2
3
4
5
6
x
4
8
12
16
20
24
28
32
36
y
f(x)=\dfrac{1}{2}\left(2\right)^x

Geometric sequences with a common ratio between 0 and 1, non-inclusive, can model exponential decay.

1
2
3
4
5
6
x
4
8
12
16
20
24
28
32
36
y
a(n)=6\left(\dfrac{1}{2}\right)^{n-1}
1
2
3
4
5
6
x
4
8
12
16
20
24
28
32
36
y
f(x)=64\left(\dfrac{1}{2}\right)^x

Geometric sequences with a negative common ratio cannot be used to model exponential growth or decay, as we can see in this graph:

1
2
3
4
5
6
x
-36
-32
-28
-24
-20
-16
-12
-8
-4
4
8
12
16
y

This is because the base of an exponential function is restricted. The base, or constant factor, must be greater than zero but not equal to one.

Geometric sequences can have a negative common ratio, but exponential functions cannot have a negative constant factor.

Explicit rule for this geometric sequence:

a_n=(-2)^{n-1}

Examples

Example 7

Tiles were stacked in a pattern as shown:

An image showing tiles stacked in a pattern with increasing stack height and corresponding number of tiles. Stack 1 has 1 tile. Stack 2 has 3 tiles. Stack 3 has 5 tiles. And stack 4 has 7 tiles
a

Describe the recursive pattern and write the explicit equation for the sequence using function notation.

Worked Solution
Create a strategy

The recursive pattern will describe the change in the number of tiles going from each stack to the next.

The explicit equation is a function that can be used to find the number of tiles for a stack of any height.

Apply the idea

The number of tiles from one stack to the next increases by 2. This tells us that the common difference is +2. We can also see that the first stack has one tile. So, the explicit equation for this pattern is a\left(n\right)=1+2\left(n-1\right)where a\left(n\right) is the number of tiles and n is the height of the stack.

Reflect and check

We can verify our rule by using it to find the number of tiles in the 4th stack.

a(4)=1+2(4-1)=7

In the image, there are 7 tiles in the 4th stack, so our rule is accurate.

b

A table of values representing the relationship between the height of the stack and the number of tiles was partially completed.

Height of stack1234510100
Number of tiles13

Complete the table of values representing the relationship between the height of the stack and the number of tiles.

Worked Solution
Create a strategy

The number of tiles in the 3rd and 4th stacks are shown, and we can use the explicit function we created in part (b) to find the number of tiles in the 5th, 10th, and 100th stacks.

Apply the idea

a(5)=1+2(5-1)=9

a(10)=1+2(10-1)=19

a(100)=1+2(100-1)=199

Height of stack1234510100
Number of tiles1357919199
Reflect and check

We can double-check that the pattern found in part (a) is correct by comparing some of the results in the table to the images of the sequence of stacks.

c

State the domain.

Worked Solution
Create a strategy

Since we are using function notation for our sequence, it is important to consider the domain. For this function, the domain represents the height of the stacks as defined in part (c). The height of the stacks begins at 1 and increases by 1 each time.

Apply the idea

The domain is the set of natural numbers.

Reflect and check

The domain of a sequence will always be a subset of the integers, and natural numbers are a subset of the integers.

Example 8

Ethan is playing a new game on his phone. After successfully playing his first game on day 1, he was awarded 25 diamonds. The game then rewards him with 3 diamonds for each consecutive day he plays after day 1.

a

Determine if the number of diamonds he has after n days of consecutive play is linear or exponential.

Worked Solution
Create a strategy

To identify whether the given problem is linear or exponential, we need to check if the situation represents an arithmetic or geometric sequence. To do this, we should determine if there is a common difference or a common ratio.

Apply the idea

The number of diamonds increases by 3 each day, so there is a common difference of d=3. Therefore, the problem involves linear growth.

b

Determine the number of diamonds Ethan will have after playing 6 consecutive days.

Worked Solution
Create a strategy

We can use the formula for the nth term of an arithmetic sequence, {a\left(n\right)=a\left(1\right) + d\left(n-1\right)} to write an explicit rule in function notation that models this scenario.

Apply the idea

First, we must find an explicit rule for the number of diamonds Ethan gets each day on his game:

\displaystyle a\left(n\right)\displaystyle =\displaystyle a\left(1\right) + d\left(n-1\right)General rule for an arithmetic sequence
\displaystyle a\left(n\right)\displaystyle =\displaystyle 25 + 3\left(n-1\right)Substitute a\left(1\right) = 25 and d=3
\displaystyle a\left(n\right)\displaystyle =\displaystyle 25+3n-3Distributive property
\displaystyle a\left(n\right)\displaystyle =\displaystyle 3n+22Combine like terms

Now we can use our explicit rule to find a\left(6\right).

\displaystyle a\left(n\right)\displaystyle =\displaystyle 3n+22Explicit rule
\displaystyle a\left(6\right)\displaystyle =\displaystyle 3\left(6\right)+22Substitute n=6
\displaystyle a\left(6\right)\displaystyle =\displaystyle 40Evaluate the multiplication and addition

Ethan will have 40 diamonds on the 6th day.

Reflect and check

The domain for this sequence, which represents the number of days Ethan plays the game, is the set of natural numbers.

c

Find the number of consecutive days Ethan will need to play to earn 160 diamonds.

Worked Solution
Create a strategy

We can use the rule from part (b), a\left(n\right)=3n+22, to find the number of days, n, it will take to get a\left(n\right)=160 diamonds.

Apply the idea
\displaystyle a(n)\displaystyle =\displaystyle 3n+22Explicit rule
\displaystyle 160\displaystyle =\displaystyle 3n+22Substitute a(n)=160
\displaystyle 138\displaystyle =\displaystyle 3nSubtract 22 from both sides
\displaystyle 46\displaystyle =\displaystyle nDivide both sides by 3
\displaystyle n\displaystyle =\displaystyle 46Symmetric property of equality

It will take Ethan 46 days to earn 160 diamonds.

Reflect and check

Alternatively, we can find the number of additional days after day 1 in a more conceptual way without using the equation:

\displaystyle 160-25\displaystyle =\displaystyle 135Number of diamonds Ethan needs to earn after day 1
\displaystyle 135 \div 3\displaystyle =\displaystyle 45Number of days Ethan needs to play after day 1
\displaystyle 45+1\displaystyle =\displaystyle 46Total number of days Ethan needs to play

Example 9

A ball is dropped onto the ground from a height of 8 \text{ m}. On each bounce, the ball reaches a maximum height of 60\% of its previous maximum height.

a

Determine if the heights of each bounce can be represented linearly or exponentially.

Worked Solution
Create a strategy

To identify whether the given problem is linear or exponential, we should determine if the situation represents an arithmetic or geometric sequence. To do this, we should determine if there is a common difference or a common ratio.

Apply the idea

In the problem, the height of the previous bounce is multiplied by 60\% or 0.6 to get the next height. So, the common ratio is r=0.6, and the sequence is geometric.

This situation can be modeled exponentially.

b

Determine the height that the ball reaches after the 4th bounce.

Worked Solution
Create a strategy

In part (a), we determined that the problem involves a geometric sequence with a common ratio of r=0.6. We also know that the ball had an initial height of 8\text{ m}, so a\left(0\right)=8. We are using a(0) since the initial height occurred before the first bounce, and n represents the number of the bounce.

To find a\left(1\right), we can multiply a\left(0\right) by the common ratio.

Apply the idea
\displaystyle a\left(1\right)\displaystyle =\displaystyle a\left(0\right)\cdot rRecursive rule
\displaystyle a\left(1\right)\displaystyle =\displaystyle 8 \cdot 0.8Substitute a\left(0\right)=8 and r=0.6
\displaystyle =\displaystyle 4.8Evaluate the multiplication

Now that we know the first term, we can use the explicit rule to find the height of the 4th bounce, where n=4:

\displaystyle a\left(n\right)\displaystyle =\displaystyle 4.8 \cdot 0.6^{n-1}Explicit rule
\displaystyle a\left(4\right)\displaystyle =\displaystyle 4.8 \cdot 0.6^{4-1}Substitute n=4
\displaystyle a\left(4\right)\displaystyle =\displaystyle 1.0368Evaluate the exponent and multiplication

The height of the 4th bounce will be 1.0368\text{ m}.

Reflect and check

To use the initial term instead of finding the first term, we can adjust the explicit rule to be:

a\left(n\right)=a\left(0\right)r^n

a\left(4\right)=8\left(0.6\right)^4=1.0368

Idea summary

Sequences are functions whose domain is a subset of the integers.

Arithmetic sequences can be written in function notation and simplified:

\displaystyle a\left(n\right)=a\left(1\right)+d\left(n-1\right)
\bm{a\left(n\right)}
value of the nth term
\bm{a\left(1\right)}
first term
\bm{n}
term number
\bm{d}
common difference, the slope of the line

Geometric sequences can be written in function notation and simplified:

\displaystyle a\left(n\right)=a\left(1\right)\cdot r^{n-1}
\bm{a\left(n\right)}
value of the nth term
\bm{a\left(1\right)}
first term
\bm{n}
term number
\bm{r}
common ratio, the constant factor

Outcomes

F.IF.A.3

Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers.

F.BF.A.1

Write a function that describes a relationship between two quantities.

F.BF.A.1.A

Determine an explicit expression, a recursive process, or steps for calculation from a context.

F.BF.A.2

Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms.

F.LE.A.1

Distinguish between situations that can be modeled with linear functions and with exponential functions.

F.LE.A.2

Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).

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