Unit 2: Lesson 3: Modeling Arithmetic Sequences with Functions
This page contains just “Launch, Teaching Time, and Collaboration & Practice” sections of the lesson. For the
complete lesson, refer back to the main webpage: https://www.relevantmathematics.com/?p=11723
Lesson-Framing Questions:



How do you write a recursive function to model an arithmetic sequence?
How do you write an explicit function to model an arithmetic sequence?
How is a linear function similar to an arithmetic sequence?
Lesson Launch:
Suppose you had $680 in your bank account at the beginning of the first month. You have to pay a monthly cell
phone bill of $40 from your bank account at the end of each month.
a. Write a rule where knowing the previous month’s balance could tell you the current month’s balance.
b. Write a function where the input is the month and the output is that month’s balance.
c. How many months will pass until you no longer have enough money in your account?
a. Write a rule where knowing the previous month’s balance could tell you the current month’s balance.
Previous month – 40 = current month
b. Write a function where the input is the month and the output is that month’s balance.
Let m = number of months
Let f(m) = account balance
f(m) = 680 – 40m
c. How many months will pass until you no longer have enough money in your account?
17 months pass and on the 18th month, you can’t pay your bill.
Teaching Time:
In your middle grades mathematics, you often examined patterns of numbers and
created mathematical rules to model those patterns.
Consider this pattern of numbers:
4, 7, 10, 13, 16, …
What are the next three numbers? How did you determine those next three
numbers?
19, 22, 25
We added 3 to each number to get the next number in the pattern.
In high school mathematics, you will study two types of patterns of numbers called
sequences.
A sequence, is a set of numbers, called terms, that has a specific pattern to derive
each term from previous terms.
The two types of sequences you will study are arithmetic sequences and
geometric sequences.
Arithmetic Sequence: a sequence of numbers such that each term is given by a
common difference, d, of the previous term.
Ex: 2, 4, 6, 8, 10, …
common difference, d, is + 2
Ex: 100, 90, 80, 70, 60, …
common difference, d, is -10
Geometric Sequence: is a sequence of numbers such that each term is given by a
common multiple, r, of the previous term.
Ex: 4, 16, 64, 256, …
common multiple, r, is × 4
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Teacher Notes:
Say
Say and write on board
Solutions in red
Say. It is already in
student notes.
Say. Already in student
notes. Students fill in
the red information.
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1 1 1
Ex: 2, 1, 2 , 4 , 8 , …
1
common multiple, r, is × 2
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Why do we need this?
Both sequences are easily recognized and will allow you to develop a deeper understanding of how to model
sequences with functions before you study more sophisticated sequences in later mathematics such as the famous
Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … which creates the golden ratio or golden number (the ratio
between the pairs of numbers approaches 1.618034)
For fun: How is the sequence made?
By adding the previous two terms together
The Fibonacci sequence and the golden ratio can be found all over the natural world.
From shells
And even to the placement of advertisements on
products.
To the ratio of bones in your fingers and arms
Source: http://www.goldennumber.net/human-handfoot/
So, let’s get ready to work with the Arithmetic and Geometric Sequences in preparation for more advanced
sequences in later mathematics.
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Example 1: Consider the sequence below
4, 7, 10, 13, 16, ...
In your middle grades mathematics, you learned to represent this pattern using
simple rules such as
“the next term is equal to the current term plus three”
Or “the current term is equal to the previous term plus three”
Using those simple NEXT/NOW representations of this sequence, we could write
an informal rule such as
NEXT = NOW + 3
or
NOW = PREVIOUS + 3
You could also think of sequences as functions. In order to show how a sequence
can be considered a function, we need an index that indicates which term of the
sequence we are talking about, and which serves as an input value of the function.
Deciding that the 4 corresponds to an index value of 1, we make a table showing
the correspondence.
Let n = the number of the term
Let f(n) = the value of term n.
n f(n)
1 4
2 7
3 10
4 13
5 16
Now that we understand functions and function notation, we can rewrite those
informal rules for the sequence as a function using function notation.
f(n+1) = f(n) + 3
or
f(n) = f(n-1) + 3
Teacher Notes:
Say. Already in student
notes.
Students fill in red.
However, these functions are not useable unless you declare the first term, n = 1,
so the function rule can be used. You also need to let people know that you can
only use integers n ≥ 2 since you don’t know how to find the 1.5 term or the 2.356
term.
f(1) = 4
State the first term, n = 1
f(n+1) = f(n) + 3 State the rule for finding each successive term
for integers n ≥ 2 State the domain of the function.
Or, changing the NOW = PREVIOUS + 3 rule,
f(1) = 4
f(n) = f(n-1) + 3 for integers n ≥ 2
Example 2:
Consider the sequence 10, 6, 2, -2, -6, …
a. Write a recursive rule to show that this sequence is a function.
b. Use your rule to find the next three terms in the sequence. Show all your
calculations
Solution:
a.
Let n = the number of the term
Let f(n) = the value of term n.
n f(n)
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1
2
3
4
5
10
6
2
-2
-6
https://www.desmos.com/calculator/lrrji9lyvh
-you will have to remind students that you can’t “connect the dots” because the
function does not define what happens when n = 1.5 or n = 2.3633… since the
domain of the function is restricted to integers n ≥ 2
b.
f(1) = 10
f(n+1) = f(n) – 4 for integers n ≥ 2
or
f(1) = 10
f(n) = f(n-1) – 4 for integers n ≥ 2
c.
To find the 6th, 7th, and 8th terms of the sequence, use the recursive rule:
6th term:
f(n+1) = f(n) – 4
f(5+1) = f(5) – 4
f(6) = f(5) – 4
f(6) = -6 – 4
f(6) = -10
or
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f(n) = f(n-1) – 4
f(6) = f(6-1) – 4
f(6) = f(5) – 4
f(6) = -6 – 4
f(6) = -10
f(7) = -14
f(8) = -18
Ex. 3:
A theater needs to place seats in rows. The function, f(r), as shown below, can be
used to determine the number of seats in each row, where r is the row number.
f(1) = 8
f(r) = f(r – 1) + 3
Use the function to complete the table indicating the number of seats in each row
for the first four rows of the theater.
Row Number Number of Seats
1
8
2
11
3
14
4
17
Ex. 4:
Consider the following sequence:
13, 8, 3, -2, …
a. Write a recursive function to model this sequence.
b. Use your recursive function to find the 6th term.
Solution:
a.
Let n = term number
Let f(n) = value of term n
n f(n)
1 13
2 8
3 3
4 -2
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Have students work this
one.
Have students work this
one
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https://www.desmos.com/calculator/p5da9eggdk
b.
f(1) = 13
f(n+1) = f(n) – 5 for integers n ≥ 2
or f(n) = f(n-1) – 5 for integers n ≥ 2
c.
f(5+1) = f(5) – 5
f(6) = f(5) – 5
f(6) = -7 - 5
f(6) = -12
or
f(6) = f(6-1) – 5
f(6) = f(5) – 5
f(6) = -7 – 5
Ex. 5:
Albert deposits $320 in a savings account at the beginning of the year. He deposits
an additional $60 every month. Write a recursive expression, where f(1) = 320 and
n ≥ 2 that can be used to represent the amount, f(n) in Albert’s account after n
months.
Have students work this
one.
Answer:
f(1) = 320
f(n+1) = f(n) + 60 for n ≥ 2
Or
f(n) = f(n-1) + 60 for n ≥ 2
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Generalizing our work:
Recursive Rule: Uses the previous term of a function (or sequence) to define the
next term.
Recursive rule for an arithmetic sequence where p is the value of the first term and
n is the term number.:
f(1) = p
f(n + 1) = f(n) + d for integers n ≥ 2
or
f(1) = p
f(n) = f(n - 1) + d for integers n ≥ 2
What if we wanted a term that was far into the sequence like the 100th or 1000th
term?
Using our sequence from Example 1:
4, 7, 10, 13, 16, …
Write a function rule that uses the term number as the input and gives the value of
the term as the output to find the 100th term.
If we used a recursive rule, we would need to know the 99th term, but, in order to
find the 99th term, we need the 98th term and the 97th term and so on. It would take
a long time to find all those terms.
Let n = the number of the term
Let f(n) = the value of term n.
n f(n)
1 4
2 7
3 10
4 13
5 16
f(n) = 3n + 1
Explicit Rule: Defines each term as a function based on the number of the term.
Explicit rule for an arithmetic sequence:
f(n + 1) = d(n) + f(1), for integers n ≥ 1
or
f(n) = d(n-1) + f(1), for integers n ≥ 1
Ex. 6:
Consider a sequence whose first five terms are
6, 12, 18, 24, 30
Write an explicit function that can be used to find any term in the sequence
without knowing the previous term.
Solution: f(n) = 6n
Ex. 7:
A sequence is shown below.
5, 15, 25, 35, 45, …
Which explicit expression can be used to determine the value of the nth number in
the sequence?
Have students
generalize their work to
arrive at this
themselves. You may
need to prompt them to
use function notation
properly.
Have students wrestle
with this one.
In student notes.
Have students wrestle
with this one.
Have students wrestle
with this one.
Solution: f(n+1) = 10n + 5 for integers n ≥ 2 or f(n) = 10(n-1) + 5
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Arithmetic Sequences and Linear Functions
Let’s go back to our first example of an arithmetic sequence:
4, 7, 10, 13, 16, …
Let them work through
this and invite a student
to share their thoughts.
If no one comes up with
the right solution, show
the solution.
And the work we did to create the recursive and explicit rules:
Let n = the number of the term
Let f(n) = the value of term n.
n f(n)
1 4
2 7
3 10
4 13
5 16
Recursive Rule:
f(1) = 4
f(n+1) = f(n) + 3 for integers n ≥ 2
Explicit Rule:
f(n+1) = 3f(n) + 4
How can you use a linear function to model the arithmetic sequence? Explain.
If you add a third column to the table that shows the change from one term to the
next, we see that, as n is increased by 1, f(n) is increased by 3 each time.
n f(n) ∆f(n)
1 4
2 7
3
3 10 3
4 13 3
5 16 3
Thus, there is a constant rate of change. Since it is a constant rate of change, it
must be a linear function.
f(x) = mx + b
x=n
m = constant rate of change = 3
b = y-intercept (function value when x = 0) which, if n = 0, f(0) = 1
f(n) = 3n + 1 would model this sequence
The linear function has a domain of all real numbers. In order for it to model this
sequence, it must have a domain of integers n ≥ 0.
Ex. 8
Using the sequence and work you did in example 2, show that the sequence 10, 6,
2, -2, -6, … can be modeled with a linear function.
Solution:
Add a third column to the table showing the rate of change:
n f(n) ∆f(n)
1 10 2 6
-4
3 2
-4
4 -2
-4
5 -6
-4
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Let students work
through this one.
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Since, as you increase the input by 1, the output increases by a constant value, this
is a linear function.
F(x) = ax + b
a = constant rate of change
x=n
b = y-intercept (function value when x = 0) which, if n = 0, f(0) = 14
f(n) = -4n + 14 would model this sequence for integers n ≥ 0.
Generalizing our work:
Let students work on
this one.
In general, how does the explicit equation of an arithmetic sequence model
repeated addition?
f(n + 1) = d(n) + f(1), for integers n ≥ 1
In general, how does the linear function model repeated addition?
A constant value, m, being multiplied to x in the linear function f(x) = mx + b is
just short-hand for adding m "x times".
How are linear functions and arithmetic sequences related?
They both represent repeated addition or constant growth by addition.
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