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SEQUENCES
Unit 2 General Maths
DESCRIBING SEQUENCES
This topic investigates different types of patterns and how they can be manipulated
mathematically.
The dictionary describes a sequence as, ‘a number of things, actions, or events
arranged or happening in a specific order or having a specific connection’.
In maths, the term sequence is used to represent an ordered set of elements.
In this topic we will examine the relationships and patterns of these sets of data.
RECOGNISING PATTERNS
For these examples:
eg1)
2, 4, 6, 8……..
Describe the pattern in words;
Describe the pattern in mathematical terms;
State the next 3 numbers in the pattern.
Increasing by 2
+2
10, 12, 14
Doubling each number
x2
80, 160, 320
eg2)
5, 10, 20, 40…..
eg3)
1000, 500, 250…… Halving each number
÷2
or
× 0.5
125, 62.5, 31.25
USING A RULE TO GENERATE A
NUMBER PATTERN
For these examples:
eg1)
Use the following rules to write down the
first five numbers of each number pattern
Start with a 4 and multiply by 2 each time.
4, 8, 16, 32, 64
GEOMETRIC SEQUENCE –
changes by a ‘common ratio’
eg2)
Start with 21 and add 6 each time.
21, 27, 33, 39, 45
ARITHMETIC SEQUENCE –
changes by a ‘common difference’
ARITHMETIC SEQUENCES
We can label the first term in the sequence
′𝑎′
We can label the ‘common difference’ between consecutive terms
Label the arithmetic sequences with their ‘a’ and ‘d’ values
𝑎=2
eg. 2, 3, 4, 5, 6, 7, 8, 9, 10
eg. 0, 3, 6, 9, 12, 15, 18, 21, 24
eg.
4, 6, 8, 10, 12, 14
𝑑=1
𝑎=0
𝑎=4
𝑑=3
𝑑=2
′𝑑′
ARITHMETIC SEQUENCES
An ARITHMETIC SEQUENCE is one in which the difference between any two
consecutive terms is the same.
Are these Arithmetic Sequences?
eg. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Yes – common difference
eg. 0, 3, 6, 9, 12, 15, 18, 21, 24
Yes – common difference
eg. 2, 4, 5, 10, 11, 16
No – the difference between consecutive terms is NOT the same
eg. 2, 4, 8, 16, 32, 64
No – the difference between consecutive terms is NOT the same
ARITHMETIC SEQUENCES
The terms at each position 𝑡𝑛 , in order can be labelled
𝑡1 , 𝑡2 , 𝑡3 , … … 𝑡𝑛
Label the terms in the following examples:
eg. 2, 3, 4, 5, 6, 7, 8, 9, 10
𝑡1 = 2,
𝑡2 = 3,
𝑡3 = 4
𝑒𝑡𝑐 …
eg. 0, 3, 6, 9, 12, 15, 18, 21, 24
𝑡1 = 0,
𝑡2 = 3,
𝑡3 = 6
𝑒𝑡𝑐 …
eg.
𝑡1 = 4,
𝑡2 = 6,
𝑡3 = 8
𝑒𝑡𝑐 …
4, 6, 8, 10, 12, 14
FINDING TERMS OF AN
ARITHMETIC SEQUENCE
If we know the first term 𝑎, and the common difference 𝑑, we can find any number
within the arithmetic sequence.
The rule for finding a term in an arithmetic sequence is:
The common
difference
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
The value of the nth
term (the term we
are trying to find)
The first term in
the sequence
The position of
the term we are
trying to find in
the sequence
Rule for Arithmetic
Sequences The value of the n
th
term (the term we
are trying to find)
eg1. Consider the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
The first term in
the sequence
The common
difference
The position of
the term we are
trying to find in
the sequence
22, 28, 34, 40, …..
a) Write a rule for the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
𝑡𝑛 = 22 + 6 𝑛 − 1
b) What number would be at the 30th position?
𝑙𝑒𝑡 𝑛 = 30:
𝑡𝑛 = 22 + 6 𝑛 − 1
𝑡30 = 22 + 6 30 − 1
𝑡30 = 22 + 6 29 = 22 + 174 = 196
Rule for Arithmetic
Sequences
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
eg2. Consider the arithmetic sequence
10, 13, 16, 19, 22, 25,…
a) Write a rule for the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
𝑡𝑛 = 10 + 3 𝑛 − 1
b) What number would be at the 16th position?
𝑡𝑛 = 10 + 3 𝑛 − 1
𝑙𝑒𝑡 𝑛 = 16:
𝑡16 = 10 + 3 16 − 1
𝑡16 = 10 + 3 15 = 10 + 45 = 55
Rule for Arithmetic
Sequences
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
eg3. Consider the arithmetic sequence
41, 37, 33, 29, 25,….
a) Write a rule for the arithmetic sequence
𝑡𝑛 = 41 − 4 𝑛 − 1
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
b) What number would be at the 11th position?
𝑙𝑒𝑡 𝑛 = 30:
𝑡𝑛 = 41 − 4 𝑛 − 1
𝑡30 = 41 − 4 11 − 1
𝑡30 = 41 − 4 10
= 41 − 40 = 1
Rule for Arithmetic
Sequences
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
eg4. Consider a sequence which starts at 14 and each number increases by 4.5.
Find the first 5 numbers in the sequence.
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
𝑡𝑛 = 14 + 4.5 𝑛 − 1
𝑙𝑒𝑡 𝑛 = 2:
𝑙𝑒𝑡 𝑛 = 3:
𝑡2 = 14 + 4.5 2 − 1
𝑡2 = 14 + 4.5 1
= 18.5
𝑡3 = 14 + 4.5 3 − 1
𝑡3 = 14 + 4.5 2
= 14 + 9
= 23
etc….
OR use your calculator
Rule for Arithmetic
Sequences
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
Eg4 (continued).
Consider a sequence which starts at 14 and each number increases by 4.5.
Find the first 5 numbers in the sequence.
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
𝑡𝑛 = 14 + 4.5 𝑛 − 1
OR use your calculator
in another way..
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
Eg4 (continued) Using the SEQUENCES function.
Choose Explicit
𝑡𝑛 = 14 + 4.5 𝑛 − 1
Change these values for
‘n’ to produce a table of
values as small or large
as you require
NOW DO
Exercise 6.2
Q1, 2, 3abc, 4abc, 5, 6, 7
FINDING OTHER VARIABLES IN AN
ARITHMETIC SEQUENCE
We can rearrange the rule
to find a, d or n:
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
Finding the first number in a sequence:
Finding the common difference:
Finding the position of the term:
𝑑=
𝑎 = 𝑡𝑛 − 𝑑 𝑛 − 1
𝑡𝑛 − 𝑎
𝑛−1
𝑛=
𝑡𝑛 − 𝑎
𝑑
+1
NOW DO
Exercise 6.2
Q1, 2, 3abc, 4abc, 5, 6, 7, 8abc, 15ac, 16ac, 17a
GEOMETRIC SEQUENCES
A geometric sequence is one where the next term is formed by multiplying the previous
term by a fixed number called the common ratio, 𝑟
The next term in the sequence is found by multiplying the current term by the common
ratio, giving the equation:
𝑡𝑛+1 = 𝑟 𝑡𝑛
Current Term
Next term
Common Ratio
Rearranging this equation allows us
to find the common ratio, using:
𝑡𝑛+1
𝑟=
𝑡𝑛
Next term
Current Term
GEOMETRIC SEQUENCES
We can decide if a sequence is a geometric sequence by calculating the ratio for
subsequent terms and checking that the ratio is consistent throughout the
sequence.
𝑡𝑛+1
𝑟=
𝑡𝑛
Next term
Current Term
eg. Are the following sequences geometric?
a) 5, 10, 20, 40, 80….
YES
𝑟=
𝑡𝑛+1
𝑡𝑛
𝑟=
𝑡𝑛+1
𝑡𝑛
=
10
5
=
20
10
=2
=2
𝑟=
𝑡𝑛+1
𝑡𝑛
𝑟=
𝑡𝑛+1
𝑡𝑛
=
40
20
=2
=
80
40
=2
GEOMETRIC SEQUENCES
We can decide if a sequence is a geometric sequence by calculating the ratio for
subsequent terms and checking that the ratio is consistent throughout the
sequence.
𝑡𝑛+1
𝑟=
𝑡𝑛
Next term
Current Term
eg. Are the following sequences geometric?
a) 100, 50, 25, 12.5….
YES
𝑟=
𝑡𝑛+1
𝑡𝑛
𝑟=
𝑡𝑛+1
𝑡𝑛
=
50
100
=
25
50
= 0.5
= 0.5
𝑟=
𝑡𝑛+1
𝑡𝑛
𝑟=
𝑡𝑛+1
𝑡𝑛
=
25
50
=
12.5
25
= 0.5
= 0.5
GEOMETRIC SEQUENCES
We can decide if a sequence is a geometric sequence by calculating the ratio for
subsequent terms and checking that the ratio is consistent throughout the
sequence.
𝑡𝑛+1
𝑟=
𝑡𝑛
Next term
Current Term
eg. Are the following sequences geometric?
a) 2, 10, 20, 60….
NO
𝑟=
𝑡𝑛+1
𝑡𝑛
𝑟=
𝑡𝑛+1
𝑡𝑛
=
10
2
=5
=
20
10
=2
𝑟=
𝑡𝑛+1
𝑡𝑛
=
60
20
=3
GEOMETRIC SEQUENCE
A geometric sequence can be written in terms of its 1st term ‘a’, and it’s common ratio ‘r’
𝑎, 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ………..etc
1st
2nd
3rd
Term Term Term
Any term n, in the sequence, can be
found if we know the 1st term and the
ratio, using:
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
eg. A geometric sequence begins with 6
and has a common ratio of 4.
Find the 3rd term using the rule above.
𝑡𝑛 = 𝑎𝑟 𝑛−1
𝑡3 = 6 × 43−1 = 6 × 42
= 6 × 16
= 96
Lets prove it using
our calculators..
Finding the
𝑛𝑡ℎ term
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
𝑎,
eg. For the sequence 4, 8, 16, 32..
Find the 10th term of the sequence.
𝑡𝑛 = 𝑎𝑟 𝑛−1
𝑡10 = 4 × 210−1
= 4 × 29
= 4 × 512
= 2048
1st
Term
𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ……..etc
2nd
Term
3rd
Term
Finding the
𝑛𝑡ℎ term
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
𝑎,
eg. For the sequence 80, 40, 20, 10, 5..
Find the 8th term of the sequence.
𝑡𝑛 = 𝑎𝑟 𝑛−1
𝑡8 = 80 × (0.5)8−1
= 80 × (0.5)7
= 80 × 0.0078125
= 0.625
1st
Term
𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 ……..etc
2nd
Term
3rd
Term
NOW DO
APPLICATIONS PROBLEMS
In this exercise, we look at solving real life, worded problems using the arithmetic and
geometric sequences and series equations from all prior exercises.
1. Read the question carefully
2. Decide whether the problem involves an arithmetic or geometric sequence
3. Write the information given, in terms of variables (ie. 𝑎, 𝑑, 𝑟, 𝑡𝑛 , 𝑒𝑡𝑐.)
4. Choose the appropriate formula to use to solve the problem.
5. Solve and answer the question being asked in the problem.
EQUATIONS
Arithmetic Sequences and Series
Geometric Sequences and Series
Finding the nth term
Finding the nth term
𝑡𝑛 = 𝑎 + 𝑑 𝑛 − 1
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
APPLICATIONS PROBLEMS
If $1000 is put into a compound interest account, at an interest rate of 8%, how much will
be in the account after 10 years?
geometric sequence
𝒂 = 𝟏𝟎𝟎𝟎
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
10−1
𝑡10 = 1000 × (1.08)
𝒓 = 𝟏. 𝟎𝟖
𝑡10 = 1000 × (1.08)9
𝒏 = 𝟏𝟎
𝑡10 = 1000 × 1.999004627
= 1999.004627
After 10 years, the account balance will be $1999.00
APPLICATIONS PROBLEMS
The seating arrangements at a concert are: first row has 10 seats, second row has 13 seats,
third row has 16 seats, following this pattern all the way through to the final 20th row.
a) How many seats are in the 20th row?
b) How many seats are there in total?
𝒂 = 𝟏𝟎
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑡20 = 10 + 20 − 1 3
𝐝=𝟑
𝑡20 = 10 + 19 × 3
Arithmetic sequence
𝒏 = 𝟐𝟎
= 10 + 57
= 67 𝑠𝑒𝑎𝑡𝑠
APPLICATIONS PROBLEMS
The seating arrangements at a concert are: first row has 10 seats, second row has 13 seats,
third row has 16 seats, following this pattern all the way through to the final 20th row.
a) How many seats are in the 20th row? b) How many seats are there in total?
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑)
Arithmetic sequence
2
𝒂 = 𝟏𝟎
𝐝=𝟑
𝒏 = 𝟐𝟎
𝑆20
20
=
( (2 × 10) + 20 − 1 3)
2
𝑆20 = 10 19 × 3
= 10 × 57
= 570
NOW DO
Use
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
to form equations for each given term
Now use the simultaneous equation solver on
your calculator to solve for our unknowns r and a
Term 3 = 100
𝑡3 = 𝑎 × 𝑟 3−1
100 = 𝑎𝑟 2
Term 5 = 400
𝑡5 = 𝑎 × 𝑟
400 = 𝑎𝑟 4
5−1
So the first term, a = 25
The common ratio is 2
Use
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
to form equations for each given term
Now use the simultaneous equation solver on
your calculator to solve for our unknowns r and a
Term 4 = 90
𝑡4 = 𝑎 × 𝑟 4−1
90 = 𝑎𝑟 3
So the first term, a = 3.33
Term 7 = 2430
The common ratio is 3
𝑡7 = 𝑎 × 𝑟
2430 = 𝑎𝑟 6
7−1
COMBINED PROBLEMS
Problems involving both Arithmetic and Geometric Sequences
Some sequences involve both a common difference and a common ratio.
These take the form:
𝒕𝒏+𝟏 = 𝒓 × 𝒕𝒏 + 𝒅,
𝒕𝟏 = 𝒂
Consider the sequence, ‘Start with 3, Multiply the number by 2 and add 4.
The recurrence relation is:
𝒕𝒏+𝟏 = 𝟐𝒕𝒏 + 𝟒,
𝒕𝟏 = 𝟑
Try these for yourself on your worksheet
ARITHMETIC SEQUENCES
GENERATED BY RECURSION
A sequence can be generated by the repeated use of an instruction. This is known as
recursion. We can form equations to model recursion:
The current term given by 𝑛 can be represented by 𝑡𝑛
The following term is 𝑡𝑛+1 & the term before it is 𝑡𝑛−1
Using this, we can form an equation to model the recursion. The recursion relation is:
𝑡𝑛+1 = 𝑡𝑛 + 𝑑 ,
The next term
The current term
𝑡1 = 𝑎
The common
difference
The first term in
the sequence
Recursion relation
The next term
𝑡𝑛+1 = 𝑡𝑛 + 𝑑,
The current term
𝑡1 = 𝑎
The common
difference
Form recursion relations for the following:
eg. 2, 3, 4, 5, 6, 7, 8, 9, 10
𝑡𝑛+1 = 𝑡𝑛 + 1,
𝑡1 = 2
eg. 0, 3, 6, 9, 12, 15, 18, 21, 24
𝑡𝑛+1 = 𝑡𝑛 + 3,
𝑡1 = 0
eg.
𝑡𝑛+1 = 𝑡𝑛 + 0.5,
𝑡1 = 4
𝑡𝑛+1 = 𝑡𝑛 − 3,
𝑡1 = 30
4, 4.5, 5, 5.5, 6, 6.5
eg. 30, 27, 24, 21, 18, 15
The first term in
the sequence
NOW DO
Chapter 5
Recursion Worksheet – Exercise 1
GEOMETRIC SEQUENCES
WRITING A RECURRENCE RELATION
So, we can write the recurrence relation for geometric sequences:
𝑡𝑛+1 = 𝑟 𝑡𝑛 ,
𝑡1 = 𝑎
1st Term
Common Ratio
eg. Given the sequence
1, 2, 4, 8, 16, 32……
a) The sequence could be described as: Start at 1 and multiply by 2 for subsequent terms
b) Form the recurrence relation for the sequence:
c) Its this growth or decay?
Geometric Growth
𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎
𝑡𝑛+1 = 2𝑡𝑛 , 𝑡1 = 1
𝑡𝑛+1 = 𝑟 𝑡𝑛 ,
recurrence relation
Common Ratio
eg. Given the sequence
𝑡1 = 𝑎
1st Term
2, 6, 18, 54, 162…
a) The sequence could be described as: Start at 2 and multiply by 3 for subsequent terms
b) Form the recurrence relation for the sequence:
c) Its this growth or decay?
Geometric Growth
𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎
𝑡𝑛+1 = 3𝑡𝑛 , 𝑡1 = 2
𝑡𝑛+1 = 𝑟 𝑡𝑛 ,
recurrence relation
Common Ratio
eg. Given the sequence
𝑡1 = 𝑎
1st Term
100, 50, 25, 12.5…
a) The sequence could be described as: Start at 100 and divide by 2 for subsequent terms
b) Form the recurrence relation for the sequence:
𝑡𝑛+1 = 𝑟 𝑡𝑛 , 𝑡1 = 𝑎
c) Its this growth or decay?
Geometric Decay
𝑡𝑛+1 =
1
𝑡
2 𝑛
, 𝑡1 = 100
NOW DO
Recursion Worksheet
Part 2