SEQUENCES AND SERIES
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
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 72 and divide by 2 each time.
72, 36, 18, 9, 4.5
eg2)
Start with 2, Multiply by 4 and subtract 3.
2, 5, 17, 65, 257
NOW DO
Chapter 5
Exercise 5A Questions 1 – 3
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 – the difference between consecutive terms is the same
eg. 0, 3, 6, 9, 12, 15, 18, 21, 24
Yes – the difference between consecutive terms is the same
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, 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
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
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
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:
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
The value of the nth
term (the term we
are trying to find)
The first term in
the sequence
The common
difference
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)
eg. Consider the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
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 + 𝑛 − 1 𝑑
𝑡𝑛 = 22 + 𝑛 − 1 6
b) What number would be at the 30th position?
𝑡𝑛 = 22 + 𝑛 − 1 6
𝑡30 = 22 + 30 − 1 6
𝑡30 = 22 + 29 6 = 22 + 174 = 196
Rule for Arithmetic
Sequences
The value of the n
th
term (the term we
are trying to find)
eg. Consider the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
The first term in
the sequence
The common
difference
The position of
the term we are
trying to find in
the sequence
10, 13, 16, 19, 22, 25,…
a) Write a rule for the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑡𝑛 = 10 + 𝑛 − 1 × 3
b) What number would be at the 16th position?
𝑡𝑛 = 10 + 𝑛 − 1 × 3
𝑡16 = 10 + 16 − 1 × 3
𝑡16 = 10 + 15 × 3 = 10 + 45 = 55
Rule for Arithmetic
Sequences
The value of the n
th
term (the term we
are trying to find)
eg. Consider the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
The first term in
the sequence
The common
difference
The position of
the term we are
trying to find in
the sequence
41, 37, 33, 29, 25,….
a) Write a rule for the arithmetic sequence
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑡𝑛 = 41 + 𝑛 − 1 × −4
b) What number would be at the 11th position?
𝑡𝑛 = 41 + 𝑛 − 1 × −4
𝑡30 = 41 + 11 − 1 × −4
𝑡30 = 41 + 10 × −4 = 41 − 40 = 1
NOW DO
Chapter 5
Exercise 5B
Questions 1, 4a, 4b, 4d 4f, 5a, 5b, 5d, 5f
ARITHMETIC SERIES
An arithmetic series is the term used for the sum of all of the terms in an arithmetic
sequence. We can find the arithmetic series using one of two different methods.
If we know the first term 𝑎, and the last term 𝑙, the sum of the 𝑛 terms is given by:
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
The first term in
the sequence
The last term in
the sequence
ARITHMETIC SERIES
An arithmetic series is the term used for the sum of all of the terms in an arithmetic
sequence. We can find the arithmetic series using one of two different methods.
If we know the first term 𝑎, the common difference 𝑑, the sum of the 𝑛 terms is given by:
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑)
2
The first term in
the sequence
The common
difference
The number of
terms in the
sequence
eg. A sequence starts at 4 and ends
with 32. If there are 8 terms in the
sequence, find the arithmetic series
(the sum) for the sequence.
𝑆𝑛 =
𝑛
2
=
8
2
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑
2
The first term in
the sequence
The first term in
the sequence
(𝑎+𝑙)
4 + 32
= 4( 36)
= 144
The common
difference
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
The last term in
the sequence
The number of
terms in the
sequence
eg. Find the sum of the first 20
values in the sequence 3, 5, 7, 9,…
𝑆𝑛 =
=
20
2
𝑛
2
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑
2
The first term in
the sequence
The common
difference
( 2𝑎 + 𝑛 − 1 𝑑
The first term in
the sequence
2 × 3 + 20 − 1 2
= 10(6 + 19 × 2 )
= 10 6 + 38
= 10 × 44 = 440
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
The last term in
the sequence
The number of
terms in the
sequence
eg. Find the sum of the first 18
values in the sequence 6, 11, 16, 21…
𝑆𝑛 =
=
18
2
𝑛
2
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑
2
The first term in
the sequence
The common
difference
( 2𝑎 + 𝑛 − 1 𝑑
The first term in
the sequence
2 × 6 + 18 − 1 5
= 9(12 + 17 × 5 )
= 9 12 + 85
= 9 × 97 = 873
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
The last term in
the sequence
The number of
terms in the
sequence
eg. Find the sum of the first 10 values
in the sequence 40, 36, 32, 28,…
𝑆𝑛 =
=
10
2
𝑛
2
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑
2
The first term in
the sequence
The common
difference
( 2𝑎 + 𝑛 − 1 𝑑
The first term in
the sequence
2 × 40 + 10 − 1 × −4
= 5(80 + 9 × −4 )
= 5 80 − 36
= 5 × 44 = 220
The number of
terms in the
sequence
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
The last term in
the sequence
NOW DO
Chapter 5
Exercise 5C
Questions 1, 4, 6, 7, 8, 9a
ARITHMETIC SERIES
We can also use the arithmetic series to find the value of a term at any position 𝑛 within
the sequence.
Value of
the term n
𝑡𝑛 = 𝑆𝑛 − 𝑆𝑛−1
Arithmetic Series
until the term n
Arithmetic
series until term
before n
Value of
the term n
𝑡𝑛 = 𝑆𝑛 − 𝑆𝑛−1
Arithmetic Series
until the term n
Arithmetic
series until term
before n
eg. Find the value of the term at position 20 of the sequence 14, 17, 20, 23,….
𝑡𝑛 = 𝑆𝑛 − 𝑆𝑛−1
𝑡20 = 𝑆20 − 𝑆19
= 850 − 779
= 71
20
2
𝑆20 =
28 + 20 − 1 3
= 10(28 + 19 × 3
= 10 28 + 57
= 10 × 85
= 850
19
2
𝑆19 =
28 + 19 − 1 3
= 9.5(28 + 18 × 3
= 9.5 (28 + 54)
= 9.5(82)
= 779
NOW DO
Chapter 5
Exercise 5C
Questions 1, 4, 6, 7, 8, 9a, 11, 12a
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 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
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
Chapter 5
Exercise 5D
Questions 1, 2, 4, 5a, 5b, 9
GEOMETRIC SERIES
A geometric series is the sum of the terms in a geometric sequence.
If we know the first term in the sequence a, and the common ratio r, the Geometric
Series can be found using:
First Term
𝑠𝑛 =
Sum of n
terms
𝑎(𝑟 𝑛 −1)
𝑟−1
Common Ratio
Remember, our
common ratio can
be found using
𝑡𝑛+1
𝑟=
𝑡𝑛
Next term
Current Term
GEOMETRIC SERIES
eg. Given the sequence 3, 6, 12,….
a) Write a geometric series equation to find the sum of n terms
b) Use the geometric series equation to find the sum of the first 10 terms of the
sequence:
a)
𝑠𝑛 =
𝑎(𝑟 𝑛 −1)
𝑟−1
𝑡𝑛+1
𝑟=
𝑡𝑛
𝒓=𝟐
𝒂=𝟑
b)
𝑠𝑛 =
3(2𝑛 −1)
2−1
𝑠𝑛 =
3(2𝑛 −1)
1
𝑠𝑛 = 3(2𝑛 − 1)
𝑠10 = 3(210 − 1)
= 3(1024 − 1)
= 3 × 1023
= 3069
GEOMETRIC SERIES
eg. Given the sequence 1, 4, 16,….
a) Write a geometric series equation to find the sum of n terms
b) Use the geometric series equation to find the sum of the first 8 terms of the
sequence:
a)
𝑠𝑛 =
𝑎(𝑟 𝑛 −1)
𝑟−1
𝑡𝑛+1
𝑟=
𝑡𝑛
𝒓=𝟒
b)
𝑠𝑛 =
1(4𝑛 −1)
4−1
𝑠𝑛 =
4 𝑛 −1
3
𝒂=𝟏
𝑠8 =
4 8 −1
3
=
65536−1
3
=
65535
3
= 21845
GEOMETRIC SERIES
eg. Given a sequence with a first term of -4 and a common ratio of 1.2,
a) 𝑆10
a)
b) 𝑆12
𝑎(𝑟 𝑛 − 1)
𝑠𝑛 =
𝑟−1
𝑡𝑛+1
𝑟=
𝑡𝑛
𝑠10 =
=
𝒓 = 𝟏. 𝟐 𝑠𝑛 =
𝒂 = −𝟒
𝑠𝑛 =
−4(1.2𝑛 −1)
1.2−1
)
−4(1.2𝑛 −1
0.2
−4(1.210 −1)
0.2
−4(6.1917−1)
0.2
−4(5.1917)
0.2
=
= −103.8
b)
𝑠12 =
=
−4(1.212 −1)
0.2
−4(8.916−1)
0.2
−4(7.916)
0.2
=
= −158.32
GEOMETRIC SERIES
eg. Given a sequence with a first term of 100 and the 10th term is 550. Find 𝑆5
Find r:
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
𝑡10 = 100 × 𝑟10−1
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
1.2085−1
550 = 100 × 𝑟 9
=
100(2.57818−1)
0.2085
550
100
=
100(1.57818)
0.2085
𝑟9
𝑎(𝑟 𝑛 − 1)
𝑠𝑛 =
𝑟−1
𝑠5 =
100(1.20855 −1)
=
𝑟 9 = 5.5
𝑟 =
9
5.5
𝑟 = 1.2085
=
157.818
0.2085
= 756.76
NOW DO
Chapter 5
Exercise 5E
Questions 1a, 1b, 1e, 2a, 2d, 3a
INFINITE SUM
What happens when a geometric sequence has a common ratio 𝑟, between −1 < 𝑟 < 1 ?
The terms in the geometric sequence are going to get smaller as 𝑛 increases,
ie the sequence is undergoing geometric decay.
This means that as 𝑛 approaches ∞ (infinity), the value of 𝑡𝑛 approaches zero.
We can calculate the geometric series (sum of the geometric sequence) in these cases, for
𝑛 = ∞ . This means indefinitely, the total sum will approach a particular number.
We call this the infinite sum, given by
𝑠∞ =
𝑎
1−𝑟
INFINITE SUM
eg. Find the sum to infinity for the geometric sequence 50, 25, 12.5,..
𝑟 = 0.5
𝑎 = 50
𝑠∞ =
𝑎
1−𝑟
𝑠∞ =
50
1−0.5
=
50
0.5
= 100
𝑠∞ =
𝑎
1−𝑟
INFINITE SUM
eg. Find the sum to infinity for the geometric sequence 10, 1, 0.1,..
𝑟 = 0.1
𝑎 = 10
𝑠∞ =
𝑎
1−𝑟
𝑠∞ =
10
1−0.1
=
10
0.9
=
100
9
≃ 11.11
𝑠∞ =
𝑎
1−𝑟
INFINITE SUM
𝑠∞ =
𝑎
1−𝑟
eg. Find the common ratio for the geometric sequence, whose first term is 10 and
𝑎
whose sum to infinity is 20
𝑠 =
∞
𝑠∞ = 20
1−𝑟
20 =
𝑎 = 10
10
1−𝑟
20(1 − r) = 10
20 − 20r = 10
−20r = −10
r=
−10
−20
1
2
= = 0.5
INFINITE SUM
𝑠∞ =
𝑎
1−𝑟
eg. Find 4th term of the geometric sequence, whose first term is 18 and whose sum to
𝑎
infinity is 45
𝑠 =
∞
𝑠∞ = 45
45 =
𝑎 = 18
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
1−𝑟
𝑡𝑛 = 𝑎 × 𝑟 𝑛−1
18
1−𝑟
𝑡4 = 18 × 0.64−1
45(1 − r) = 18
𝑡4 = 18 × 0.63
45 − 45r = 18
𝑡4 = 18 × 0.216
−45r = −27
r=
−27
−45
3
5
= = 0.6
= 3.88
NOW DO
Chapter 5
Exercise 5E
Questions 9a, 9b, 10b
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 × 𝑟 𝑛−1
𝑡𝑛 = 𝑎 + 𝑛 − 1 𝑑
Finding the sum of ‘n’ terms
𝑛
𝑆𝑛 = ( 2𝑎 + 𝑛 − 1 𝑑)
2
𝑛
𝑆𝑛 = ( 𝑎 + 𝑙 )
2
Finding the sum of ‘n’ terms
𝑠𝑛 =
𝑎(𝑟 𝑛 −1)
𝑟−1
𝑠∞ =
𝑎
1−𝑟
if 𝑟 > 1
if −1 < 𝑟 < 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
Chapter 5
Exercise 5E
Questions 1, 3, 4, 5, 7, 9, 13, 15
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