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C1:Ch5  Arithmetic Sequences and Series
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TweetPeter Scott
on 2 February 2018Transcript of C1:Ch5  Arithmetic Sequences and Series
Chapter 3
Using the nth term
buy milk
pick up kids
with
Mr Scott
when you know a formula for the nth term of a sequence then you can find any term in the sequence
Core Mathematics 1
Sequences and Series
in this chapter we will look at how a sequence of numbers is created and how we can use this knowledge to find an individual term and the sum of an arithmetic series.
5, 8, 11, 14, 17
algebraic!!
you can see that the rule is
"add 3 to the previous term"
Sequence
a list of numbers following a set rule
for example 3, 7, 11, 15, 19, ...
Series
the terms of a sequence added together
for example 3 + 7 + 11 + 15 + 19 + ...
each number in a sequence is called a
term
What are the next 3 terms in the following sequences and the rule?
1) 14, 11, 8, 5, ...
2) 1, 2, 4, 8, ...
3) 1, 3, 7, 15, ...
here's an example:
U = 3n  1
an nth term is given by
n
Find the 6th term
U = 3(6)  1
6
= 17
a sequence is generated by formula
U = an + b
n
U = 5
and
3
U = 20
8
where a and b are constants to be found
find the constants
a
and
b
given that
U = 5 so
3
3a + b = 5
We know
U = 20 so
8
8a + b = 20
and

gives
5a = 15
a = 3
substitute a=3 into
9 + b = 5
b = 4
constants are a= 3 b= 4
try this on for size!
try Exercise 6B
Questions
a c e g
a d e j
and
Recurrence Relationship
when you know the rule to get from one term to the next you can use this information to produce a recurrence relationship (or recurrence formula)
so
U
2
= + 3
U
1
U
3
= + 3
U
2
U
4
= + 3
U
3
etc
this sequence can also be described by the recurrence formula:
U
k+1
= + 3 (k> 1)
U
k
_
Riddle me this! Why do you need to also give the first term when describing a recurrence formula?
try Exercise 6C
Questions
a c e g
a d e j
a c e
and
a sequence that increases by a constant amount each time is called an
arithmetic sequence
U
k+1
= + n (k> 1)
U
k
_
it will have a recurrence relationship of the form
n
C


/

/

An arithmetic
series
is formed by adding together the terms of an arithmetic sequence
Arithmetic Series
U + U + U + ... + U
2
n
3
1
In an arithmetic series the next term is found by adding (or subtracting) a constant number.
This number is called the
common difference
d.
The first term is represented by
a.
This means all arithmetic series can be put in the form
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)
6th term
5th term
4th term
3rd term
2nd term
1st term
The nth term of an
arithmetic sequence is
a + (n1)d
where a = the first term
and d = the common difference
@budmath
You will need to be able to find the sum of an arithmetic series
Apparently, whilst in primary school, Carl Friedrich Gauss (17771855) was asked to add up all the numbers from 1 to 100. He solved it in less than a minute with the following method:
S = 1 + 2 + 3 + 4 + ... + 99 + 100
Let
Reverse the sum
S = 100 + 99 + 98 + 97 + ... + 2 + 1
Add the sums
2S = 101 + 101 + 101 + ... + 101 + 101
2S = 100 x 101
S = (100 x 101)  2
.
.
S = 5050
so in general
S = a + (a+d) + (a+2d) + ... + (a+(n2)d) + (a+(n1)d)
Reverse the sum
Add the sums
2S = 2a + (n1)d + 2a +...+ (n1)d + 2a + (n1)d
2S = n[2a + (n1)d]
S = (a+(n1)d) + (a+(n2)d) + ... + (a+2d) + (a+d) + a
S = [2a + (n1)d]
n
2

n
n
n
n
n
S = [2a + (n1)d]
n
2

n
the formula for the sum of an arithmetic series is
or
S = [a + L]
n
2

n
where
a
= the first term,
d
= the common difference
n
= the number of terms and
L
= the last term
Exercise
6F
SIGMA ("
the sum of
")
You can (and will) use the symbol
to mean
"the sum of"
for example
n = 1
10
U
n
= U + U + U + . . . + U
r = 5
10
3
2
1
15
(10  2r) =
0 + 2 + 4 + ... + 20
means the sum of (10  2r) from r=5 to r=15
r = 1
20
(4r + 1)
= 5 + 9 + 13 + ... + 81
Calculate
S = [2a + (n1)d]
n
2

n
a = d = n =
5 4 20
= [2
(5)
+ (
20
1)
(4)
]
20
2
= [
10
+
76
]
10
=
860
Q 1 bgh
Q 5, 8, 9
Exercise
6G
Full transcriptUsing the nth term
buy milk
pick up kids
with
Mr Scott
when you know a formula for the nth term of a sequence then you can find any term in the sequence
Core Mathematics 1
Sequences and Series
in this chapter we will look at how a sequence of numbers is created and how we can use this knowledge to find an individual term and the sum of an arithmetic series.
5, 8, 11, 14, 17
algebraic!!
you can see that the rule is
"add 3 to the previous term"
Sequence
a list of numbers following a set rule
for example 3, 7, 11, 15, 19, ...
Series
the terms of a sequence added together
for example 3 + 7 + 11 + 15 + 19 + ...
each number in a sequence is called a
term
What are the next 3 terms in the following sequences and the rule?
1) 14, 11, 8, 5, ...
2) 1, 2, 4, 8, ...
3) 1, 3, 7, 15, ...
here's an example:
U = 3n  1
an nth term is given by
n
Find the 6th term
U = 3(6)  1
6
= 17
a sequence is generated by formula
U = an + b
n
U = 5
and
3
U = 20
8
where a and b are constants to be found
find the constants
a
and
b
given that
U = 5 so
3
3a + b = 5
We know
U = 20 so
8
8a + b = 20
and

gives
5a = 15
a = 3
substitute a=3 into
9 + b = 5
b = 4
constants are a= 3 b= 4
try this on for size!
try Exercise 6B
Questions
a c e g
a d e j
and
Recurrence Relationship
when you know the rule to get from one term to the next you can use this information to produce a recurrence relationship (or recurrence formula)
so
U
2
= + 3
U
1
U
3
= + 3
U
2
U
4
= + 3
U
3
etc
this sequence can also be described by the recurrence formula:
U
k+1
= + 3 (k> 1)
U
k
_
Riddle me this! Why do you need to also give the first term when describing a recurrence formula?
try Exercise 6C
Questions
a c e g
a d e j
a c e
and
a sequence that increases by a constant amount each time is called an
arithmetic sequence
U
k+1
= + n (k> 1)
U
k
_
it will have a recurrence relationship of the form
n
C


/

/

An arithmetic
series
is formed by adding together the terms of an arithmetic sequence
Arithmetic Series
U + U + U + ... + U
2
n
3
1
In an arithmetic series the next term is found by adding (or subtracting) a constant number.
This number is called the
common difference
d.
The first term is represented by
a.
This means all arithmetic series can be put in the form
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)
6th term
5th term
4th term
3rd term
2nd term
1st term
The nth term of an
arithmetic sequence is
a + (n1)d
where a = the first term
and d = the common difference
@budmath
You will need to be able to find the sum of an arithmetic series
Apparently, whilst in primary school, Carl Friedrich Gauss (17771855) was asked to add up all the numbers from 1 to 100. He solved it in less than a minute with the following method:
S = 1 + 2 + 3 + 4 + ... + 99 + 100
Let
Reverse the sum
S = 100 + 99 + 98 + 97 + ... + 2 + 1
Add the sums
2S = 101 + 101 + 101 + ... + 101 + 101
2S = 100 x 101
S = (100 x 101)  2
.
.
S = 5050
so in general
S = a + (a+d) + (a+2d) + ... + (a+(n2)d) + (a+(n1)d)
Reverse the sum
Add the sums
2S = 2a + (n1)d + 2a +...+ (n1)d + 2a + (n1)d
2S = n[2a + (n1)d]
S = (a+(n1)d) + (a+(n2)d) + ... + (a+2d) + (a+d) + a
S = [2a + (n1)d]
n
2

n
n
n
n
n
S = [2a + (n1)d]
n
2

n
the formula for the sum of an arithmetic series is
or
S = [a + L]
n
2

n
where
a
= the first term,
d
= the common difference
n
= the number of terms and
L
= the last term
Exercise
6F
SIGMA ("
the sum of
")
You can (and will) use the symbol
to mean
"the sum of"
for example
n = 1
10
U
n
= U + U + U + . . . + U
r = 5
10
3
2
1
15
(10  2r) =
0 + 2 + 4 + ... + 20
means the sum of (10  2r) from r=5 to r=15
r = 1
20
(4r + 1)
= 5 + 9 + 13 + ... + 81
Calculate
S = [2a + (n1)d]
n
2

n
a = d = n =
5 4 20
= [2
(5)
+ (
20
1)
(4)
]
20
2
= [
10
+
76
]
10
=
860
Q 1 bgh
Q 5, 8, 9
Exercise
6G