# Mathematics

### Linear, quadratic, arithmetic, geometric and Fibonacci Sequences

In this tutorial, you will learn the following:

• Finding the nth term of a linear sequence
• Finding the nth term of a quadratic sequence
• Arithmetic sequences
• Geometric sequences
• Lots of practice questions for each above
• A programme to generate random linear, quadratic and Fibonacci sequences - for practice
• Sequences for GCSE, IGCSE, AS, A-Level, GCE OL and GCE AL

#### Linear Sequences

If the difference between two consecutive terms is a constant, it is called a linear sequence.

E.g.
2, 4, 6, 8, 10,...
5, 3, 1, -1, -3,...

E.g.1

Find the nth term of the sequence, 3, 5, 7, 9, ...

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = 3 => 3 = a + b
n = 2; N = 5 => 5 = 2a + b
Solving the two simultaneous equations, a = 2 and b = 1.
So, the nth term, N = 2n + 1.

E.g.2

Find the nth term of the sequence, 8, 5, 2, -1, ...

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = 8 => 8 = a + b
n = 2; N = 5 => 5 = 2a + b
Solving the two simultaneous equations, a = -3 and b = 11.
So, the nth term, N = -3n + 11.

E.g.3

Find the nth term of the sequence, 4, 7, 10, 13... Hence, determine whether 200 is a term of this sequence.

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = 4 => 4 = a + b
n = 2; N = 7 => 7 = 2a + b
Solving the two simultaneous equations, a = 3 and b = 1.
So, the nth term, N = 3n + 1.
If 200 is a term of this sequence,
200 = 3n + 1
n = 199/3 = 66.7
Since n is not a whole number, 200 is not a term of this sequence.

E.g.4

Find the nth term of the sequence, 125, 123, 121, 119... Hence, determine the first number that is negative.

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = 125 => 125 = a + b
n = 2; N = 123 => 123 = 2a + b
Solving the two simultaneous equations, a = -2 and b = 127.
So, the nth term, N = -2n + 127.
When the term is negative, N < 0,
-2n + 127 < 0 => 2n > 127
n > 63.5
It is the 64th term that becomes negative first.

E.g.5

Find the nth term of the sequence, -6, 3, 12, 21... Hence, determine the first number that exceeds 1000.

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = -6 => -6 = a + b
n = 2; N = 3 => 3 = 2a + b
Solving the two simultaneous equations, a = 9 and b = -15.
So, the nth term, N = 9n - 15.
When the term exceeds 1000, N > 1000,
9n - 15 > 1000
9n > 1015 => n > 112.6
It is the 113th term that exceeds 1000 first.

E.g.6

Find the nth term of the sequence, 8, 11, 14, 17... Hence, determine whether this sequence and the sequence of, N = 2n -5 share a term.

Let the nth term, N = an + b, where a and b are two constants to be determined.
Since there are two unknowns, we need to make simultaneous equations.
n = 1; N = 8 => 8 = a + b
n = 2; N = 11 => 11 = 2a + b
Solving the two simultaneous equations, a = 3 and b = 5.
So, the nth term, N = 3n + 5.
If the two sequences share a term,
3n + 5 = 2n - 5
n = -10
5 Since negative terms do not exist, the two sequences do not have as common term.

If the difference of two consecutive terms is a constant at the 'second' level as follows, it is a quadratic sequence:

E.g.1
2, 3, 5, 8, 12,...
The differences between the consecutive terms of the above: 1, 2, 3, 4
The differences between the consecutive terms: 1, 1, 1 of the above
E.g.2
4, 6, 9, 13, 18,...
The differences between the consecutive terms of the above: 2, 3, 4, 5
The differences between the consecutive terms: 1, 1, 1 of the above

E.g.1

Find the nth term of the sequence, 3, 4, 6, 9, 13...

The differences between the consecutive terms of the above: 1, 2, 3, 4
The differences between the consecutive terms: 1, 1, 1 of the above
Let the nth term, N = an2 + bn + c, where a, b and c are constants to be found.
Since there are three unknowns, we need to make three equations.
n = 1; N = 3 => 3 = a + b + c
n = 2; N = 4 => 4 = 4a + 2b + c
n = 3; N = 6 => 6 = 9a + 3b + c
From the first two, we get:
1 = 3a + b
From the last two, we get:
2 = 5a + b
By solving the simultaneous equations, we get,
a = 1/2 and b = -1/2; sub them in the first equation,
c = 3
So, N = (1/2)n2 - (1/2)n + 3.

E.g.2

Find the nth term of the sequence, 2, 5, 10, 17, 26...

The differences between the consecutive terms of the above: 3, 5, 7, 9
The differences between the consecutive terms: 2, 2, 2 of the above
Let the nth term, N = an2 + bn + c, where a, b and c are constants to be found.
Since there are three unknowns, we need to make three equations.
n = 1; N = 2 => 2 = a + b + c
n = 2; N = 5 => 5 = 4a + 2b + c
n = 3; N = 10 => 10 = 9a + 3b + c
From the first two, we get:
3 = 3a + b
From the last two, we get:
5 = 5a + b
By solving the simultaneous equations, we get,
a = 1 and b = 0; sub them in the first equation,
c = 1
So, N = n2 + 1.

E.g.3

Find the nth term of the sequence, 2, 6, 12, 20, 30... Which term is 156?

The differences between the consecutive terms of the above: 4, 6, 8, 10
The differences between the consecutive terms: 2, 2, 2 of the above
Let the nth term, N = an2 + bn + c, where a, b and c are constants to be found.
Since there are three unknowns, we need to make three equations.
n = 1; N = 2 => 2 = a + b + c
n = 2; N = 6 => 6 = 4a + 2b + c
n = 3; N = 12 => 12 = 9a + 3b + c
From the first two, we get:
4 = 3a + b
From the last two, we get:
6 = 5a + b
By solving the simultaneous equations, we get,
a = 1 and b = 1; sub them in the first equation,
c = 0
So, N = n2 + n.
n2 + n = 156
n2 + n - 156 = 0
(n + 13)(n - 12) = 0
n = -13 or n = 12
It's the twelfth term.

#### Fibonacci Sequences

A Fibonacci number sequence is formed when a term is produced by adding the two preceding terms.

E.g.1

1, 2, 3, 5, 8, 13, ...
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13

E.g.2

0, 2, 2, 4, 6, 10, ...
0 + 2 = 2
2 + 2 = 4
2 + 4 = 6
4 + 6 = 10

E.g.3

-1, 1, 0, 1, 1, 2, ...
-1 + 1 = 0
1 + 0 = 1
0 + 1 = 1
1 + 1 = 2

E.g.4

x, y, x + y, x + 2y, 2x + 3y, 3x + 5y, ...
x + y = x + y
y + x + y = x + 2y
x + y + x + 2y = 2x + 3y
x + 2y + 2x + 3y = 3x + 5y

E.g.5

The first two terms of a Fibonacci sequence are p and q. Show that the seventh term is 5p + 8q. If the 4th term is 8 and the 7th term is 34, find p and q.

The sequence progresses as follows:
p, q, p + q, p + 2q, 2p + 3q, 3p + 5q, 5p + 8q
p + 2q = 8; 5p + 8q = 34
p = (8 - 2q)
Substitute this in 5p + 8q = 34
5(8-2q) + 8q = 34
40 - 10q + 8q = 34
40 -2q = 34
2q = 6 => q = 3
p + 6 = 8 => p = 2.

#### Geometric Sequences - geometric progression

If the consecutive terms of a sequence have a common ratio, it is called a geometric sequence.

E.g.1

2,4,8,16,32,64,...
Each term, in this case, can be obtained by multiplying the previous term by 2

E.g.2

32,16,8,4,2,1,...
Each term, in this case, can be obtained by multiplying the previous term by 1/2

E.g.3

2,-2, 2,-2,2,2,...
Each term, in this case, can be obtained by dividing the previous term by -1

#### The nth term of a geometric sequences

Let the first term be a and the common ration be r. The sequence takes the following form:
a, ar, ar2, ar3, ar4, ar5, ar6, ...
So, a pattern emerges: ar3 is the forth term; ar4 is the fifth term and so on.
On that basis, if the nth term is N,
N = arn-1

3, 6, 12

E.g.1

The first term and the common ratio of a geometric sequence are 3 and 2 respectively. Find the nth term. Hence, find the 5th and 7th terms.
N = arn-1
a = 3; r = 2
N = 3(2n-1)
If n = 5 => N = 3(25-1) = 3(24) = 48
If n = 7 => N = 3(27-1) = 3(26) = 192

E.g.2

The third term and the sixth term of a geometric sequence are 16 and 128 respectively. Find the nth term. Hence, find the ninth term.
N = arn-1
If n = 3 => 16 = ar3-1) = ar2
If n = 6 => 128 = ar6-1) = ar5
Dividing the second equation by the first gives,
r3 = 8 => r = 2
Sub this in the first equation,
16 = a(22) = 4a => a = 4.
First term = 4; common ratio = 2.
N = 4(2n-1)
n = 9 => N = 4(29-1) = 4(28) = 1024

E.g.3

The first three terms of a geometric sequence are x, x + 3 and 2x + 6. Find x, if it is positive.
Since they form a geometric sequence, they have a common ratio.
So, (x + 3)/x = (2x + 6)/(x + 3)
By cross multiplying, you get,
(x + 3)2 = x(2x + 6)
x2 + 6x + 9 = 2x2 + 6x
x2 = 9
x = ± 3
Since x is positive, x = 3.

With the following programme, you can generate unlimited number of sequences for you to practice.

#### Generate a Sequence

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