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

In this tutorial, you will learn the following:

- Finding the n
^{th} term of a linear sequence
- Finding the n
^{th} 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 n^{th} term of the sequence, 3, 5, 7, 9, ...

Let the n^{th} 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 n^{th} term, N = 2n + 1.

**E.g.2**

Find the n^{th} term of the sequence, 8, 5, 2, -1, ...

Let the n^{th} 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 n^{th} term, N = -3n + 11.

**E.g.3**

Find the n^{th} term of the sequence, 4, 7, 10, 13... Hence, determine whether 200 is a term of this sequence.

Let the n^{th} 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 n^{th} 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 n^{th} term of the sequence, 125, 123, 121, 119... Hence, determine the first number that is negative.

Let the n^{th} 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 n^{th} term, N = -2n + 127.

When the term is negative, N < 0,

-2n + 127 < 0 => 2n > 127

n > 63.5

It is the 64^{th} term that becomes negative first.

**E.g.5**

Find the n^{th} term of the sequence, -6, 3, 12, 21... Hence, determine the first number that exceeds 1000.

Let the n^{th} 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 n^{th} term, N = 9n - 15.

When the term exceeds 1000, N > 1000,

9n - 15 > 1000

9n > 1015 => n > 112.6

It is the 113^{th} term that exceeds 1000 first.

**E.g.6**

Find the n^{th} 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 n^{th} 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 n^{th} 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.

#### Quadratic Sequences

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

So, the sequence is quadratic.

**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

So, the sequence is quadratic.

**E.g.1**

Find the n^{th} 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

So, the sequence is quadratic.

Let the n^{th} term, N = an^{2} + 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)n^{2} - (1/2)n + 3.

**E.g.2**

Find the n^{th} 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

So, the sequence is quadratic.

Let the n^{th} term, N = an^{2} + 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 = n^{2} + 1.

**E.g.3**

Find the n^{th} 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

So, the sequence is quadratic.

Let the n^{th} term, N = an^{2} + 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 = n^{2} + n.

n^{2} + n = 156

n^{2} + 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 4^{th} term is 8 and the 7^{th} 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 n^{th} 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, ar^{2}, ar^{3}, ar^{4}, ar^{5}, ar^{6}, ...

So, a pattern emerges: ar^{3} is the forth term; ar^{4} is the fifth term and so on.

On that basis, if the n^{th} term is N,

**N = ar**^{n-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 n^{th} term. Hence, find the 5^{th} and 7^{th} terms.

N = ar^{n-1}

a = 3; r = 2

N = 3(2^{n-1})

If n = 5 => N = 3(2^{5-1}) = 3(2^{4}) = 48

If n = 7 => N = 3(2^{7-1}) = 3(2^{6}) = 192

**E.g.2**

The third term and the sixth term of a geometric sequence are 16 and 128 respectively. Find the n^{th} term. Hence, find the ninth term.

N = ar^{n-1}

If n = 3 => 16 = ar^{3-1}) = ar^{2}

If n = 6 => 128 = ar^{6-1}) = ar^{5}

Dividing the second equation by the first gives,

r^{3} = 8 => r = 2

Sub this in the first equation,

16 = a(2^{2}) = 4a => a = 4.

First term = 4; common ratio = 2.

N = 4(2^{n-1})

n = 9 => N = 4(2^{9-1}) = 4(2^{8}) = 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)

x^{2} + 6x + 9 = 2x^{2} + 6x

x^{2} = 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