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AS-Level Maths: Core 2for Edexcel

C2.3 Sequences and series

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Geometric sequences

Geometric series

The sum to infinity of a geometric series

Examination-style questions

Geometric sequences


Geometric sequences

In a geometric sequence (or geometric progression) each term is produced by multiplying the previous term by a constant value called the common ratio.

For example, the sequence

3, 6, 12, 24, 48, 96, …

is a geometric sequence that starts with 3 and has a common ratio of 2.

We could write this sequence as

3, 3 × 2, 3 × 2 × 2, 3 × 2 × 2 × 2, 3 × 2 × 2 × 2 × 2, …

or

3, 3 × 2, 3 × 22, 3 × 23, 3 × 24, 3 × 25, …


Geometric sequences

If we call the first term of a geometric sequence a and the common ratio r we can write a general geometric sequence as:

Also:

The nth term of a geometric sequence with first term a and common ratio r is:

un = arn–1

The inductive definition of a geometric sequence with first term a and common ratio r is:

u1 = a, un+1 = run

a, ar, ar2, ar3, ar5, …ar4,


Geometric sequences

This is a geometric sequence with first term a = 7 and common ratio r = 12 ÷ 8 = 1.5.

The nth term is given by arn–1 so the 7th term is:

u7 = 8 × 1.56

= 91.125

What is the 7th term of the geometric sequence 8, 12, 18, 27, …?

Using the 3rd term: ar2 = 36Using the 6th term: ar5 = 972

The 3rd term in a geometric sequence is 36 and the 6th term is 972. What is the value of the 1st term and the common ratio?


Geometric sequences

Dividing these gives:

Substituting this into ar2 = 36 gives:

5

2

972=

36

ar

arSo: r3 = 27

r = 3

a × 32 = 36

9a = 36

a = 4

So the first term of the sequence is 4 and the common ratio is 3.

The general term of this sequence is un = 4 × (3)n–1.


Find the missing terms


Convergent and divergent sequences

Geometric sequences either converge or diverge depending on the value of the common ratio r.

8, 2, 12 , 1

8 , 132 , 1

128 , ...

Suppose the first value of a geometric sequence is 8 and the common ratio is . This gives the following sequence:1

4

This sequence converges to 0.

6, –3, 32 , 3

4 , 38 , 3

16 , ...

If the common ratio is negative the signs of the terms will alternate. For example, if the first value of a geometric sequence is 6 and the common ratio is , we have:1

2

This sequence also converges to 0.



In general, if the common ratio r of a geometric sequence is between 0 and 1 or between –1 and 0, the terms of the sequence will converge to 0.

For a geometric sequence, if 0 < |r| < 1 the sequence will converge.

For a geometric sequence, if 0 < |r| < 1 the sequence will converge.

We use |r| to represent the modulus of r, mod r.

This is the numerical value of r, regardless of whether it is positive or negative.

So, for example:

|0.6| = 0.6 and |–0.6| = 0.6

In general:



For example, if the first value of a geometric sequence is 0.2 and the common ratio is –5, the sequence will be:

This means that the terms will get larger and larger without limit.

For a geometric sequence, if |r| > 1 the sequence will diverge.

For a geometric sequence, if |r| > 1 the sequence will diverge.

0.2, –1, 5, –25, 125, –625, …

This sequence is divergent.

The only geometric sequences that neither converge nor diverge are those where the common ratio is –1. For example:

This sequence oscillates between two values.

4, –4, 4, –4, 4, –4, …


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Geometric series

Geometric sequences

Geometric series




Geometric series

The sum of all the terms of a geometric sequence is called a geometric series.

We can write the sum of the first n terms of a geometric series as:

When n is large, a more systematic approach for calculating the sum of a given number of terms is required.

Sn = a + ar + ar2 + ar3 + … + arn–1 Sn = a + ar + ar2 + ar3 + … + arn–1

For example, the sum of the first 5 terms of the geometric series with first term 2 and common ratio 3 is:

S4 = 2 + (2 × 3) + (2 × 32) + (2 × 33) + (2 × 34)

= 2 + 6 + 18 + 54 + 162

= 242


The sum of a geometric series

Start by writing the sum of the first n terms of a general geometric series with first term a and common ratio r as:

Multiplying both sides by r gives:

Sn = a + ar + ar2 + ar3 + … + arn–1

rSn = ar + ar2 + ar3 + … + arn–1 + arn

Now if we subtract the first equation from the second we have:

rSn – Sn= arn – a

Sn(r – 1) = a(rn – 1)

( 1)=

1

n

n

a rS

r


The sum of a geometric series

If we multiply the numerator and the denominator by –1 we can also write the sum of the first n terms as:

(1 )=

1

n

n

a rS

r

This form is more useful when |r| < 1 since it avoids the use of negative numbers.

For example:

Find the sum of the first 8 terms of the geometric series that starts 4 – 3 + 2 –, …1

4


Using Σ notation

71

=1

5(2)i

i

Find

This is a geometric series with first term 5 and common ratio 2.

71 6

=1

5(2) = 5 +10 + 20 +...+ 5(2)i

i

There are 7 terms in this sequence so:77

1

=1

5(2 1)5(2) =

2 1i

i

= 635

We can write the sum of a geometric series using Σ notation as:

1

=1

( 1) (1 )= =

1 1

n nni

i

a r a rar

r r


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Geometric sequences

Geometric series





When the common ratio of a geometric series is between –1 and 1, the sum of the series will tend to a particular value as more terms are added.

For example, the geometric series1 1 1 12 4 8 161 + + + + + ...

tends to 2 as the number of terms increases.

We can show this diagrammatically as follows:

11

12

14

2

18

116 1

2



If we use the formula Sn = with a = 1 and r = we have

(1 )

1

na r

r

12

12

12

1(1 ( ) )=

1

n

nS

12

12

1 ( )=

n

112= 2 ( )n

1

1= 2

2n

As n 1

1, 0.

2n

So: = 2S where = lim nnS S



(1 )=

1

n

n

a rS

r

But if |r| < 1,

=1

aS

r

In general, the sum of the first n terms of a geometric series is:

0 as nr n

In this case, we can write the sum to infinity as:

For example:

Find the sum to infinity of the geometric series with first term 6 and common ratio –0.2.

6=

1+ 0.2S = 5



The first term of a geometric series is 20 and the sum to infinity is 15. What is the common ratio?

Using we have=1

aS

r 2015 =

1 r15(1 ) = 20r

201 =

15r

20=1

15r

1=

3r


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Geometric sequences

Geometric series




Examination-style question 1

a) Using the 2nd term: ar = 40

Using the 5th term: ar4 = 20.48

The 2nd term of a geometric series is 40 and the 5th term is 20.48.

a) Find the value of the first term and the common ratio.

b) Calculate the sum of the first 10 terms of the series.

c) Calculate the sum to infinity.

Dividing these gives:4 20.48

=40

ar

ar

So: r3 = 0.512

r = 0.8



Substituting this into ar = 40 gives:

a × 0.8 = 40

a = 50

b) Using with n = 10, a = 50 and r = 0.8 gives:( 1)

=1

n

n

a rS

r

c) Using with a = 50 and r = 0.8 gives:=1

aS

r

10

10

50(0.8 1)=

0.8 1S

= 223.16 (to 2 d.p.)

50=

1 0.8S

= 250



a) Write down the first four terms in the expansion of (1 + ax)13 in ascending powers of x, where a > 0.

b) Given that in the expansion of (1 + ax)13 the coefficient of x is –b and the coefficient of x2 is 12b, find the value of a and b.

a) (1 + ax)13 =

= 1 + 13ax + 78a2x2 + 286a3x3 + …

2 313×12 13×12×111+13 + ( ) + ( ) +...

2 3×2ax ax ax

b) 13a = –b 1

78a2 = 12b 2

78a2 = 12 × –13a

Substituting into :1 2 78a = –156

a = –2

b = 26

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