18 18.1Introduction to Sequences 18.2Arithmetic Sequence 18.3Geometric Sequence Chapter Summary Case Study Arithmetic and Geometric Sequences 18.4Summing

18

18.1 Introduction to Sequences

18.2 Arithmetic Sequence

18.3 Geometric Sequence

Chapter Summary

Case Study

Arithmetic and Geometric Sequences

18.4 Summing an Arithmetic Sequence

18.5 Summing a Geometric Sequence

P. 2

Eric and Andy find that the ball bounces up to 75% of its previous height in each rebound.Thus, the height that the ball reaches in the first rebound 2 m 75%

1.5 m

Case StudyCase Study

The height that the ball reaches in the second rebound 1.5 m 75% 1.123 m

If I drop a ball from 2 m above the ground, how high will it bounce up?

This problem belongs to the topic ‘geometric sequence’.

Let’s try it.

P. 3

Each number in the sequence is called a term.

18.1 18.1 Introduction to SequencesIntroduction to Sequences

We usually use the notation T(n) to represent the nth term of the sequence.

A sequence is a list of numbers which are arranged in a certain order.

For the sequence above, we have T(1) 1, T(2) 4, T(3) 9, T(4) 16 and T(5) 25.

For example, consider the sequence of square numbers: 1, 4, 9, 16, 25, ...

The sequence of square numbers can also be expressed as 12, 22, 32, 42, 52, ...

Hence, T(1) 12, T(2) 22, T(3) 32, ...

The nth term of the sequence is T(n) n2, which is called the general term of the sequence.

P. 4


Remarks:When we consider the pattern of a sequence and find the subsequent terms, there can be more than one possible solution.

For example, we are going to find the fourth and the fifth terms of the sequence 1, 2, 4, ... .

Solution 1:

The first term = 20 = 1The second term = 21 = 2The third term = 22 = 4The fourth term = 23 = 8The fifth term = 24 = 16

Solution 2:

The first term = 0 + 1 = 1 The second term = 0 + 1 + 1 = 2 The third term = 0 + 1 + 1 + 2 = 4 The fourth term = 0 + 1 + 1 + 2 + 3 = 7 The fifth term = 0 + 1 + 1 + 2 + 3 + 4 = 11

P. 5

Example 18.1T

Solution:


The general term of a sequence is T(n) = 5n – 2. (a) Find the first and the fourth terms. (b) Find the value of n when T(n) = 48.

(a) T(1) = 5(1) – 2 3

T(4) = 5(4) – 2 18

(b) T(n) = 48

5n – 2 = 4810n

P. 6

Example 18.2T


Solution:

Find the general terms of the following sequences.(There are more than one possible solution.)

(a) 2, , , , ...

(b) –2, 4, 10, 16, ... 125

2

25

2

5

2

(a) T(1) = 2; ;5

12

5

2)2(

T ;

5

12

25

2)3(

2

T

3

5

12

125

2)4(

T

(b) T(1) = –2;

1

5

12)(

n

nT (or other reasonable answers)

T(2) = 4 = –2 + 6(2 – 1);

T(3) = 10 = –2 + 6(3 – 1); T(4) = 16 = –2 + 6(4 – 1)

T(n) = –2 + 6(n – 1)(or other reasonable answers) 86 n

P. 7

Example 18.3T


Solution:

Suppose the general term of a sequence is and the

fourth term is .

(a) Find the value of k. (b) Using the result in (a), find the value of n when .

kn

nnT

)(

5

2

8

5)( nT

(a) When n = 4, we have

5

2

4

4

k6k

(b) When T(n) = , we have 8

5

8

5

6

n

n

3058 nn303 n

10n

P. 8

Consider the following arithmetic sequence:2, 7, 12, 17, 22, …

In this sequence, the common difference is 5.

18.2 Arithmetic Sequence18.2 Arithmetic Sequence

In previous section, we learnt some kinds of sequences. Now we are going to study a particular kind of sequence which has a special pattern.

We call a sequence with the same difference between any two successive terms and arithmetic sequence and the difference is known as the common difference.

P. 9

For an arithmetic sequence, if we let the first term be a and the common difference be d, then we have


The general term T(n) of an arithmetic sequence is given by T(n) = a + (n 1)d,

where a is the first term and d is the common difference.

For the sequence in the previous page, the general term is given byT(n) = 2 + (n 1)(5)

= 5n 3

T(1) = a T(2) = T(1) + d = a + d

T(3) = T(2) + d = a + d + d = a + 2d

T(n) = a + (n 1)d. From the above pattern, we can observe that

T(4) = T(3) + d = a + 2d + d = a + 3d

…

P. 10

Example 18.4T

Solution:


Consider an arithmetic sequence with the first term 4 and the second term 1. (a) Find the general term of the sequence. (b) Find the 15th term.

(a) Common difference d = T(2) – T(1)

Since a = 4 and d = –3, we have T(n) = 4 + (n – 1)(–3)

= 4 – 3n + 3n37

(b) Substituting n = 15 into the general term, we have T(15) = 7 – 3(15)

38

= 1 – 4= –3

P. 11

Example 18.5T

Solution:


Let a be the first term and d be the common difference. We have –20 as the first term and 4 as the fifth term. T(1) = a = –20 and T(5) = a + 4d = 4 –20 + 4d = 4 4d = 24 d = 6The three numbers are;

T(2) = –20 + 6 14T(3) = –20 +

6(2) 8

T(4) = –20 + 6(3)

2

Find the missing terms in the arithmetic sequence –20, , , , 4.

P. 12

Example 18.6T

Solution:


Consider the arithmetic sequence 110, 104, 98, 92, .... (a) Find the general term. (b) How many positive terms are there in the sequence? (c) Find the first negative term in the sequence.

(a) First term a = 110 Common difference d = 104 – 110

= –6T(n) = 110 + (n – 1)(–6)

= 110 – 6n + 6n6116

P. 13

Example 18.6T

Solution:


Consider the arithmetic sequence 110, 104, 98, 92, .... (a) Find the general term. (b) How many positive terms are there in the sequence? (c) Find the first negative term in the sequence.

(b) Let the number of positive terms be k, we have T(k) = 116 – 6k > 0

6k < 116 k < 19.333

There are 19 positive terms in the sequence.

(c) Since T(19) is the last positive term, T(20) will be the first negative term. T(20) = 116 – 6(20) = –4

The first negative term is –4.

P. 14

Example 18.7TConsider an arithmetic sequence with T(10) = 88 and T(3) + T(5) = 92. (a) Find the general term. (b) Find the value of k if T(k) = 123.

Solution:


(a) Let the general term be T(n) = a + (n – 1)d, we have T(10) = a + (10 – 1)d = 88

a + 9d = 88…………………(1)T(3) + T(5) = a + (3 – 1)d + a + (5 – 1)d = 92

2a + 6d = 92 a + 3d = 46…………………………...(2)

(1) – (2): 6d = 42 d = 7

Substituting d = 7 into (1), we have a + 9(7) = 88a = 25

The general term T(n) = 25 + (n – 1)(7) n718

P. 15

Example 18.7TConsider an arithmetic sequence with T(10) = 88 and T(3) + T(5) = 92. (a) Find the general term. (b) Find the value of k if T(k) = 123.

Solution:


(b) T(k) = 18 + 7k = 1237k = 105

15k

P. 16

Example 18.8TMr. Chan gives some candies to a group of students, one student at a time. Each student will get 4 candies more than the preceding one. If the sixth student gets 40 candies, find the number of candies the first student gets.

Solution:


Since each student will get 4 candies more than the preceding one, the common difference is 4.

Let T(n) = a + 4(n – 1) be the number of candies that the nth student gets. We have

T(6) = a + 4(5) = 40 a = 20

The first student gets 20 candies.

P. 17


For any arithmetic sequence, we have the following properties.

1.

2. If T(1), T(2), T(3), … is an arithmetic sequence, then kT(1) + a, kT(2) + a, kT(3) + a, … is also an arithmetic sequence, where k and a are constants.

.1 where,)1()1(2

1)( nnTnTnT

P. 18

Consider the following sequence: 2, 6, 18, 54, ...

The first term is 2, and starting from the second term, each term is three times of the preceding term.

18.3 Geometric Sequence18.3 Geometric Sequence

Thus, the ratio between any two successive terms is 3, which is a constant.

We call a sequence with the same ratio for any two successive terms a geometric sequence, and the ratio is called the common ratio.

Apart from arithmetic sequence, there is another kind of sequence that has a special characteristic. This is called geometric sequence.

In the above example, the common ratio is 3.

P. 19


The general term T(n) of a geometric sequence is given by T(n) = arn 1,

where a is the first term and r is the common ratio.

For a geometric sequence, if we let the first term be a and the common ratio be r, then we have

T(1) = a

T(2) = T(1) r = ar

T(3) = T(2) r = ar2

T(4) = T(3) r = ar2 r = ar3

From the above pattern, we can observe that

T(n) = arn 1.

P. 20

Example 18.9T


Solution:

Consider a geometric sequence with the first term 5 and the second term 10. (a) Write down the general term. (b) Find the 8th term.

(a) Common ratio r = T(2) ÷ T(1) = 10 ÷ 5 = 2

Since a = 5 and r = 2, we have1)2(5)( nnT

(b) When n = 8, we have T(8) = 5(2)8 – 1

640

P. 21

Example 18.10T

Solution:

Insert two numbers between –2 and –54 such that they form a geometric sequence.

Let a be the first term and r be the common ratio. We have –2 as the first term and –54 as the fourth term. T(1) = a = –2 and T(4) = ar3 = –54

–2r3 = –54

r3 = 27 r = 3The two numbers to be inserted are:

632)2( T

1832)3( 2 T


P. 22

Example 18.11T

Solution:

Consider a geometric sequence , 2, , 4, ... (a) Find the general term. (b) Find the tenth term. (c) Which term is equal to 128?

2 22

(a) First term and common ratio 2a 222 rSince the general term is given by T(n) = arn – 1, we have

nnnT )2()2(2)( 1

(b) When n = 10, 10)2()10( T 3225

(c) Since T(n) = 128,128)2( n

72 22 n

72

n

14n

The 14th term is equal to 128.


P. 23

Example 18.12TIn a geometric sequence, the second term and the fifth term are –64 and 8 respectively. (a) Find the first term a and the common ratio r. (b) Find the seventh term. Solution:

(a) Let a be the first term and r be the common ratio, we have

)2..(..........8

)1........(644

ar

ar

:)1()2(

8

13 r

2

1r

When 128,2

1 ar

First term common ratio ,1282

1

(b) The seventh term = ar6

6

2

1128

2


P. 24

Example 18.13TThe population in a city is 250 000 at the beginning of 2008. It is estimated that the population grows at 4% per annum. (a) What will the population of the city at the beginning of 2013 be? (Give

the answer correct to 3 significant figures.) (b) In which year will the population start exceed 400 000?

Solution:


(a) The populations form a geometric sequence with a = 250 000 and r = 1 + 4% = 1.04.

T(6) = 250 000(1.04)6 – 1

304 163 = 304 000 (cor. to 3 sig. fig.)

The population of the city at the beginning of 2013 will be 304 000.

P. 25

Example 18.13TThe population in a city is 250 000 at the beginning of 2008. It is estimated that the population grows at 4% per annum. (a) What will the population of the city at the beginning of 2013 be? (Give

the answer correct to 3 significant figures.) (b) In which year will the population start exceed 400 000?

Solution:


(b) T(n) = 250 000(1.04)n – 1 > 400 0001.04n – 1 > 1.6

log 1.04n – 1 > log 1.6

The population will start exceed 400 000 at the beginning of 2020.

(n – 1)log 1.04 > log 1.6

04.1log

6.1log1 n

n > 12.98

P. 26

For any geometric sequences, we have the following properties.


1. T(n)2 = T(n 1) T(n + 1), where n > 1.2. If T(1), T(2), T(3), … is a geometric sequence, then kT(1), kT(2), kT(3), … is also a geometric sequence, where k is a constant.

P. 27

The evaluated value of the series, which is denoted by S(n), is called the sum of the series, that is,

S(n) = T(1) + T(2) + T(3) + … + T(n).

Consider a sequence T(1), T(2), T(3), …, T(n). The sum of the first n terms of the sequence, that is T(1) + T(2) + T(3) + … + T(n), is called a series.

18.4 Summing an Arithmetic Sequence18.4 Summing an Arithmetic Sequence

If T(1), T(2), T(3), …, T(n) is an arithmetic sequence, then T(1) + T(2) + T(3) + … + T(n) is an arithmetic series.

Sum of an arithmetic sequence with n terms:

or

where a is the first term, l is the last term and d is the common difference.

dnan

nS )1(22

)(

)(2

)( lan

nS

P. 28

Example 18.14TFind the sum of the arithmetic series 10 + 5 + 0 + … + (20).

Solution:Since a = 10, d = 5 – 10 = 5 and T(n) = 20, we have

20 = 10 + (n – 1)(5)

30 = 5n + 5

5n = 35

n = 7

)]20(10[2

7)7( S

35


P. 29

Example 18.15THow many terms of the arithmetic series 29 + 26 + 23 + … must be included such that the sum is 148?

Solution:a = 29, d = 26 – 29 = 3

)]3)(1()29(2[2

148 nn

)361(296 nn 2361296 nn

0296613 2 nn

0)373)(8( nn

(rejected) 3

37or 8n


P. 30

Example 18.16TIn an arithmetic sequence, the 9th term is 76 and the 4th term is smaller than twice of the 19th term by 1.(a) Find the first term and the common difference.(b) Find the sum of the first 19 terms. Solution:

(a) T(9) = a + (9 – 1)d = 76a + 8d = 76…………….(1)

T(4) = a + (4 – 1)d = a + 3dT(19) = a + (19 – 1)d = a + 18d∵ T(4) + 1 = 2T(19)

a + 3d + 1 = 2(a + 18d)a + 33d = 1…………………(2)

(2) – (1): 37525

dd


P. 31

Example 18.16TIn an arithmetic sequence, the 9th term is 76 and the 4th term is smaller than twice of the 19th term by 1.(a) Find the first term and the common difference.(b) Find the sum of the first 19 terms.

Solution:Substituting d = –3 into (1), we have

a + 8(–3) = 76100a

(b) The sum of the first 19 terms

)]3)(119()100(2[2

19

1387


P. 32

Example 18.17T

Solution:

(a) Find the sum of all natural numbers between 100 and 1000 inclusively that are divisible by 6. (b) Hence find the sum of all natural numbers between 100 and 1000 inclusively that are not divisible by 6.

(a) The numbers that are divisible by 6 are 102, 108, 114, … . These form an arithmetic sequence with first term a = 102 and common difference d = 6.

The smallest and the largest numbers between 100 and 1000 that are divisible by 6 are 102 and 996 respectively.

1496

102996

150 numbers between 100 and 1000 are divisible by 6.

)996102(2

150)150( S

350 82


P. 33

Example 18.17T

Solution:

(a) Find the sum of all natural numbers between 100 and 1000 inclusively that are divisible by 6. (b) Hence find the sum of all natural numbers between 100 and 1000 inclusively that are not divisible by 6.

(b) For the series 100 + 101 + … + 1000, we have a = 100, l = 1000 and n = 901.

The sum of all natural numbers between 100 and 1000

inclusively that are not divisible by 6

)1000100(2

901)901( S

550495

35082550495 200 413


P. 34

Multiplying both sides of (1) by r, we have

18.5 Summing a Geometric Sequence18.5 Summing a Geometric Sequence

In addition to the arithmetic series, we are also interested in calculating the sum of a geometric sequence, that is, the geometric series.

Consider a geometric sequence with the first term a and the common ratio r (r 1). The sum of the first n terms can be written as:

S(n) = a + ar + ar2 + ... + arn – 2 + arn – 1……..(1)

A. Geometric SeriesA. Geometric Series

rS(n) = ar + ar2+ ar3 + ... + arn – 1 + arn.……..(2) (2) – (1):

rS(n) – S(n) = arn – a

(r – 1)S(n) = a(rn – 1)

1

)1()(

r

ranS

n

P. 35


The sum to n terms of a geometric series with the first term a and the common ratio r is:

S(n) = , where r 1.

If r < 1, we may use the following formula to calculate the sum.

S(n) = , where r 1. r

ra n

1

)1(

1

)1(

r

ra n


P. 36

Example 18.18T

Solution:



Find the sum of the geometric series .243

1...

9

1

3

11

From the series, we have a = 1 and .3

1r

Let n be the number of terms of the above series.

243

1)( 1 narnT

243

1

3

1 1

n

51

3

1

3

1

n

651

nn

243

364

3

11

3

111

)6(

6

S

P. 37

Example 18.19T

Solution:

In a geometric series, the first term is 4 and the last term is –512. If the sum of the series is –340, find (a) the common ratio; (b) the number of terms.

(a) Let r be the common ratio and n be the number of terms. For the last term T(n), we have 512)(4 1 nr

)1(....................128rr n

For the sum of the series, we have 3401

)1(4

r

r n

)2..().........1(851 rr n

Substituting (1) into (2), we have )1(851128 rr85851128 rr

8643 r2r



P. 38

Example 18.19T

Solution:

In a geometric series, the first term is 4 and the last term is –512. If the sum of the series is –340, find (a) the common ratio; (b) the number of terms.

(b) Substituting r = –2 into (1), we have



82

256)2(128)2(

n

8 n

P. 39

Example 18.20T

Solution:

In a geometric sequence, the sum of the first three terms is and the sum

of the next three terms is 7. Find the first term and the common ratio.

8

7

For r 1,

)1.........(..........8

7

1

)1()3(

3

r

raS

78

7

1

)1()6(

6

r

raS

)2.......(..........8

63

(2) (1): 97

63

1

13

6

r

r

)1(91 36 rr0919 36 rr089 36 rr

Let y = r3, we have0892 yy0)8)(1( yy

8or 1yWhen y = 1, r3 = 1.

r = 1 (rejected) When y = 8, r3 = 8.

Substituting r = 2 into (1),



8

7

12

)12( 3

a

8

1 a

2 r

P. 40

Example 18.21T

Solution:

Mr. Cheung has deposited some money into a bank on the first day of a year for ten consecutive years. In the first year, he deposited $10 000. Starting from the second year, he deposited 8% less than that of the preceding year. Suppose the interest was compounded yearly and the interest rate is 5% per annum. Find the total amount that he can get after ten years. (Give the answer correct to 3 significant figures.)



Let T(n) be the amount obtained from the nth deposit after ten years.1010 )05.1(00010%)51(00010)1( T

99 )05.1)(92.0(00010%)51%)(81(00010)2( T8282 )05.1()92.0(00010%)51(%)81(00010)3( T

)05.1()92.0(00010%)51(%)81(00010)10( 99 T

T(1), T(2), …, T(n) form a geometric sequence with

a = 10 000(1.05)10 and 05.1

92.0r

P. 41

Example 18.21T

Solution:

Mr. Cheung has deposited some money into a bank on the first day of a year for ten consecutive years. In the first year, he deposited $10 000. Starting from the second year, he deposited 8% less than that of the preceding year. Suppose the interest was compounded yearly and the interest rate is 5% per annum. Find the total amount that he can get after ten years. (Give the answer correct to 3 significant figures.)



Total amount

05.1

92.01

05.1

92.01)05.1(00010

$

1010

50096$ (cor. to 3 sig. fig.)

P. 42

Thus, the sum to infinity of a geometric series with the first term a and the common ratio r (where 1 < r < 1) is given by


For 1 < r < 1, arn tends to 0 when n is getting large infinitely. As a result, the second term vanishes.

B. Sum to InfinityB. Sum to Infinity

where ‘’ represents ‘infinity’ and S() represents the sum to infinity of a geometric series when n tends to infinity.

We have just learnt that the sum of the first n terms of a

geometric sequence is , where r 1. r

ranS

n

1

)1()(

.11

)1()(

ra

ar

r

a

r

ranS

nn

We can express the sum of the geometric series as

P. 43

Example 18.22T

Solution:



0007.0007.007.01.0...17777.071.0

Express the recurring decimal as a fraction. 71.0

We can regard the part 0.07 + 0.007 + 0.0007 + … as a sum to infinity

of a geometric series with the first term 0.07 and the common ratio .10

1

10

11

07.01.071.0

45

890

7

90

9

P. 44

Example 18.23T

Solution:

The sum to infinity of a geometric sequence is three times that of the first term, and the third term is 8. Find the first term and the common ratio.



Let a be the first term and r be the common ratio.

ar

a3

1

3

11 r

3

2r

8)3( T

Substituting into (1), we have3

2r

83

2 2

a

18a

)1(82 ar

P. 45

Example 18.24TConsider a square A1B1C1D1 with side 8 cm. The points A2, B2, C2, D2 divide A1B1, B1C1, C1D1, D1A1 respectively in the ratio 1 : 3. (a) Find the length of the side of the square A2B2C2D2.(b) The above step is repeated infinitely and the squares A3B3C3D3, A4

B4C4D4, … are formed. (i) Show that the perimeters of the squares form

a geometric sequence. (ii) Find the sum of the perimeters of all the squares.



Solution:

cm 2cm 4

1821

AA cm 6cm

4

3821

DA

By Pythagoras’ theorem, we have cm 102cm )62( 2222 AD

Since A2B2C2D2 is a square, we have 22222222 ADDCCBBA

(a)

cm 102

A2

B2

D2

C1

A1 B1

D1 C1

P. 46

Example 18.24TSolution:



= 32cmThe perimeter of A2B2C2D2 cm 108cm )4102(

cm, 2

10cm

4

110232

AA

cm. 102

3cm

4

310232

DA

By Pythagoras’ theorem, we have

cm 5cm 2

1010

2

322

33

DA

= 20cm

4

10

32

108

ofPerimeter

ofPerimeter

1111

2222 DCBA

DCBA

(b)(i) The perimeter of A1B1C1D1 = (8 4)cm

The perimeter of A3B3C3D3 = (5 4)cm

A2

B2

D2

C1

A1 B1

D1 C1

P. 47

Example 18.24TSolution:



4

10

108

45

ofPerimeter

ofPerimeter

2222

3333 DCBA

DCBA

The perimeters of the squares form a geometric sequence

with the first term 32 and common ratio .4

10

(ii) The sum of perimeters

cm

4

101

32

cm 104

128

A2

B2

D2

C1

A1 B1

D1 C1

P. 48

18.1 Introduction to Sequences

A sequence is a list of numbers which are arranged in a certain order. T(n) is used to represent the nth term in a sequence.

Chapter Chapter SummarySummary

P. 49

1. An arithmetic sequence is arranged in the form:a, a + d, a + 2d, …,

where a is the first term and d is the common difference.


2. The general term of an arithmetic sequence is T(n) = a + (n 1)d.

3. Properties of an arithmetic sequence:

(a) , where n > 1.

(b) If T(1), T(2), T(3), … is an arithmetic sequence, then kT(1) + a, kT(2) + a, kT(3) + a, … is also an arithmetic sequence, where k and a are constants.

)1()1(2

1)( nTnTnT

18.2 Arithmetic Sequence

P. 50

1. A geometric sequence is arranged in the form:a, ar, ar2, …,

where a is the first term and r is the common ratio.


2. The general term of a geometric sequence isT(n) = arn 1.

3. Properties of geometric sequences: (a) T(n)2 = T(n 1) T(n + 1), where n > 1. (b) If T(1), T(2), T(3), … is a geometric sequence, then kT(1), kT(2), kT(3), … is also a geometric sequence, where k is a constant.

18.3 Geometric Sequence

P. 51


Sum of an arithmetic sequence with n terms:

or

where a is the first term, l is the last term and d is the common difference.

dnan

nS )1(22

)(

)(2

)( lan

nS

18.4 Summing an Arithmetic Sequence

P. 52

Chapter Chapter SummarySummary18.5 Summing a Geometric Sequence

1. The sum to n terms of a geometric series with the first term a and the common ratio r is:

S(n) = for r 1 or

S(n) = for r 1. r

ra n

1

)1(

1

)1(

r

ra n

2. The sum to infinity of a geometric series is given by

for 1 < r < 1,

where a is the first term, r is the common ratio. r

aS

1)(

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18 18.1Introduction to Sequences 18.2Arithmetic Sequence 18.3Geometric Sequence Chapter Summary Case Study Arithmetic and Geometric Sequences 18.4Summing