1. 18 Arithmetic and Geometric Sequences Case Study
18.1 Introduction to Sequences
18.2 Arithmetic Sequence
18.3 Geometric Sequence
18.4 Summing an Arithmetic Sequence
18.5 Summing a Geometric Sequence
Chapter Summary

2. Case Study
If I drop a ball from 2 m above the ground, how high will it bounce up?
Let’s try it.
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
The height that the ball reaches in the second rebound  1.5 m  75%  m
This problem belongs to the topic ‘geometric sequence’.

3. 18.1 Introduction to Sequences
A sequence is a list of numbers which are arranged in a certain order.
For example, consider the sequence of square numbers:
1, 4, 9, 16, 25, ...
Each number in the sequence is called a term.
We usually use the notation T(n) to represent the nth term of the sequence.
For the sequence above, we have
T(1)  1, T(2)  4, T(3)  9, T(4)  16 and T(5)  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.

4. 18.1 Introduction to Sequences
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:
Solution 2:
The first term = 20 = 1
The first term = = 1
The second term = 21 = 2
The second term = = 2
The third term = 22 = 4
The third term = = 4
The fourth term = 23 = 8
The fourth term = = 7
The fifth term = 24 = 16
The fifth term = = 11

5. Example 18.1T 18.1 Introduction to Sequences 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.
Solution:
(a) T(1) = 5(1) – 2
T(4) = 5(4) – 2
(b) T(n) = 48
5n – 2 = 48

6. Example 18.2T 18.1 Introduction to Sequences Solution:
Find the general terms of the following sequences.
(There are more than one possible solution.)
2, , , , ...
(b) –2, 4, 10, 16, ...
Solution:
(a) T(1) = 2;
\ (or other reasonable answers)
(b) T(1) = –2;
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)

7. Example 18.3T 18.1 Introduction to Sequences 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
Solution:
(b) When T(n) = , we have
(a) When n = 4, we have

8. 18.2 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.
Consider the following arithmetic sequence:
2, 7, 12, 17, 22, …
In this sequence, the common difference is 5.

9. 18.2 Arithmetic Sequence
For an arithmetic sequence, if we let the first term be a and the common difference be d, then we have    
T(1) = a  
T(2) = T(1) + d = a + d  
 T(3) = T(2) + d = a + d + d = a + 2d
T(4) = T(3) + d = a + 2d + d = a + 3d …
T(n) = a + (n  1)d.  
From the above pattern, we can observe that
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 by
T(n) = 2 + (n  1)(5)
= 5n  3

10. Example 18.4T 18.2 Arithmetic Sequence 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.
Solution:
(a) Common difference d = T(2) – T(1)
= 1 – 4
= –3
Since a = 4 and d = –3, we have
T(n) = 4 + (n – 1)(–3)
= 4 – 3n + 3
(b) Substituting n = 15 into the general term, we have
T(15) = 7 – 3(15)

11. Example 18.5T 18.2 Arithmetic Sequence Solution:
Find the missing terms in the arithmetic sequence –20, , , , 4.
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 = 6
The three numbers are;
T(2) = –20 + 6
T(3) = –20 + 6(2)
T(4) = –20 + 6(3)

12. Example 18.6T 18.2 Arithmetic Sequence 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.
Solution:
(a) First term a = 110
Common difference d = 104 – 110
= –6
\ T(n) = (n – 1)(–6)
= 110 – 6n + 6

13. Example 18.6T 18.2 Arithmetic Sequence 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.
Solution:
(b) Let the number of positive terms be k, we have
T(k) = 116 – 6k > 0
6k < 116
k <
 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.

14. Example 18.7T 18.2 Arithmetic Sequence Solution:
Consider 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) = 88
\ a = 25
The general term T(n) = 25 + (n – 1)(7)

15. Example 18.7T 18.2 Arithmetic Sequence Solution:
Consider 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) = k = 123
7k = 105

16. Example 18.8T 18.2 Arithmetic Sequence Solution:
Mr. 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.

17. 18.2 Arithmetic Sequence
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.

18. Consider the following sequence:
18.3 Geometric Sequence
Apart from arithmetic sequence, there is another kind of sequence that has a special characteristic.
This is called geometric sequence.
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.
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.
In the above example, the common ratio is 3.

19. 18.3 Geometric Sequence
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.
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.

20. Example 18.9T 18.3 Geometric Sequence 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.
Solution:
(a) Common ratio r = T(2) ÷ T(1)
= 10 ÷ 5 = 2
Since a = 5 and r = 2, we have
(b) When n = 8, we have
T(8) = 5(2)8 – 1

21. Example 18.10T 18.3 Geometric Sequence Solution:
Insert two numbers between –2 and –54 such that they form a geometric sequence.
Solution:
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 = 3
The two numbers to be inserted are:

22. Example 18.11T 18.3 Geometric Sequence Solution:
Consider a geometric sequence , 2, , 4, ...
(a) Find the general term.
(b) Find the tenth term.
(c) Which term is equal to 128?
Solution:
(a) First term and common ratio
Since the general term is given by T(n) = arn – 1, we have
(b) When n = 10,
(c) Since T(n) = 128,
 The 14th term is equal to 128.

23. Example 18.12T 18.3 Geometric Sequence Solution:
In 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
(b) The seventh term
= ar6
When
 First term common ratio

24. Example 18.13T 18.3 Geometric Sequence Solution:
The population in a city is at the beginning of 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 ?
Solution:
(a) The populations form a geometric sequence with a = and r = 1 + 4% = 1.04.
 T(6) = (1.04)6 – 1 
= (cor. to 3 sig. fig.)
 The population of the city at the beginning of 2013 will be

25. Example 18.13T 18.3 Geometric Sequence Solution:
The population in a city is at the beginning of 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 ?
Solution:
(b) T(n) = (1.04)n – 1 >
1.04n – 1 > 1.6
log 1.04n – 1 > log 1.6
(n – 1)log 1.04 > log 1.6
n > 12.98
 The population will start exceed at the beginning of 2020.

26. 18.3 Geometric Sequence
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.

27. 18.4 Summing an Arithmetic Sequence
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.
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).
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.

28. Example 18.14T 18.4 Summing an Arithmetic Sequence Solution:
Find the sum of the arithmetic series … + (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

29. Example 18.15T 18.4 Summing an Arithmetic Sequence Solution:
How many terms of the arithmetic series … must be included such that the sum is 148?
Solution:
a = 29, d = 26 – 29 = 3

30. Example 18.16T 18.4 Summing an Arithmetic Sequence Solution:
In 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 = 76
a + 8d = 76…………….(1)
T(4) = a + (4 – 1)d = a + 3d
T(19) = a + (19 – 1)d = a + 18d
∵ T(4) + 1 = 2T(19)
a + 3d + 1 = 2(a + 18d)
a + 33d = 1…………………(2)
(2) – (1):

31. Example 18.16T 18.4 Summing an Arithmetic Sequence Solution:
In 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) = 76
(b) The sum of the first 19 terms

32. Example 18.17T 18.4 Summing an Arithmetic Sequence Solution:
(a) Find the sum of all natural numbers between 100 and inclusively that are divisible by 6.
(b) Hence find the sum of all natural numbers between 100 and inclusively that are not divisible by 6.
Solution:
(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.
\ 150 numbers between 100 and 1000 are divisible by 6.

33. Example 18.17T 18.4 Summing an Arithmetic Sequence Solution:
(a) Find the sum of all natural numbers between 100 and inclusively that are divisible by 6.
(b) Hence find the sum of all natural numbers between 100 and inclusively that are not divisible by 6.
Solution:
(b) For the series … , we have a = 100, l = and n = 901.
\ The sum of all natural numbers between 100 and 1000
inclusively that are not divisible by 6

34. 18.5 Summing a Geometric Sequence
A. Geometric Series
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 1). The sum of the first n terms can be written as:
S(n) = a + ar + ar arn – 2 + arn – 1……..(1)
Multiplying both sides of (1) by r, we have
rS(n) = ar + ar2+ ar arn – 1 + arn.……..(2)
(2) – (1):
rS(n) – S(n) = arn – a
(r – 1)S(n) = a(rn – 1)

35. 18.5 Summing a Geometric Sequence
A. Geometric Series
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 1.
If r < 1, we may use the following formula to calculate the sum.

36. Example 18.18T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.18T
Find the sum of the geometric series
Solution:
From the series, we have a = 1 and
Let n be the number of terms of the above series.

37. Example 18.19T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.19T
In a geometric series, the first term is 4 and the last term is –512. If the sum of the series is –340, find
the common ratio;
the number of terms.
Solution:
(a) Let r be the common ratio and n be the number of terms.
For the last term T(n), we have
For the sum of the series, we have
Substituting (1) into (2), we have

38. Example 18.19T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.19T
In a geometric series, the first term is 4 and the last term is –512. If the sum of the series is –340, find
the common ratio;
the number of terms.
Solution:
(b) Substituting r = –2 into (1), we have

39. Example 18.20T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.20T
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.
Solution:
For r  1,
Let y = r3, we have
When y = 1, r3 = 1.
r = 1 (rejected)
When y = 8, r3 = 8.
(2)  (1):
Substituting r = 2 into (1),

40. Example 18.21T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.21T
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 $ 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.)
Solution:
Let T(n) be the amount obtained from the nth deposit after ten years.
 T(1), T(2), …, T(n) form a geometric sequence with
a = (1.05)10 and

41. Example 18.21T 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Example 18.21T
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 $ 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.)
Solution:
Total amount
(cor. to 3 sig. fig.)

42. 18.5 Summing a Geometric Sequence
B. Sum to Infinity
We have just learnt that the sum of the first n terms of a
geometric sequence is , where r ¹ 1.
We can express the sum of the geometric series as
For 1 < r < 1, arn tends to 0 when n is getting large infinitely. As a result, the second term vanishes.
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
where ‘’ represents ‘infinity’ and S() represents the sum to infinity of a geometric series when n tends to infinity.

43. Example 18.22T 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Example 18.22T
Express the recurring decimal as a fraction.
Solution:
We can regard the part … as a sum to infinity
of a geometric series with the first term 0.07 and the common ratio

44. Example 18.23T 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Example 18.23T
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.
Solution:
Let a be the first term and r be the common ratio.
Substituting into (1), we have

45. Example 18.24T 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Example 18.24T
Consider 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, A4B4C4D4, … 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. A2 B2 D2 C1
A B1
D C1
Solution:
(a)
By Pythagoras’ theorem, we have
Since A2B2C2D2 is a square, we have

46. Example 18.24T 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Example 18.24T
Solution:
(b)(i) The perimeter of A1B1C1D1 = (8  4)cm
= 32cm
The perimeter of A2B2C2D2 A2 B2 D2 C1
A B1
D C1
By Pythagoras’ theorem, we have
The perimeter of A3B3C3D3 = (5  4)cm
= 20cm

47. Example 18.24T 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Example 18.24T
Solution:
The perimeters of the squares form a geometric sequence
with the first term 32 and common ratio A2 B2 D2 C1
A B1
D C1
(ii) The sum of perimeters

48. Chapter Summary 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.

49. Chapter Summary 18.2 Arithmetic Sequence
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.

50. Chapter Summary 18.3 Geometric Sequence
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 is
T(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.

51. Chapter Summary 18.4 Summing an Arithmetic Sequence
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.

52. Chapter Summary 18.5 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.
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.

53. Follow-up 18.1 18.1 Introduction to Sequences Solution:
The general term of a sequence is T(n) = 7n + 3.
(a) Find the second and the fifth terms.
(b) Find the value of n when T(n) = 108.
Solution:
(a) T(2) = 7(2) + 3
T(5) = 7(5) + 3
(b) T(n) = 108
7n + 3 = 108

54. Follow-up 18.2 18.1 Introduction to Sequences Solution:
Find the general terms of the following sequences. (There are more than one possible solution.)
(a) 1, , , , ... (b) 40, 35, 30, 25, ...
Solution:
(a) T(1) = 1;
(b) T(1) = 40
T(2) = 35 = 40 – 5 = 40 – 5(2 – 1)
T(3) = 30 = 40 – 10 = 40 – 5(3 – 1)
T(4) = 25 = 40 – 15 = 40 – 5(4 – 1)
\ T(n) = 40 – 5(n – 1)
(or other reasonable answers)
(or other reasonable answers)

55. Follow-up 18.3 18.1 Introduction to Sequences Solution:
Suppose the general term of a sequence is T(n) = k + 2n2 and the fifth term is 54.
(a) Find the value of k.
(b) Using the result in (a), find the value of n when T(n) = 102.
Solution:
(a) When n = 5, we have
(b) When T(n) = 102, we have
For any sequences, n must be a positive integer.
or –7 (rejected)

56. Follow-up 18.4 18.2 Arithmetic Sequence Solution:
Consider an arithmetic sequence with the first term –5 and the second term –1.
(a) Find the general term of the sequence.
(b) Find the 15th term.
Solution:
(a) Common difference d = T(2) – T(1)
= –1 – (–5)
= 4
Since a = –5 and d = 4, we have
T(n) = –5 + (n – 1)  4
= –5 + 4n – 4
(b) Substituting n = 15 into the general term, we have
T(15) = 4(15) – 9

57. Follow-up 18.5 18.2 Arithmetic Sequence Solution:
Find the missing terms in the arithmetic sequence –2, , , , , –17.
Solution:
Let a be the first term and d be the common difference. We have –2 as the first term and –17 as the sixth term.
T(1) = a = –2 and T(6) = a + 5d = –17
\ –2 + 5d = –17
5d = –15
d = –3
The four numbers are:

58. Follow-up 18.6 18.2 Arithmetic Sequence Solution:
Consider an arithmetic sequence 90, 86, 82, 78, ..., 14.
(a) Find the general term.
(b) Find the number of terms in the above sequence.
(c) Find the greatest term that is less than 40.
Solution:
(a) First term a = 90
(c) Let m be an integer such that T(m) < 40.
Common difference d = 86 – 90 = –4
\ T(n) = 90 + (n – 1)(–4)
= 90 – 4n + 4
(b) Let the number of terms be k.
 The smallest value of m is 14.
We have T(k) = 94 – 4k = 14
–4k = –80
The greatest term that is less than 40 is
k = 20
There are 20 terms in the above sequence.

59. Follow-up 18.7 18.2 Arithmetic Sequence Solution:
The fourth term and the eighth term of an arithmetic sequence are 163 and 127 respectively.
(a) Find the general term.
(b) Find the 30th term.
Solution:
(a) Let the general term be T(n) = a + (n – 1)d.
We have
The general term T(n)
(2) – (1):
Substituting d = –9 into (1), we have
(b) T(30) = 199 – 9(30)

60. Follow-up 18.8 18.2 Arithmetic Sequence Solution:
In a theatre, the number of seats in each row form an arithmetic sequence. The fourth row has 15 seats and each row has 3 seats less than the succeeding one. Find the row which has 54 seats.
Solution:
Since each row has 3 seats less than the succeeding one, the common difference is 3.
Let T(n) = a + 3(n – 1) be the number of seats in the nth row. We have
Suppose there are 54 seats in the mth row. We have
 The 17th row has 54 seats.

61. Follow-up 18.9 18.3 Geometric Sequence Solution:
Consider a geometric sequence with the first term 128 and the second term 64.
(a) Write down the general term.
(b) Find the sixth term.
Solution:
(a) Common ratio r = T(2) ÷ T(1)
(b) When n = 6, we have
= 64 ÷ 128
Since a = 128 and we have

62. Follow-up 18.10 18.3 Geometric Sequence Solution:
Insert two numbers between –32 and such that they form a geometric sequence.
Solution:
Let a be the first term and r be the common ratio. We have –32 as
the first term and as the fourth term.
T(1) = a = –32 and
The two numbers to be inserted are:

63. Follow-up 18.11 18.3 Geometric Sequence Solution:
Consider the geometric sequence 486, 162, 54, 18, ... .
(a) Find the general term.
(b) Find the eighth term.
(c) Which term is equal to ?
Solution:
(a) First term a = 486 and common ratio
Since the general term is given by T(n) = arn – 1, we have
(b) When n = 8,

64. Follow-up 18.11 18.3 Geometric Sequence Solution:
Consider the geometric sequence 486, 162, 54, 18, ...
(a) Find the general term.
(b) Find the eighth term.
(c) Which term is equal to ?
Solution:
(c) Since
\ The 11th term is equal to

65. Follow-up 18.12 18.3 Geometric Sequence Solution:
In a geometric sequence, the third term and the sixth term are 8 and –64 respectively.
(a) Find the first term and the common ratio.
(b) Find the ninth term.
Solution:
(a) Let a be the first term and r be the common ratio. We have
(2) ÷ (1):
(b) The ninth term = ar8
= (2)(–2)8
When r = –2,
 First term common ratio

66. Follow-up 18.13 18.3 Geometric Sequence Solution:
At 1:00 pm, the number of bacteria of a specimen is 300. If the bacteria increase at a rate of 15% per hour, when will the number of the bacteria be doubled? (Give the answer correct to the nearest hour.)
Solution:
The numbers of bacteria form a geometric sequence with a = 300 and r = 115% = 1.15.
 The number of bacteria will be doubled after 6 hours, that is, at 7:00 pm.

67. Follow-up 18.14 18.4 Summing an Arithmetic Sequence Solution:
Find the sum of the arithmetic series –10 + (–7) + (–4)
Solution:
Since a = –10, d = –7 – (–10) = 3 and T(n) = 32, we have
32 = –10 + (n – 1)(3)
42 = 3(n – 1)
n = 15

68. Follow-up 18.15 18.4 Summing an Arithmetic Sequence Solution:
How many terms of the arithmetic series with the first term –2 and the common difference 5 must be included such that the sum is 568?
Solution:
Since a = –2 and d = 5, we have
or –14.2 (rejected)

69. Follow-up 18.16 18.4 Summing an Arithmetic Sequence Solution:
In an arithmetic sequence, the fifth term and the tenth term are 24 and 39 respectively. Find the sum of the first 15 terms.
Solution:
(2) – (1):
Substituting d = 3 into (1), we have
The sum of the first 15 terms

70. Follow-up 18.17 18.4 Summing an Arithmetic Sequence Solution:
(a) Find the sum of all natural numbers between 1 and 500 inclusively that are divisible by 5.
(b) Hence find the sum of all natural numbers between 1 and 500 inclusively that are not divisible by 5.
Solution:
(a) The numbers that are divisible by 5 are 5, 10, 15, … . These form an arithmetic sequence with first term a = 5 and common difference d = 5. The largest number between 1 and 500 that is divisible by 5 is 500.
 There are 100 numbers between 1 and 500 that are divisible by 5.
(b) For the series … + 500, we have a = 1, l = 500 and n = 500.
 The sum of all natural numbers between 1 and 500 inclusively that are not divisible by 5

71. Follow-up 18.18 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.18
Find the sum of the geometric series 2 – – – 256.
Solution:
From the series, we have a = 2 and
Let n be the number of terms of the above series.

72. Follow-up 18.19 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.19
In a geometric series, the first term is 1458 and the last term is 2. If the sum of the series is 2186, find
(a) the common ratio; (b) the number of terms.
Solution:
(a) Let r be the common ratio and n be the number of terms.
For the last term T(n), we have
Substituting (1) into (2),
we have
For the sum of the series, we have

73. Follow-up 18.19 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.19
In a geometric series, the first term is 1458 and the last term is 2. If the sum of the series is 2186, find
(a) the common ratio; (b) the number of terms.
Solution:
(b) Substituting into (1), we have

74. Follow-up 18.20 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.20
In a geometric sequence, the sum of the first three terms is 48, and the sum of the next three terms is –6. Find the first term and the common ratio.
Solution:
For r  1,
Let y = r3, we have
When y = 1, r3 = 1.
 r = 1 (rejected)
When

75. Follow-up 18.20 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.20
In a geometric sequence, the sum of the first three terms is 48, and the sum of the next three terms is –6. Find the first term and the common ratio.
Solution:
Substituting into (1),

76. Follow-up 18.21 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.21
Miss Wong has opened a savings account. She will make a monthly contribution of $2500 on the first day of each month for 2 years. The interest is compounded monthly at an interest rate of 9% per annum. Find the total amount that she can get after two years. (Give the answer correct to 3 significant figures.)
Solution:
Let T(n) be the amount obtained from the nth deposit after 2 years.
 T(1), T(2), …, T(n) form a geometric sequence with a = 2500(1.0075) and r =

77. Follow-up 18.21 18.5 Summing a Geometric Sequence Solution:
A. Geometric Series
Follow-up 18.21
Miss Wong has opened a savings account. She will make a monthly contribution of $2500 on the first day of each month for 2 years. The interest is compounded monthly at an interest rate of 9% per annum. Find the total amount that she can get after two years. (Give the answer correct to 3 significant figures.)
Solution:
Total amount
(cor. to 3 sig. fig.)

78. Follow-up 18.22 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Follow-up 18.22
Express the recurring decimal as a fraction.
Solution:
We can regard it as a sum to infinity of a geometric series with the
first term 0.35 and the common ratio

79. Follow-up 18.23 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Follow-up 18.23
If the sum to infinity of a geometric sequence is 54 and the second term is 12, find the fifth term.
Solution:
Let a be the first term and r be the common ratio.
When
Substituting (1) into (2), we have
When

80. Follow-up 18.24 18.5 Summing a Geometric Sequence Solution:
B. Sum to Infinity
Follow-up 18.24
The figure shows an equilateral triangle A1B1C1 with side
12 cm. The mid-points of each side are joined to form another equilateral triangle. If the process is repeated infinitely, find the sum of the perimeters of all the triangles.
Solution:
Since A2, B2 and C2 are mid-points,
As A1B1C1 is equilateral, A2B2C2 is also equilateral, that is A1B1C1 ~ A2B2C2.
Similarly, A2B2C2 ~ A3B3C3 ~ A4B4C4…
Ratio of the perimeters of two successive triangles
Perimeter of A1B1C1 = (3  12) cm = 36 cm
Common ratio
 Sum of the perimeters