# Notes on arithmetic and geometric sequences, suitable for HKCEE

```Carto Wong
Notes for HKCEE Mathematics
Arithmetic and Geometric Sequence/Series
Recall that a sequence is a function defined on {1, 2, 3, } and usually denoted by one
of the following notations:
{xn }n 1 ,
{xn } ,
{x1 , x2 , x3 , } .
For example, {2n  1}n 1 , {2n  1} and {1, 3, 5, } represent the same sequence.
Definition. (AS and GS)
A sequence {xn } is called an arithmetic sequence（算術數列） if
( n  1, 2, 3, ),
xn1  xn  d
where d is a constant. Arithmetic sequence is abbreviated to “A.S.”.
A sequence {xn } is called a geometric sequence（幾何數列） if
( n  1, 2, 3, ),
xn1  rxn
where r is a constant. Geometric sequence is abbreviated to “G.S.”.

d is called common difference（公差） and r is called common ratio（公比）.
General Term and Partial Sum.
The general term of an A.S. {xn } with common difference d is
xn  x1  (n  1)d .
The nth partial sum is
x1 
 xn 
n( x1  xn ) n
 [2 x1  (n  1)d ] .
2
2
The general term of a G.S. { yn } with common ratio r is
yn  y1 r n 1 .
The nth partial sum is
y1 
 yn 
y1 (1  r n )
.
1 r
If 1  r  1 , the sum to infinity is
y1  y2 

y1
.
1 r
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Carto Wong
Basic Techniques and Typical Examples
Example 1.
Let {xn } be an A.S. with x4  3 and common difference 5. Find the general term.
Solution.
The general term is xn  x1  5(n  1) . Put n  4 , we have
3  x1  5  3  x1  12 .
Hence, the general term is xn  12  5(n  1)  5n  17 .
The general term of any A.S. is of the form xn  pn  q (p, q are constants).
TIP.
Moreover, the constant p should be the common difference. If the answer you
obtained is not of this form, then it must be wrong.
Example 2.
Let {xn } be an A.S. with the nth partial sum S n  n 2  3n . Find the general term.
Solution.
The general term is xn  Sn  Sn 1  [(n  1) 2  3(n  1)]  [n 2  3n]  2n  2 .

TIP.
The nth partial sum S n of an A.S. must be a quadratic expression of n with constant
term zero. This allows us to adopt the convention S0  0 .
By examples 1 and 2, we know that the following information of an A.S. are
equivalent: (in the sense that if we know one then we know the others)
(i.) Particular term and common difference;
(ii.) Partial sum;
(iii.) General term.
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Carto Wong
Example 3.
Let { yn } be a G.S. with y4  2 and common ratio 3. Find the general term.
Solution.
The general term is yn  3n 1 y1 . Put n  4 , we have
2  33 y1  y1 
Hence, the general term is yn 
2
.
27
2
 3n 1 .
27
Example 4.
Let { yn } be a G.S. with the nth partial sum Tn 
3n  1
. Find the general term.
2
Solution.
The general term is yn  Tn  Tn 1 

TIP.
3n  1 3n 1  1 3n  3n 1 2  3n 1



 3n 1 .
2
2
2
2
Again, we adopt the convention T0  0 here.
In the examination, the question may ask you to find the nth partial sum S n (of an
A.S. or a G.S.). Once you get your answer, it is a good habit to put n  0 and
check if S0  0 . For example, if your answer is Sn  n 2  n  2 (note that
S0  2  0 ) then it must be wrong.
A similar figure in which we saw before:
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Carto Wong
Interesting Example.
Let ABC be an equilateral triangle with area 1 cm 2 and A , B  ,
C  are respectively the midpoints of BC, CA, AB. The midpoints
are joined to form ABC as shown in the figure.
A
C'
B'
(a) Find the area of ABC .
B
In each of the remaining triangles ACB , CBA , BAC .
We draw their medial triangles (which are shaded in the figure).
A
(b) Find the area of each shaded smaller triangle.
C'
(c) Repeat the process indefinitely. Find the total area of
C
A'
B
B'
C
A'
Solution.
(a) Since ABC is similar to ABC ,
 AB  
AB C   
  ABC
 AB 
1
  ABC
4
1
 cm 2
4
2
(b) By similarity again, the area of each shaded smaller triangle is
1 3
9
 

(c) Total area is
4 16 64


1
4
3
1
4
1 1 1
 
cm 2 .
4 4 16
 1 cm 2 .
The figure in the right is called Sierpinski Triangle（西

modern mathematics. The most interesting property of
Sierpinski Triangle is that it has non-integral dimension
log3 log 2  1.585 .
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Carto Wong
A sequence {xn } is defined recursively by
 x1  1

 xn1  2 xn  1 (n  1, 2, )
Find the general term of this sequence.

We write down the first few terms:
n
xn
1
2
3
4
5
6
7
1
3
7
15
31
63
127
This is neither A.S. nor G.S.!
Solution.
Let yn  xn  1 , then y1  2 and the recurrence relation becomes
yn 1  xn 1  1
 (2 xn  1)  1
 2( xn  1)
 2 yn
Note that { yn } is a G.S.! Its general term is yn  2n 1 y1  2n . Hence, xn  2n  1 is the
general term of the original sequence.
Further Investigation.
More generally, suppose a sequence is defined by
 x1  a

 xn1  pxn  q (n  1, 2, )
where a, p, q are given constants. (Note that A.S. and G.S. are sequences of this type.)
It is natural to ask: how to find the general term of this sequence?
(Think yourself and then see the solution in next page)
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Carto Wong
Solution.
If p  1 then the sequence is an A.S., this case is easy. We now consider the case p  1 .
q
Let yn  xn 
. Then { yn } is a G.S. with common ratio p. This is because
p 1
q
yn 1  xn 1 
p 1
q
 ( pxn  q ) 
p 1

q 
q
 p  yn 
q
p 1
p 1

 pyn
Hence, yn  p n 1 y1  xn  p n 1 y1 

q
q 
q
 p n 1  a 
.

p 1
p 1 p 1

Exercise.
Find the general term of the sequence {xn } defined by
 x1  2

 xn1  3xn  5 (n  1, 2, )
The first few terms are:
n
xn
1
2
3
4
5
6
7
2
1
2
11
38
119
362
5 3n1
.

2
2
Appendix: why 0.9 = 1 ?
What is the value of 0.999 ? May be somebody told you before the answer is 1. This
surprising result can be proved as follows:
The number 0.9 is, by definition, the sum of the series
0.9  0.09  0.009 
This is a geometric series with common ratio 0.1, the sum is

0.9
 1.
1  0.1
This example shows that the decimal representation（十進制表達式）of a number is not
unique in general.
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