S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Chapter 2 Series: Arithmetic and Geometric Series
2.1
Series (級數)
Definition & Notation
Sequence: T(1), T(2), T(3), …, T(n)
Series:
2.2
Arithmetic Series
A. Definition
Arithmetic Sequence:
a, a  d , a  2d , a  3d ,...
Arithmetic Series:
B. Sum to n terms
Let
S(n) = the sum to n terms of the series
a
= the first term
d
= the common difference
l
= the last term
1
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
S (n) 
Example 1:
Find the sum of the first 100 natural numbers.
Solution:
Also, l  a  (n  1)d ,
S (n) 
Example 2:
Find the sum of the first 14 terms of the arithmetic sequence 2, 5, 8, 11, 14, 17, …
Solution:
Example 3:
A cheerleader pyramid has 15 girls on the bottom row, 14 on the next row, and so on
until there is 1 girl on top. How many cheerleaders are in the pyramid?
Solution:
2
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 4: (Ex 2A Qn 7a)
Find the number of terms in the following arithmetic series with the given sum.
15+9+3+… = -81
Solution:
Example 5:
Find the sum of all numbers divisible by five whose numerals have two digits.
Solution:
Example 6:
The sum to n terms of an arithmetic series is S n  2n 2  5n . Find
(a) the first term and the common difference of the corresponding arithmetic
sequence,
(b) the sum of all those terms in the sequence that are less than 1000.
Solution:
3
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 7:
Given an arithmetic sequence 33, 36, 39, …, 393.
(a)
(b)
(c)
(d)
(e)
How many terms of the arithmetic sequence are even?
Find the sum of all the terms of the arithmetic sequence that are even.
How many terms of the arithmetic sequence are divisible by 7?
Find the sum of all the terms of the arithmetic sequence that are divisible by 7.
Find the sum of all the terms of the sequence that are neither divisible by 2 nor
divisible by 7.
Solution:
4
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
2.3
Geometric Series
A. Definition
Geometric Sequence:
a, ar, ar 2 , ar 3 ,...
Geometric Series:
B. Sum to n terms
Let
S(n) = the sum to n terms of the series
a
= the first term
r
= the common ratio
S (n) 
Alternatively,
S (n) 
5
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 8:
Find the sum of the first 6 terms of the geometric sequence 3, 6, 12, 24, …
Solution:
Example 9: (Ex 2B Qn 2)
Find the number of terms in the geometric series 1+3+9+…, which has the sum is 3280.
Solution:
Example 10: (Ex 2B Qn 8)
(a) Find the first term and the common ratio of the geometric series whose fifth term is
27 and whose eighth term is 729.
(b) Hence find the sum of the first 9 terms of the series. (Give the answer correct to the
nearest whole number.)
Solution:
Example 11:
Suppose someone offered you a job for a 30-day month under the following conditions.
You would be paid $.01 for the first day, $.02 for the second, $.04 for the third, and so
on, doubling your previous day’s salary. Would you take the job? How much would you
earn?
Solution:
6
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 12: Regular Investment (e.g. Insurance, MPF, etc.)
A man invests $100 on the same date each year. How much stands to his credit just after
the 10th deposit of $100 if compound interest 3% is added per annum?
Solution:
Example 13:
Mr. Lee has to pay Tax $1045 at the end of the year. What fixed amount of money
should be invested on the first day of each month of that year (at 2% per month
compound interest) so that he can just pay his tax?
Solution:
7
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 14:
If 1, b, c are in arithmetic sequence and 1, b – 1, c – 1 are in geometric sequence.
(a) Show that b 
1
(c  1) and b  12  c  1 .
2
(b) Find b and c.
(c) Find the general term of the arithmetic sequence and geometric sequence.
(d) What is the difference between the sum of the first 10 terms of the given arithmetic
sequence and geometric sequence?
Solution:
(a) Given 1, b, c are in A.P.,
 b 1  c  b
 2b  c  1
1
 b  (c  1)
2
(1)
Given 1, b – 1, c – 1 are in G..P.
b 1 c 1


1
b 1
 b  12  c  1
(2)
(b) From (1), c  2b  1
(3)
b  12  (2b  1)  1
 b 2  2b  1  2b  2
 b 2  4b  3  0
 b  1b  3  0
 b  1 or 3
When b  1, c  2(1)  1  1 (rej)
When b  3, c  2(3)  1  5
(c) When b  3, c  5
The A.P. is 1, 3, 5, …  A(n)  1  (n  1)( 2)  2n  1
The G.P. is 1, 2, 4, …  G(n)  2 n1
(d) Sum of 10 A.P. =
10
2(1)  9(2)  100
2
1(210  1)
 1023
2 1
Difference = 100 – 1023 = -923
Sum of 10 G.P. =
8
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
2.4
Sum to Infinity of a Series
A. Basic Idea
Definition
S   T (1)  T (2)  T (3)  ...
B.
Sum to Infinity of Arithmetic Series
Example:
Consider the following arithmetic sequences:
Type (A)
Infinite no.
of terms
(i) 1, 3, 5, 7, 9, …
(ii) -8, -4, 0, 4, 8, …
(iii) 2, 2 12 , 3, 3 12 , 4, 4 12 , …
Type (B)
(i) 5, 3, 1, -1, -3, …
(ii) 34, 27, 20, 14, 7, 0, -7, -14, …
(iii) -2, -3, -4, -5, -6, -7…
(Type A)
(Type B)
When d > 0, S ()  
When d < 0, S ()  
C. Sum to Infinity of Geometric Series
Consider the following geometric series:
Geometric Series
Common
Ratio (r)
3  6  12  24  
1
2
1   ( 1 )n  
 14  18  16
2
1    (1) n ( 1 ) n  
1  12  14  18  16
2
1  (1)  1  (1)  
2
0.5
Sum to n
terms
( Sn )
3(2 n  1)
2 1
1
(1  0.5 n )
2
1  0.5
-0.5
1(1  (0.5) n )
1  0.5
-1
1(( 1) n  1)
11
Sum to Infinity
( S )

To be explored
indeterminate
Note: S  depends on r.
9
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Recall, the sum to n terms is given by:
S (n) 
n
1
2
3
4
10
100
200
500
a (1  r n )
1 r
(0.5)n
0.5
0.25
0.125
0.0625
0.000976…
7.89 x 10-31
6.22 x 10-61
0
Theorem:
If  1  r  1 ,
When  1  r  1 ,
r n  0 as n  
S ( ) 
a
1 r
Example 15:
Find the sum of the infinite geometric series
5  52  54  85  
Solution:
a  5, r 
1
2
S ( ) 
5
1
1
2
 10
Example 16:
Find a fractional numeral for (a) 0.2 7 and (b) 5.3 6 .
Solution:
(a)
(b)
0.2 7  0.27  0.0027  0.000027  ...
0.27

1  0.01
0.27

0.99
27

99
3

11
5.3 6  5  0.36  0.0036  0.000036  ...
0.36
1  0.01
0.36
 5
0.99
4
5
11
 5
10
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 17:
The sum to infinity of a geometric series is 36 and the second term of the series is 8.
Find the sequence.
Solution:
 a
 36  (1)

1  r
ar  8 (2)
Solving the simultaneous equation,
r
When r 
1
2
or
3
3
1
1
, a  24 , therefore the sequence is 24 
3
 3
2
2
When r  , a  12 , therefore the sequence is 12 
3
3
n1
n1
Example 18:
A ball is dropped from a height of 4m above the ground. After each bounce, it reaches to
2
3
of the height from which it falls. Calculate the total distance traveled by the ball
before coming to rest on the ground.
Solution:
Total distance:
2
2
2 2 2
2
 4  4   4   4   4   ...
3 3 3
3
  2   2 2

 4  2 4   4   ...
  3   3 

  2 
 4 3  
 4  2   
1 2 

3 

…
 20m
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S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 19:
The positive numbers 1, k,
1
,… are in geometric sequence.
2
(a) Find the value of k, leaving the answer in surd form.
(b) Express the nth term T(n) in terms of n.
(c) Find the sum to infinity, expressing the answer in the form of p  q , where p
and q are integers.
(d) Express the product T (1)  T (3)  T (5)    T (2n  1) in terms of n.
Solution:
(a) Given positive numbers 1, k,
1
,… are in G.P., therefore
2
k 1/ 2
1

 k2 
1
k
2
 2

(b) T (n)  

2


(c)
S ( ) 
k
2
2
or -
2
(rej)
2
n 1
1
2
1
2

2
2 2
 2 2
(d)
T (1)  T (3)  T (5)  ...  T 2n  1
1 1
1
 1      
2 4
2
n 1
1 23... ( n 1)
1
 
2
1
 
2
1
 
2
n 1
(1 n 1)
2
n ( n 1)
2
12
S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 20:
The sides of a square are each 16cm. A second square is
inscribed by joining the midpoints of the sides,
successively. In the second square we repeat the process,
inscribing a third square. If this process is continued
indefinitely, what is the sum of the perimeters of all the
squares?
Solution:
16cm
Side Length of the 1st square = 16 cm
8 2  8 2  8 2 cm
Side Length of the 2nd square =
 Common ratio =
8 2
2

16
2
Let P(n) be the perimeter of the nth square
P(1) = 64 cm
P(2) = 64 
2
cm
2
2
 2
 cm
P(3) = 64  

2



Sum of all perimeters
= 64 + 64 
=
2
+ 64 
2
2
 2


 2  +…


64
1
2
2
= 218.51 cm (corr. to 2 decimal places)
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S.5 Math: Chapter 2 Series: Arithmetic and Geometric Series
Example 21:
A chain is made by joining an infinite number of squares. The diagonals of the squares
M1M 2 , M 2 M 3 , … lie on a straight line. If the lengths of the diagonals form a
geometric sequence such that M1M 2 = 12cm and M 2 M 3 =8cm, find
(a) the length of the chain,
(b) the length of wire in making the
M1
chain,
(c) the total area bounded by the
squares.
(Give the answers in surd form if necessary.)
Solution:
(a)
M1M 2 = 12 cm
M 2 M 3 = 8 cm
M3
M2

Common ratio =
M4 M5
8 2

12 3
2
12
2
2
Total length = 12  12   12   ... 
= 36 cm
2
3
3
1
3
(b) Let a cm be the side length of the 1st square
 a 2  a 2  12 2
a  6 2

Perimeter of 1st square = 24 2
2
Perimeter of 2nd square = 24 2   
3
2
Perimeter of
Total length of wire =
3rd
2
square = 24 2    and so on…
3
 2  2 2

24 2 1      ...
 3  3 

=
24 2
= 72 2 cm
2
1
3
(c) Area of the 1st square = 72
4
Area of the 2nd square = 72   
9
2
Area of the
3rd
4
square = 72    and so on…
9
 4  4 2

72
Total Area Bounded = 721      ... 
 129.6 cm 2
 9  9 
 1  4
9
14