SERIES AND SEQUENCE

INTRODUCTORY PURE MATHEMATICS 1 - MATH 161

M. A. BOATENG (Ph.D., MIMA)
 TABLE OF CONTEXT
0

▶ INTRODUCTION

▶ ARITHMETIC PROGRESSION

▶ GEOMETRIC PROGRESSION

▶ APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 1/29

 Table of Contents
1 INTRODUCTION

▶ INTRODUCTION

▶ ARITHMETIC PROGRESSION

▶ GEOMETRIC PROGRESSION

▶ APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 2/29

 DEFINITIONS
1 INTRODUCTION

SEQUENCE
A sequence is a set of numbers (terms) written in a defined order with a rule or formula
for obtaining the terms. For example, 1,3,5,7,... and 2,4,6,8,... are all sequences. Each
term is obtained by adding 2 to the preceding term.

SERIES
A series is formed when the terms of the sequence are added. For example, 2+4+6+8+...
and 1+3+5+7+... are all series. If a series stops after a finite number, it’s is called finite
series. However, if the series does not stop, then it’s called infinite series.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 3/29

We will consider two types of sequences in our studies. They are
1. Arithmetic Progression (AP), or Linear Progression or Linear Sequence

2. Geometric Progression (GP), or Exponential Progression,or Exponential Sequence

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 4/29

 Table of Contents
2 ARITHMETIC PROGRESSION

▶ INTRODUCTION

▶ ARITHMETIC PROGRESSION

▶ GEOMETRIC PROGRESSION

▶ APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 5/29

 ARITHMETIC PROGRESSION
2 ARITHMETIC PROGRESSION

Definition
An Arithmetic Progression (AP) is a sequence in which any term differs from the preceding
term by a constant called the common difference which may be positive or negative. Eg.
2,4,6,8,... is a linear sequence with a common difference of 2.
The first term of a linear sequence is denoted by a and the common difference by d.
The general term of or the formula for generating a linear sequence is given by:

Un = l = a + (n − 1)d

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 6/29

where
Un = terms of the series

l = last term

a = first term

d = common difference

n = number of terms

d = Un − Un−1

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 7/29

 EXAMPLE
2 ARITHMETIC PROGRESSION

1. Find the 11th term of the sequence 4,9,14,19,...

Solution
a=4
d = 9 − 4 = 14 − 9 = 5
Un = l = a + (n − 1)d
U11 = 4 + (11 − 1)5
U11 = 54

2. The fourth term of a linear sequence is 19 and the eleventh term is 54. Find the eighth
term.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 8/29

Solution

Un = l = a + (n − 1)d
U4 = a + 3d = 19————(1)
U11 = a + 10d = 54———–(2)
solving them simultaneously yields
Eq(2)-Eq(1)
7d = 35, d = 5
substituting d into Eq(1)
a + 15 = 19, a = 4
Un = 4 + (n − 1)5
U8 = 4 + 7(5)
U8 = 39

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 9/29

 TRIAL
2 ARITHMETIC PROGRESSION

• Find a formula for the nth term of the linear sequence 10,6,2,-2,...

• The first, twelth and last term of an AP is 4, 31 21 , and 376 12 respectively. Determine.
a. the number of terms in the sequence.
b. the 80th term.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 10/29

 SUM OF AN AP
2 ARITHMETIC PROGRESSION

The sum of a linear progression is denoted by the following formulae

n
Sn = [2a + (n − 1)d] (1)
2
n
Sn = (a + l) (2)
2
The symbols denote their usual meanings.
The first formula is used when a and d are known and the second is used when a and l are
known but not the common difference.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 11/29

 EXAMPLES
2 ARITHMETIC PROGRESSION

1. Find the sum of the first 15 terms of the sequence 5, 9, 13, 17, ...

Solution
a=5
d=9−5=4
n = 15
Sn = 2n [2a + (n − 1)d]
15
S15 = 2 [2(5) + (14)4]
15
S15 = 2 [66]
S15 = 495

2. Find the number of terms of the linear sequence 5, 8, 11, 14, ... which has sum 1025.
M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 12/29
Solution

Sn = 2n [2a + (n − 1)d]
solving for n,
2Sn = n2 d + n(2a − d)
a=5
d=8−5=3
Sn = 1025
2(1025) = 3n2 + (2(5) − 3)
2050 = 3n2 + 7n
n1 = 25, n2 = −27.333, number of terms cannot be negative hence we take the positive
one
=⇒ n = 25

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 13/29

 TRIAL
2 ARITHMETIC PROGRESSION

1. Find the number of terms of an arithmetic progression 4, 6 12 , 9, 11 12 , ... needed to

make a sum of 126.

2. The sixth term and eleventh term of an AP is 23 and 48 respectively. Calculate the sum
of the first twenty terms of the sequence.

3. The fifth term of an arithmetic progression is -1 and the sum of the first twenty terms is
-240. Find the third term.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 14/29

 Table of Contents
3 GEOMETRIC PROGRESSION

▶ INTRODUCTION

▶ ARITHMETIC PROGRESSION

▶ GEOMETRIC PROGRESSION

▶ APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 15/29

 GEOMETRIC PROGRESSION
3 GEOMETRIC PROGRESSION

Definition
A Geometric Progression (GP) is a sequence where each term is obtained from the
preceding term by multiplying it by a constant factor. The constant factor is called
common ratio and is denoted by r. Eg. 2, 4, 8, 16, ... is a GP with a common ratio of 2.
The general term or formula for generating a GP with first term a and common ratio r is
given by,
Un = arn−1
where
r = UUn−1
n

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 16/29

 EXAMPLES
3 GEOMETRIC PROGRESSION

1. Find the seventh term of the exponential sequence 5,10,20,...

Solution
a=5
r = 10
5 =2
Un = arn−1
U7 = 5 × 26
U7 = 5 × 64
U7 = 320

2. Find the 12th term of the GP 128, 64, 32, ...

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 17/29

Solution
a = 128
64
r = 128 = 12
Un = arn−1
U12 = 128 × ( 21 )11
U12 = 27 × 2−11
= 2−4 = 16 1

EXERCISE
1. Three consecutive terms of a geometric series have product 343 and sum 2 .
49
Find the
terms.

2. Find the nth term of the sequence 12 , 14 , 81 , ...

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 18/29

 SUM OF A GP
3 GEOMETRIC PROGRESSION

The general formula of the first n terms of an exponential sequence is given by the
following formulae:
a(1 − rn )
Sn =
1−r
This formula is used when r < 1 and

a(rn − 1)
Sn =
r−1
This formula is used when r > 1 and

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 19/29

 EXAMPLE
3 GEOMETRIC PROGRESSION

1. Find the sum of the first 13 terms of the GP 21 , 14 , 18 , ...

Solution
a = 12
r = 41 ÷ 12 = 12
n)
Sn = a(1−r
1−r , r < 1
1
(1−( 21 )13 )
S13 = 2
1− 12
S13 = 0.9998779296875

2. The second term and fourth terms of an exponential sequence (GP) are 9 and 4
respectively. Find the common ratio and first term.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 20/29

Solution
U2 = ar = 9.................(1)
U4 = ar3 = 4................(2)
dividing eqn(2) by eqn(1)
U4 ar3
U2 = ar
= r2 = 49
r = 23
Substituting r into eqn(1) yields
a = 92
3
a = 13.5
Hence a = 13.5, r = 2
3

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 21/29

 TRIAL
3 GEOMETRIC PROGRESSION

1. The first three terms of a GP are (k − 3), (2k − 4), (4k − 3), in that order. Find the
value of k and the sum of the the first eight terms.

2. In an exponential sequence, the sixth term is eight times the third term and the sum
of the seventh and eighth term is 192. Find
a. the common ratio.
b. the first term.
c. the sum of the fifth term to the eleventh term, inclusive.

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 22/29

 SUM TO INFINITY OF A GEOMETRIC PROGRESSION
3 GEOMETRIC PROGRESSION

When the common

n)
ratio is less than one, the sum to n terms is given by
arn
Sn = a(1−r
1−r , r < 1 which can be written as Sn = 1−r a
− 1−r . As n approaches ∞, rn = 0.
n
Hence 1−r
ar
= 0 as n approaches ∞.
a
S∞ =
1−r

Example
Find the sum to infinity of 7
10 + 7
100 + 7
1000 + ...

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 23/29

 Solution
7
10
7
+ 100 7
+ 1000 + ... is a geometric series with
a = 10 and r = 100
7 7
÷ 10 7 1
= 10
a
S∞ = 1−r
7 7
S∞ = 1
10
1− 10
= 10
9
10
7
S∞ = 9

TRIAL
Find the sum to infinity of the following exponential sequences:
1. 72, 24, 8, ...
2. 8, 2, 12 , ...

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 24/29

 Table of Contents
4 APPLICATIONS OF SEQUENCE

▶ INTRODUCTION

▶ ARITHMETIC PROGRESSION

▶ GEOMETRIC PROGRESSION

▶ APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 25/29

 APPLICATION OF SEQUENCE
4 APPLICATIONS OF SEQUENCE

A man’s salary increases by $8, 000.00 each year. If his total salary at the end of 12years is
$18, 525, 000.00. Find
a. his initial salary
b. his salary in the twelth year
Solution
From the question we can see that its an AP since the common difference is $8, 000.00,
hence Sn = $18, 525, 000.00.
n = 12
Sn = 2n [2a + (n − 1)d]
18, 525, 000 = 6[2a + (11)$8, 000.00]
3, 087, 500 = 2a + 88, 000
2a = 2, 999, 500
a = 1, 499, 750
M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 26/29
The initial salary is $1, 499, 750
b. His salary in the twelth year is
U12 = a + 11d
= 1, 499, 750 + 11(8000)
= 1, 499, 750 + 88, 000
= $1, 587, 750

EXERCISE
The value of a lathe originally valued at $30, 000, 000 depreciate 15% per annum.
Calculate its value after 5 years. The machine is sold when its value is less than
$5, 500, 000. After how many years is the lathe sold?

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 27/29

 SERIES AND SEQUENCE
Thank you for listening!
Any questions?

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 28/29

 End of Lesson
4 APPLICATIONS OF SEQUENCE

M. A. BOATENG (Ph.D., MIMA) SERIES AND SEQUENCE 29/29