Sets, Sequences & Series: Comprehensive Summary of Module 11

This assignment provides a summary of Module 11, focusing on sets, sequences, and series. It begins by defining set notation using positive integers and differentiating sequences from sets, highlighting that sequences allow repetition and are ordered. The document explains how to identify a linear sequence based on a constant difference between consecutive terms and demonstrates finding the common difference and nth rule for a given sequence. It then transitions to series, differentiating between arithmetic and geometric series, and provides the corresponding series for a given sequence. The assignment includes sigma notation for summing terms and formulas for calculating the nth term and sum of the first n terms in a geometric series, illustrated with examples and relevant references.

1.
a
Positive integers (n) = 1, 2, 3, 4, 5, 6, . . .
Let,
A = Set of positive integers
A is then represented as:
A = {1, 2, 3, 4, 5, . . . }
2.
c
Set: A collection ofunique objects (elements),so elements cannot be re-
peated.
eg.A = {1, 2, 3} is a set but A = {1, 2, 2, 3} is not a set.
Sequence:An ordered list of elements which can also be infinite.
In a sequence the elements can be repeated unlike in sets.
eg. {0, 1, 1, 2, 3, 5, 8, . . . } is a sequence (Fibonacci) but it cannot be termed
as set.
d
Linear sequence:A sequence which increases or decreases by the same amount
each time is called a linear sequence.Alternatively, if the difference between
the consecutive terms in a sequence remains same it is a linear sequence.
Let, A = (a0, a1, a2, a3, a4, a5, . . . , an, an+1, . . . ) be any sequence.If the dif-
ference between consecutive terms:|a1 − a0| = |a2 − a1| = · · · = |an+1 − an|,
is constant and equalto the same amount d,where d is any realnumber,
then A is a linear sequence.
Eg. (1, 2, 3, 4), (1, 3, 5, 7, . . . ) etc.
e
5, 8, 11, 14, . . .
Difference between the consecutive termsof the sequence is:|8 − 5| =
|11 − 8|= |14 − 11| . . . ,which is 3. Therefore,the common difference is
1

3.
f
5, 8, 11, 14, . . .
5 is the first term n = 1, 8 is the second term n = 2 and so on.
Let the nth term in the sequence be represented as Un = an+b.To determine
the nth rule, we need to determine a and b.
For n = 1:U1 = 5 = a · 1 + b, implies a + b = 5
For n = 2:U2 = 8 = a · 2 + b, implies 2a + b = 8
Therefore, we have:
a + b = 5 (1)
2a + b = 8 (2)
Solving (1) and (2) simultaneously, we get:a = 3 and b = 2
Implies,
Un = 3n + 2
is the nth rule for the sequence.
g
The 20th term (n = 20) in the sequence is determined using the nth rule:
Un = 3n + 2
Therefore, the 20th term is:
U20 = 3(20) + 2 = 62
3
h
Series:Results from adding all the terms of the sequence.
Given sequence is:
5, 8, 11, 14, . . .
2

Its corresponding series is:
5 + 8 + 11 + 14 + . . .
i
In an arithmetic series, the difference between consecutive terms of the series
remains constant and equal.As it was already determined that the common
difference is 3 for the given sequence, it implies that the series obtained in h
is an arithmetic series.
j
The nth term of the series/sequence is:
Un = 3n + 2
Summation of the first 20 terms of the sequence is given by:
S =
20X
i=1
Un =
20X
i=1
3n + 2
k
The given series is:
3 + 9 + 27
Common ratio (r) for a series is the ratio of consecutive terms:
r = 9
3 = 27
9 = 3
Therefore, the common ratio is 3.
l
The nth of a geometric series is determined using formula:
Un = a0(r)n−1
where, a0 is the leading term of the series.Since the series starts from 3, we
have a0 = 3 and as determined earlier r = 3.
Therefore, the nth term in the series is given by:
Un = 3(3)n−1
3

m
Formula for sum of first n terms of geometric series:
Sn = a0
rn − 1
r − 1
For the given series:a0 = 3 and r = 3.Therefore,the sum of first 6 terms
(n = 6) of the series is:
S6 = 3 ×36 − 1
3 − 1= 3 ×728
2 = 1092
References:
1. Scocco,D. (2012)‘Sets,Sequencesand Tuples.’Programming Logic.
www.programminglogic.com
https://www.programminglogic.com/sets-sequences-and-tuples/
2. Stapel, E. (2019) ‘Geometric Series.’ Series.Purple Math.
https://www.purplemath.com/modules/series5.htm
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