MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
https://youtu.be/VgBSzYBMbh4
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
https://youtu.be/m5Yn4BdpOV0
EVALUATION:
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
SPECIFIC TOPIC: SERIES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are
3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
https://youtu.be/r614AD021n0
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
https://youtu.be/Tj89FA-d0f8
EVALUATION:
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
https://youtu.be/ZZaUO_35hdk
EVALUATION:
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
https://youtu.be/SyCsmjvOODo
EVALUATION:
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
SPECIFIC TOPIC: CALCULATION INVOLVING SUM OF AN A.P
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Solve problems on sum of an A.P
(ii) Find the nth term of an A.P
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
https://youtu.be/0m8GKU-eiyQ
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
TOPIC: ARITHMETIC PROGRESSION (A. P).
CONTENTS
• Sequence
• Definition of Arithmetic Progression
• Denotations in Arithmetic progression
• Deriving formulae for the term of A. P.
• Sum of an arithmetic series
SEQUENCE
Examples:
Find the next two terms in each of the following sets of number and in each case state the rule which gives the term.
(a) 1, 5, 9, 13, 17, 21, 25 (any term +4 = next term)
(b) 2, 6, 18, 54, 162, 486, 1458 (any term x 3 = next term)
(c) 1, 9, 25, 49, 81, 121, 169, (sequence of consecutive odd no)
(d) 10, 9, 7, 4, 0, -5, -11, -18, -26, (starting from 10, subtract 1, 2, 3 from immediate no).
In each of the examples below, there is a rule which will give more terms in the list. A list like this is called a SEQUENCE in many cases; it can simply matter if a general term can be found for a sequence e.g.
1, 5, 9, 13, 17 can be expressed as
1, 5, 9, 13, 17 ……………. 4n – 3 where n = no of terms
Check: 5th term = 4(5) -3
20 – 3 = 17
10th term = 4(10) – 3
40 – 3 = 37
Example 2
Find the 6th and 9th terms of the sequence whose nth term is
(a) (2n + 1)
(b) 3 – 5n.
Solution
(a) 2n + 1
6th term = 2(6) + 1
=12 + 1 = 13
9th term = 2 (9) + 1
=18 + 1 = 19
(b) 3 – 5n
6th term = 3 – 5 (6)
=3 – 30 = -27
9th term = 3 – 5 (9)
=3 – 45 = -42
https://youtu.be/vlPHnGbKi9w
Evaluation
For each of the following sequence, find the next two terms and the rules which gives the term.
1. 1, , , , , ____, ____
2 100, 96, 92, 88, _____, ____
3. 2, 4, 6, 8, 10, ____, _____
4. 1, 4, 9, 16, 25, ____, _____
(i) Arrange the numbers in order of increasing give to a sequence.
(ii) Find the next two terms in the sequence
5. 19, 13, 16, 22, 10
6. -21/2, 51/2, 31/2, 11/2, -1/2
7. Find the 15th term of the sequence whose nth term is 3n - 5
4
Reading Assignment
New General mathematics SS II by G. B. Channon & Co. exercise 18a.
DEFINITION OF ARITHMETIC PROGRESSION
This is a sequence in which the terms either increase or decrease in equal steps is called an Arithmetic Progression.
The sequence 9, 12, 15, 18, 21, ____, _____, _____ has a first term of 9 and a common difference of +3 between the terms.
Denotations in A. P.
a = 1st term
d = common difference
n = no of terms
Un = nth term
Sn = Sum of the first n terms
Formula for nth term of Arithmetic Progression.
e.g. in the sequence 9, 12, 15, 18, 21.
a = 9
d = 12 – 9 or 18 – 15 = 3.
1st term = U1 = 9 = a
2nd term = U2 = 9 + 3 = a + d
3rd term = U3 = 9 + 3 + 3 = a + 2d
10th term = U10 = 9 + 9(3) = a + 9d
nth term = Un = 9+(n-1)3 = a + (n-1)d
nth term = Un = a + (n-1)d
Example:
Given the A.P, 9, 12, 15, 18 …… find the 50th term.
a = 9
d = 3
n = 50
Un = U50
Un = a + (n-1) d
U50 = 9 + (50-1) 3
9 + (49) 3
9 + 147
=156.
2. The 43rd term of an AP is 26, find the 1st term of the progression given that its common difference is ½ and also find the 50th term.
U43 = 26
d = ½
a = ?
n = 43
Un = a + (n-1) d
26 = a + (43-1) ½
26 = a + 42(1/2)
26 = a + 21
26 – 21 = a
5 = a
a = 5
(b) a = 5
d = ½
n = 50
U50 =?
Un = a + (n-1) d
U50 = 5 + (50-1)1/2
= 5 + 49(1/2)
U50 = 5 + 241/2
U50 = 291/2
https://youtu.be/YqP5TuwtEdg
Evaluation
1. Find the 37th term of the sequence 20, 10, 0, -10…
2. 1, 5… 69 are the 1st, 2nd, and last term of the sequence; find the common difference between them and the number of terms in the sequence.
Reading Assessment
New General Mathematics for SS 2, Exercise 18b, no 8, 10, 11, 12.
SUM OF AN ARITHMETIC SERIES
When the terms of a sequence are added, the resulting expression is called series e.g. in the sequence 1, 3, 5, 7, 9, 11.
Series = 1 + 3 + 5 + 7 + 9 + 11
When the terms of a sequence are unending the series is called infinite series, it is often impossible to find the sum of the terms in an infinite series.
e.g. 1 + 3 + 5 + 7 + 9 + 11 + …………………. Infinite
Sequence with last term or nth term is termed finite series.
e.g.
Find the sum of
1, 3, 5, 7, 9, 11, 13, 15
If sum = 2, n = 8
Then
S = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
Or S = 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
Add eqn1 and eqn 2
2s = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16
= 48 = 8(16)
2 2 = S = 64
Deriving the formula for sum of A. P. The following represent a general arithmetic series when the terms are added.
S = a + (a+d) + a + 2d + …………………………… + (L-2d) + (L-d) + L – eqn
S = L + (L-d) + L – 2d + ……………………………… a + 2d + (a+d) + a – eqn
2s = (a + L) + (a + L) + (a + L) + …………………… (a + L) + (a + L) + (a + L)
2s = n(a + L)
2
S = n(a+L)
2
L => Un = a + (n-1)d
Substitute L into eq**
S = n(a + a+(n-1)d
2
S = n(2a + (n-1)d = n ( 2a+ (n-1)d
2 2
S = n(a+L) when Un (last term L) is given
2
or
S = n(2a + (n-1)d when d is given or obtained
2
Example 2
Find the sum of the 20th term of the series 16 + 9 + 2 + …………………
a = 16
d = 9 – 16 = -7
n = 20
S = n(2a + (n-1)d)
2
S = 20 (2x16) + (20-1)(-7)
2
20 (32 + 19(-7)
2
S =10 (32 - 133) = 10(-101)
S = -1010
https://youtu.be/TtJ4dphwh0U
Evaluation
1. Find the sum of the arithmetic series with 16 and -117 as the first and 20th term respectively.
2. The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
Reading Assignment
New General Mathematics for SS 2, Page 181 – 184. Exercise 18d.
Evaluations (Exercise)
1. An A. P. has 15 terms and a common difference of -3, find its first and last term if its sum is 120.
2. On the 1st of January, a student puts N10 in a box, on the 2nd she puts N20 in the box, on the 3rd she puts N30 and so on putting on the same no. of N10 notes as the day of the month. How much will be in the box if she keeps doing this till 16th January.
Evaluation
The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
ASSIGNMENT
Theory
1. Eight wooden poles are to be used for pillars and the length of the poles form an arc Arithmetic Progression (A. P.) if the second pole is 2m and the 6th pole is 5m, give the lengths of the poles in order and sum up the lengths of the poles.
2a. Write down the 15th term of the sequence.
2_, 3 , 4 , 5
1x3 2x4 3x5 4 x6
b. An arithmetic progression (A. P.) has 3 as its term and 4 as the common difference.
c. Write an expression in its simplest form for the nth term.
d. Find the least term of the A. P. that is greater than 100.
Objectives
1. Find the 4th term of an A. P. whose first term is 2 and the common difference is 0.5 (a) 4 (b) 4.5 (c) 3.5 (d) 2.5
2. In an A. P. the difference between the 8th and 4th term is 20 and the 8th term is 11/2 times the 4th term, find the common difference
(a) 5 (b) 7 (c) 3 (d) 10
3. Find the first term of the sequence in no. 2
(a) 70 (b) 45 (c) 25 (d) 5
4. The next term of the sequence 18, 12, 60 is
(a) 12 (b) 6 (c) -6 (d) -12
5. Find the no. of terms of the sequence 1/2 , ¾, 1, ……………….. 51/2
(a) 21 (b) 43/4 (c) 1 (d) 22
