FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
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FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
SUBJECT: FURTHER MATHEMATICS
CLASS: SS1
SCHEME OF WORK
WEEK
TOPIC
1
Arithmetic Progression (AP)
2
Geometric Progression (GP)
3
Linear inequalities in one variable
4
Inequalities in two variables (Graph of inequalities)
5
Introduction to the concept of functions.
6
Review of half term work.
7
Functions (one – to – one, onto, composite and inverse functions)
8
Trigonometric ratio: Graph of Sine, Cosine and tangent of angles, deviation of trigonometric ratio of special angles (300, 450 and 600). Application of trigonometric ratios.
9
Logical reasoning: Simple True and False statement, Negation, Converse and Contra positive of statement,
10
Logical reasoning continues: Compound statement, connectives and their symbols, conditional statements and symbols.
11
Revision of Second Term’s lesson
12
Examination
FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
REFERECES
Ø FutherMaths Project 1 and 2 by TuttuhAdegun (main text).
Ø Additional Mathematics by Godman
Ø Further Mathematics by E. Egbe et al.
WEEK ONE DATE……………
TOPIC: SEQUENCE & SERIES
CONTENT
· Sequence and series
· Arithmetic Progression (AP)
· Arithmetic Mean
· Sum of terms in an AP
Sequence & Series
A sequence is a pattern of numbers arranged in a particular order. Each of the number in the sequence is called a term. The terms are related to one another according to a well defined rule.
Consider the sequence 1, 4, 7, 10, 13 …., 1 is the first term,(T1) 4 is the second term(T2), 7 is the third term (T3).
The sum of the terms in a sequence is regarded as series. The series of the above sequence is
1 + 4 + 7 + 10 + 13 = 35
The nth term of a Sequence
The nth term of a sequence whose rule is stated may be represented by Tnso that T1, T2, T3etc represent the first term, second term, third term … etc respectively.
Consider the sequence 5, 9, 13, 17, 21 ……..
T1 = 5 + 4(0)
T2 = 5 + 4(1)
T3 = 5 + 4 (2)
T4 = 5 + 4 (3)
Tn = 5 + 4 (n – 1)
Tn = 5 + 4n – 4 = 4n +1
when n = 30
T30 = 4(30) + 1
T30 = 121
Find the nth term of these sequences:
(i) 3, 5, 7, 9 …… 2n + 1
(ii) 0, 1, 4, 9 ……… (n -1)2
(iii) 1/3, 3/4, 1, 7/6 ………………2n - 1
n + 2
Examples
Write down the first four terms of the sequence whose general term is given by:
(i) Tn = n+1 (ii) Tn = 5 x (1/2)n-2
3n +2
Solution
i. Tn = n+1
3n + 2
T1 = 1 + 1 = 2/5
3(1) + 2
T2 = 2 + 1 = 3/8
3(2) +2
T3 = 3 + 1 = 4/11
3(3) + 2
T4 = 4 + 1 = 5/14
(ii) Tn = 5 x (1/2)n-2
T1 = 5 x (1/2)1-2 = 5(1/2)-1 = 5(2-1)-1 = 5 x 2 = 10
T2 = 5 x (1/2)2-2 = 5(1/2)0 = 5 x 1 = 5
T3 = 5 x (1/2)3-2 = 5 x (1/2) = 5/2
T4 = 5 x (1/2)4-2 = 5(1/2)2 = 5/4
The sequence is 10, 5, 5/2, 5/4 ………
3(4) + 2
The sequence is 2/5, 3/8, 4/11, 5/14 ……..
Evaluation
Find the first term of the sequence whose general term is given by
(i) 50 – (½)n (ii) 2 + 3/2(n+1)
Arithmetic Progression (A.P) or Linear Sequence
An arithmetic progression (A.P) is generated by adding or subtracting a constant number to a preceding term to get a term. This constant number is called the common difference designated by the letter d. The first term is designated by a.
Ex: A.P d (common difference) a (first term)
6½, 5, 3½, 2 …. -1½ 6½
-2, -3/4, ½, 1 ¾ 1¼ -2
T1 T2 T3 T4 T5
a a + d a + 2d a + 3d a + 4d
So for any A.P, the nth term (Tn = Un) is given by
Tn =Un= a + (n – 1) d. Tn= Un= nth term
a = first term
d = common difference
n = no of terms
Examples
1. What is the 10th term of the sequence 10, 6, 2, -4 …..
2. Find the term of the A.P 3½, 7, 10½ ….. Which is 77.
3. The fist term of an A.P is 3 and the 8th term is 31. Find the common difference.
Solution
(1.) The A.P = 10, 6, 2, -4
a = 10, d = 6 – 10 = - 4, n = 10
Tn = a + (n – 1) d
T10 = 10 + (10 – 1) (-4)
T10 = 10 +9(-4) = 10 – 36
T10 = -26.
(2.) A.P = 3 ½, 7, 10½ ……………… 77
a = 3½, d = 7 – 3½ = 3½, n =? Tn = 77
Tn = a + (n-1)d
77 = 3½ + (n-1)3½
77 = 3½ + 3½n - 3½
77 = 3½ n
n = 77/3½ = 77/7/2
n = 77 x 2/7 = 22
FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
(3) a = 3, T8 = 31, d = ? n = 8
Tn = a + (n-1) d
31 = 3 + (8-1) d
31 – 3 = 7d
d = 28/7 = 4
Evaluation
(i) Find the 15th term of the A.P 5, 2, -1, -4 …………
(ii) Find the term of the A.P 1, 6, 11, 16…. which is 66.
Arithmetic Mean
If a, b, c are three consecutive terms of an A.P, then the common difference, d, equals
b – a = c – b = common difference.
b + b = a +c
2b = a + c
b = ½(a +c)
Examples
(i) Insert four arithmetic means between -5 and 10.
(ii) The 8th term of a linear sequence is 18 and the 12th term is 26. Find the first term, the common difference and the 20th term.
Solution
(i) Let the sequence be -5, a, b, c, d, 10.
a = -5, T6 = 10, n =6.
Tn = a + (n-1) d
10 = -5 + (6 – 1) d
15 = 5d
d = 15/5 = 3
a = -5 + 3 = -2
b = -2 + 3 = 1
c = 1 + 3 = 4
d = 4 + 3 = 7
The numbers will be -5, -2, 1, 4, 7, 10.
(ii) T8 = a + 7d = 18, T12 = a +11d = 26
a + 7d = 18 ……………….. (i)
a + 11d =26 ……………….. (ii)
Subtract (i) from (ii)
4d = 8
d = 2
Substitute for d = 2 in (i)
a + 7 (2) = 18
a = 18 – 14
a = 4
T20 = a + (n – 1) d = a + 19d
T20 = 4 + (20 – 1) 2
= 4 + 19 x 2
T20 = 42
Evaluation
(1) Given that 4, p, q, 13 are consecutive terms of an A.P, find the values of p and q.
(2) The sum of the 4th and 6th terms of an A.P is 42. The sum of the 3rd and 9th terms of the progression is 52. Find the first term, the common difference and the twentieth term of the progression.
Sum of terms in an A.P
To find an expression for the sum of n terms of a linear sequence, Let Sn be the sum, then
Sn = a + (a + d) + (a + 2d) + ……. + Tn ………….. (i)
Also
Sn = Tn+ (Tn- d) + (Tn- 2d) + ……… a ………. (ii)
Adding (1) and (2)
2Sn = (a + Tn) + (a + Tn) + (a + Tn) + ………… + (a + Tn)
\2Sn = n (a + Tn)
\Sn = n/2 (a + Tn)
But Tn = a + (n-1) d
Sn = n/2 (2a + (n-1) d)
FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
Examples
(1) Find the sum of the first 25 terms of the A.P 3, 10, 17 ……….
(2) Find the sum of the first eight terms of a linear sequence whose first term is 6 and last term is 46.
(3) The sum of the first ten terms of an arithmetic progression is 255. Find the sum of the next twenty
term of the A.P if the sum of the first twenty terms is 1010.
Solution
1. A.P = 3, 10, 17 …………..
a = 3, d = 7, n = 25
Sn = n/2 (2a + (n-1) d)
= 25/2 (2 x3 + (25 – 1) 7)
Sn = 25/2 (6 +24 x 7)
S25 =25/2 x174 = 2175
2. A.P , a = 6, Tn = 46, n = 8
Sn = n/2 (a + Tn)
= 8/2 (6 + 46)
Sn = 4 (52) = 208.
3. S10 = 10/2 (2a + (10 – 1) d) = 255
S20 = 20/2 (2a + (20 – 1) d) = 1010
5 (2a + 9d) = 255
10 (2a + 19d) = 1010
2a + 9d = 51 …………….(i)
2a +19d = 101 ……………… (ii)
Subtract (i) from (ii)
10d = 50
d = 5
Substitute for d = 5 in (i)
2a + 9 x 5 = 51
2a = 51 – 45
2a = 6
a = 3
Sum of the next 20 terms = S30 – S10
S30 = 30/2 (2 x 3 + (30 -1) 5)
= 15 (6 + 29 x 5)
S30 = 2265
S30 – S10 = 2265 – 255
= 2010
Evaluation:
The sum of the first ten term of a linear sequence is -60 and the sum of the first fifteen term of the sequence is -165. Find the 18th term of the sequence.
General Evaluation
1. The sum of the first four terms of a linear sequence (A.P) is 26 and that of the next four terms is 74.
Find the values of (i) the first term (ii) the common difference.
2. Calculate the (i) common difference (ii) the 20th term of the arithmetic progression;
100, 96, 92, 88, 86...
3. Solve the equation: log4(x2 + 6x + 11) = ½
4. Express 1 in the formm√5 + n√3 where m and n are rational numbers
3√5 + 5√3
Reading Assignment: Further Mathematics Project Book 1(New third edition).Chapter 28 -33 & 36 – 37
Weekend Assignment
1. Find T9 of the sequence -1, 2, 5, 8 ……………. A. 21 B. 22 C. 23 D. 24
2. The 10th term of an A.P is 68 and the common difference is 7, find the first term of the sequence.
A. 3 B. 5 C. 7 D. 9
3. Find the sum of the first twelve term of the sequence 2, 5, 8, 11... A. 202 B. 212 C.222 D. 232
4. What is the general term of the sequence 31, 26, 21, 16, 11…………
A. 1 + 4n B. 3 x 2n-1 C. 36 – 5n D. 5(½)n-2
5. Find the sum of n terms of the A.P 3 + 6 +9 + 12 + ……….
A. 3n (n+1) B. 5n + 3/2 (n+1) C. n + 3(n-1) D. 2n + n (n-3)
2 2 3
Theory
1. The first three terms of an A.P are x, 2x+1, 4x+1, find x and the sum of the first 18 terms.
2. The sum of the first twenty –one terms of an A.P is 28, and the sum of the first twenty-eight terms is 21. Find which terms of the sequence is o and also the sum of the term proceeding it.
WEEK TWO DATE……………
TOPIC: SEQUENCE AND SERIES (II)
CONTENT
· Nth term of a G.P
· Geometric Mean
· Sum of n terms of a G.P
· Sum to infinity of a G.P
FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
Nth term of a G.P
A Geometric Progression is a sequence generated by multiplying or dividing a preceding term by a constant number to get a term. This constant number is called common ratio designated by letter r.
Examples: r a
4, 8, 16, 32, …………….. 8/4 = 2 4
8, 4, 2, 1, ½ 4/8 = ½ 8
3,-1, 11/3, -1/9 -1/3 = -1/3 3
For any G.P, the nth term is given by
Tn = arn-1
Tn = nth term
a = first term
r = common ratio
n = number of terms
Examples:
1. Find the 9th term of the sequence G.P 2, -10, 50 ………….
2. Find the number of term of the G.P 27, 81, 243 …………. 320
3. If 7, x, y, 189 are in G.P, find x and y
Solution
(1) G.P = 2, -10, 50 ………………
a = 2, r = -5, n = 9, T9 = ?
Tn = arn-1
T9 = 2(-5)9-1
= 2 x 390625
T9 = 781250
(2) G.P = 27, 81, 243 ………320
a = 27, r = 3, n =?, Tn = 320
Tn = arn-1
320 = 27(3)n-1
320 = 33(3)n-1
320 = 33+n-1
320 = 32+n
2+n = 20
n = 20 – 2 = 18
(3) The G.P = 7, x, y, 189.
a = 7, n = 4, T4 = 189
Tn = arn-1 = 189
7(r)4-1 = 189
r3 = 189/7 = 27 = 33
r = 3
T2 = x = ar = 7x 3 = 21
T3 = y ar2 = 7x3x3 = 63
Evaluation
(1) Find T9 of the G.P 5, 2½, 1¼, 5/8 …………..
(2) If 3, p, q, 24 are consecutive term of an exponential sequence, find the values of p and q.
Geometric Mean
Suppose x, y, z are consecutive terms of a geometric progression, then the common ratio r can be written as:
r = y/x = z/y
\y/x = z/y
y2 = xz
y = xz
y is the geometric mean of x and z.
Examples:
(1) Insert two geometric mean between 12 and 324.
(2) The 2nd term of an exponential sequence is 9 while the 4th term is 81. Find the common ratio and the first term of the G.P
Solution
(1) Let the G.P = 12, x, y, 324.
a = 12, T4 = 324, n = 4
Tn = arn-1
324 = 12(r)4-1
r3 = 324/12 = 27 = 33
r = 3
x = T2 = ar = 12 x 3 = 36
y = T3 = ar2 =12 x 3 x3 = 108
The geometric means are 36 and 108
2) T2 = 9, T4 = 81
T4 = ar3 = 81 ……………………(i)
T2 = ar = 9 …………………… (ii)
Divide (i) by (ii)
ar3 = 81/9
ar1
r2 = 9 Þ r = + √ 9 = +√3
ar = 9
a(+3) = 9
a = 9 = + 3
+3
The first term = + 3, the common ratio = + 3
FURTHER MATHEMATICS SS1 SECOND TERM LESSON NOTE
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