Mathematics lesson note for SS2 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.

Mathematics lesson note for SS2  First Term has been provided in detail here on schoolgist.ng

For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.

To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.

Mathematics Lesson note for SS2 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.

The SS2 Mathematics lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.

The sudden increase in the search for SS2 Mathematics lesson note for First Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.

This post is quite a lengthy one as it provides in full detail, the Mathematics-approved lesson note for all topics and sub-topics in Mathematics as a subject offered in SS2.

Please note that Mathematics lesson note for SS2 provided here for First Term is approved by the Ministry of Education based on the scheme of work.

I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.

SS2 Mathematics Lesson Note (First Term) 2024

FIRST TERM: E-LEARNING NOTES

SS 2 MATHEMATICS

SCHEME FIRST TERM

WEEK 1:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC:  logarithm 1

Content:

• Comparison of characteristics of logarithms and standard form of numbers.
• Revision of logarithm numbers greater than 1.

Comparison of characteristics of logarithms and standard form of numbers

There is a relationship between the standard form and the logarithm of a number.

For instance, (a) 189.7 =

This shows that the logarithm of a number is the power to which the base 10 is raised. Hence, Logarithm of 189.7 = 2.2781, where 189.7 =

(b)  850.9 = 8.509 (standard form)

Log850.9 = 2.9299

The integer (characteristics) is the same with the power 10

CLASS ACTIVITY

Show how the characteristics following Logarithms are related to standard form

1. 82000
2. 9
3. 6895
4. 8

Revision of logarithm numbers greater than 1

Logarithm of numbers is the power to which 10 is raised to give that number. Logarithms used in calculations are normally expressed in base 10.

• Rules for the use of Logarithms

1. Multiplication: find the logarithms of the numbers andadd them together
2. Division: find the logarithm of each number. Then subtract the logarithm of the denominator from that of the numerator

• Powers: find the logarithm of the number and then multiply it by the power or the index

1. Roots:find the logarithm of the number and then divide it by the root

Example1: Evaluate using logarithm tables

Example 2: Evaluate using logarithm tables

CLASS ACTIVITY

Use logarithm tables to evaluate correct to 4 s.f.

PRACTICE EXERCISE

Use log tables to find the value of

(SSCE 1991)

ASSIGNMENT

Use log tables to find the value of

KEYWORDS: base, logarithm, integer, antilogarithm, mantissa, characteristics etc.

WEEK 2:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC:  logarithm 2

Content:

• Logarithm of numbers less than one, involving: Multiplication, Division, Powers and roots.
• Solution of simple logarithmic equations.

Logarithm of numbers less than one

Simple logarithm operations

To find the logarithms of numbers less than 1, (i.e. numbers between 0 and 1), we use negative powers of 10.

For example, 0.08356 = 8.356  (standard form)

0.08356 =    (from log tables)

=

So Log0.08356 = -2+0.9220

Characteristics (i.e. power of 10) = -2

Mantissa = 0.9220

Note: -2 is called bar 2  i.e.

Example 1: Work out the following giving the answers in bar notation

SOLUTION

(a)

(b)

⇒

Example 2: Work out the following giving the answers in bar notation

SOLUTION

CLASS ACTIVITY

Work out the following in bar notation form

•                ii.       iii.     iv.      v.

•                   + 5.6          –          –        –

(b)

 Logarithm of numbers less than one, involving: Multiplication, Division, Powers and roots.

Example 1: a. Evaluate, using logarithm tables  0.9807 x 0.007692  

Solution:     0.9807 x 0.007692

antilog of = 0.00754 to 3s.f

Evaluate the following using logarithm tables

1. 00889 204.6

                                     antilog of  = 0.0000435 to 3s.f

Note: In Logarithm, powers take multiplication while roots take division.

Example 2: a. Evaluate

                       Solution:

0.000315

CLASS ACTIVITY

Evaluate the following using Logarithm tables

(1)i.

(2) i.

SOLUTION OF SIMPLE LOGARITHMIC EQUATIONS

In this lesson, the equations have to be solved first, then tables used to evaluate

If

Then

RULES OF LOGARITHM

• note that any logarithm to the same base is 1, that is, .

e.g

e.g     , e.t.c

Example 1: Solve the logarithmic equations

Solution:

from the formula above, we have that

this implies that because the powers will cancel each other

EXAMPLE 2: Solve the logarithmic equations

CLASS ACTIVITY

1. Solve for

2. Evaluate the following

PRACTICE EXERCISE

1. Use logarithm tables to evaluate

2. Evaluate using logarithm tables, correct to 3 significant figures
3. Evaluate using tables

(SSCE 1993)

4. Evaluate using tables, leaving your answer in standard form

Where P =3.6 × 10^-3 and Q = 2.25 × 10⁶

5. Evaluate using logarithm table, correct to 1 decimal place

ASSIGNMENT

1. Evaluate using logarithm table, leaving your answer in standard form

2. Use logarithm tables to evaluate

(SSCE 1991)

3. Use logarithm table to Evaluate, correct to 3 significant figures;

(SSCE 1994)

4. Use logarithm tables to Evaluate, correct to 3 significant figures;

(SSCE 1997)

5. Evaluate using logarithm tables
6. Evaluate using logarithm tables
7. Given that log₁₀p = 1, find the value of p.
8. B. 3    C. 10    D. 100    E.1000
9.  Evaluate log₅0.04
10. 0.008 B.  – 1.4 C. – 2 D. 1

KEYWORDS: base, logarithm, integer, antilogarithm, mantissa, characteristics etc.

WEEK 3:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC: Sequence and Series 1

Content:

• Meaning and types of sequence.
• Example of an A. P.
• Calculation of: (i) first term (ii) common difference (iii) nth term (iv) Arithmetic mean (v) sum of an A. P.
• Practical problems involving real life situations.

Meaning and types of sequence

SEQUENCES:

A sequence is an ordered list of numbers whose subsequent values are formed based on a definite rule. The numbers in the sequence are called terms and these terms are normally separated from each other by commas.

Examples:

2,  4,  6,  8,  10,……

 Rule: Addition of 2 for subsequent terms.

70,  66,  62,  58,  54,……

Rule: Subtraction of 4 for subsequent   terms.

3,  -6,  12,  -24,……

Rule: Multiply each term by –2.

There are many types of sequences. We shall be considering the Arithmetic Progression and Geometric progression.

 Finite and Infinite Sequences

A finite sequence is a sequence whose terms can be counted. i.e. it has an end. These types of sequences are usually terminated with a full stop. e.g.   (i) 3,5,7,9,11,13.         (ii)  -7,-10,-13,-16,-19,-21.

If however, the terms in the sequence have no end, the sequence is said to be infinite. These types of sequences are usually ended with three dots, showing that it is continuous. e.g.  (i)    5,8,11,14,17,20…    (ii) -35,-33,-31,-29,-27,… 

ARITHMETIC PROGRESSION (AP) {LINEAR SEQUENCE}

If in a sequence of terms T₁, T₂, T_3, …T_n-1, T_n the difference between any term and the one preceding it is constant, then the sequence is said to be in arithmetic progression (A.P) and the difference is known as the common difference, denoted by d.

\d = T_n – T_n-1, where n = 1, 2, 3, 4, …

i.e d = T₂ – T₁ = T₃ – T₂ = T₄ – T₃ and so on.

Examples of A.P

(i) 1, 3, 5,7, 9, …

T_n – T_n-1 Þ 5 – 3 = 2

7 – 5 = 2

9 – 7 = 2

• d = 2

The difference is common, hence it is an A.P.

(ii) 2, 4, 8, 16, 32, …

T_n – T_n-1 Þ 4 – 2 = 2

8 – 4 = 4

16 – 8 = 8

32 – 16 = 16

The difference is NOT common; therefore it is not an A.P.

(iii) 70, 66, 62, 58, 54, …

T_n – T_n-1 Þ 66 – 70 = -4

62 – 66 = -4

58 – 62 = -4

\  d = -4

The difference is common; hence it is an A.P.

(iv) –2, -5, -8, -11, …

T_n – T_n-1 Þ (-5) – (-2) = -5 + 2  = -3

(-8) – (-5) = -8 + 5  = -3

(-11) – (-8) = -11 + 8 = -3.

The difference is common; hence it is an A.P.

CLASS ACTIVITY

Which of the following are arithmetic progressing sequence?

• 4,6,8,10,…
• 3,7,9,11,..
• 1,6,11,16,21,26…
• 100,96,92,88,84,…
• 20,17,15,11,…
• 45,42,39,36,…

 THE nth TERM OF AN A.P

If the first term of an A.P is 3 and the common difference is 2. The terms of the sequence are formed as follows.

1^st term   = 3

2nd term = 3+2                 = 3 +  (1)2

3rd term = 3+2+2             = 3 +  (2)2

4th term = 3+2+2+2         = 3 +  (3)2

5th term = 3+2+2+2+2     = 3 +  (4)2

nth term = 3+2+2+2+ …  = 3 +  (n – 1)2

Hence, the nth term (T_n) of an A.P whose first term is “a” and the common difference is “d” is given as

Example 1:

a.Find the 21^st term of the A.P 3, 5, 7, 9, …

Solution

a = 3

d = 2

n = 21

T_n  = a + (n  – 1)d

T₂₁ = 3 + (21 – 1)2

= 3 + 20 x 2

= 3 + 40

= 43.

b.Find the 27th term of the A.P

100, 96, 92, 88, …

Solution

a = 100

d = -4

n = 27

T_n   = a + (n – 1)d

T₂₇  = 100 + (27 – 1)(-4)

= 100 + 26 x –4

= 100 – 104

= -4.

Example 2:     

a.Find the value of n given that 77 is the nth term of an A.P 3½, 7, 10½, …

Solution:

a = 3½

d = 7 – 3½

\    d = 3½

T_n = 77

T_n = a + (n – 1)d

77 = 3½ + (n – 1)3½

77 = 3½ + 3½n – 3½

77 = 3½n

77 = ⁷/₂n

7n = 77 x 2

n = 77 x 2

7

n = 11 x 2

\n = 22.

1. What is the first term of an A.P whose 21^stterm is 43 and the common difference is 2   ?

Solution:

T₂₁ = 43

n = 21

d = 2

T_n = a + (n –1)d

43 = a + 20 x 2

43 = a + 40

a = 43 – 40

\a = 3

3. Find the common difference of an A.P given that 43is the21^st term of the sequence and the first term is 3.

Solution:

a = 3

T₂₁ = 43

n = 21

T_n = a + (n – 1)d

43 = 3 + (21 – 1) d

43 = 3 + 20d

43 –3 = 20d

20d = 40

d = 40

20

\ d = 2.

CLASS ACTIVITY

• Find the 31^stterm of the sequence       –7, -10, -13, -16, …

(2)   What is the 26th term of the A.P 5, 10, 15, 20, …?

FURTHER Example 1:

The first three terms of an A.P are

x, 3x + 1, and (7x – 4). Find the

(i) Value of x

(ii) 10th term

Solution:

(i) Recall that given an A.P T₁, T₂, T₃

T₂ – T₁ = T₃ – T₂

Hence for,   x, (3x + 1), (7x – 4)

(3x + 1) – x = (7x – 4) – (3x + 1)

3x+1 – x = 7x – 4 – 3x – 1

2x + 1 = 4x – 5

1+ 5 = 4x – 2x

2x = 6

x = 6

2

\ x = 3.

The sequence x, (3x + 1), (7x – 4) is

= 3, (3×3 + 1), (7×3 – 4)

= 3, 10, 17.

(ii)  a = 3

n = 10

d = 7

T_n = a + (n – 1)d

T₁₀ = 3 + (10 – 1)7

= 3 + 9×7

= 3 + 63

= 66.

Example 2:

The 6th term of an A.P is –10 and the 9th term is –28.

Find the (i) Common difference

  (ii) First term

(iii) 26th term of the sequence.

Solution:

(i) T₆ = -10       T_n = a + (n – 1)d

n = 6        -10 = a + (6 – 1)d

-10 = a + 5d ———– (1)

T₉ = -28      -28 = a + (9 – 1)d

n = 9         -28 = a + 8d ———- (2)

Solve equation (1) and (2) simultaneously.

Eqn. (1):   -10 = a + 5d

Eqn. (2):   -28 = a + 8d

18 =     -3d

d = 18

-3

\  d = -6.

(ii)      Put    d = -6 in equation (1)

-10 = a + 5(-6)

-10 = a – 30

-10+30 = a

\ a = 20.

(iii) To find the 26th term of the sequence.

a = 20

d = -6

n = 26

T_n = a + (n – 1)d

T₂₆ = 20 + (26 – 1)(-6)

= 20 + 25(-6)

= 20 – 150

= -130.

CLASS ACTIVITY

• The 6th term of an AP is –10 and the 9th term is 18 less than the 6th term. Find the

(a) common difference (b) first term

(c) 26th term of the sequence.

(2) The 7th term of an AP is 17 and the 13th term is 12 more than the 7th     term. Find the (i) common difference (ii) first term (iii) 21st term of the AP.

Arithmetic Series:

These are series formed from an arithmetic progression. e.g.

1 + 4 + 7 + 10 + …

In general, if S_n is the sum of n terms of an arithmetic series then

S_n = a + (a + d) + (a + 2d) + … + (l – d) +

l — (1)

Where l is the nth term, a is the first term and d is the common difference.

Rewriting the series above starting with the nth term, we have.

S_n = l +(l – d) + (l – 2d) +… + (a + d) + a —– (2)

Adding equation (1) and (2) we have

2S_n = (a + l) + (a + l) + … + (a + l) + (a + l) in n places

2S_n = n(a + l)

\S_n = ⁿ/₂(a + l)

But l is the nth term i.e a + (n – 1) d

S_n =  ⁿ/₂{a +a + (n – 1)d}

\ S_n = ⁿ/₂ {2a + (n – 1)d}

Example 1 :

Find the sum of the first 20 terms of the series 3+5+7+9+ …

Solution:

 a = 3

d = 2

n = 20

S_n = ⁿ/₂ {2a +(n – 1)d}

S₂₀ = ²⁰/₂ {2×3 +(20 – 1)2}

S₂₀ = 10{6 + 19 x 2}

= 10{6 + 38}

= 10{44}

\ S₂₀ = 440

Example 2:

Find the sum of the first 28 terms of the series –17 + (-14) + (-11) + (-8) + …

Solution:

a = -17

d = 3

n = 28

S_n = ⁿ/₂{2a + (n – 1)d}

S₂₈ = ²⁸/₂ {2 (-17) + (28 – 1) 3}

= 14 {-34 + 27 x 3}

= 14 { -34 + 81}

= 14 {47}

\ S₂₈  = 658

CLASS ACTIVITY

(1) Find the sum of the numbers from 1 to 100.

(2) Find the sum of the first 26 terms of the A.P –18, -15, -12, -9, …

FURTHER Example :

The sum of the first 9 terms of an A.P is 117 and the sum of the next 4 terms is 104.

Find the(i) Common difference

      (ii) First term

(iii)25th term of the A.P

                                        (WAEC)

Solution:

T₁, T₂, T₃, T₄ … T₈, T₉,               T₁₀, T₁₁, T₁₂, T₁₃

117                104

S₉ = 117   ————————–(*)

n = 9

since a sequence is normally summed from the first term

S_{9 + 4} = 117 + 104

\ S₁₃ = 221               —————(**)

n = 13

S_n = ⁿ/₂ {2a + (n – 1) d}

From (*)  above;

117 = ⁹/₂ {2a + (9 – 1) d}

117 = ⁹/₂ x 2a + ⁹/₂ x 8d

117 = 9a + 36d

Divide through by 9 to have;

13 = a + 4d ———————-(1)

From  (**) above;

221 = ¹³/₂ {2a + (13 – 1) d}

221 = ¹³/₂ x 2a + ¹³/₂ x 12d

221 = 13a + 78d

Divide through by 13 to have;

17 = a + 6d ———————-(2)

From equation (1) and (2)

Eqn. (1): 13 = a + 4d

Eqn. (2): 17 = a + 6d

-4  =     -2d

d = -4

-2

\ d = 2.

(ii) From equation (1) we have

13 = a + 4 x 2

13 = a + 8

a = 13 – 8

\  a = 5

(iii)a = 5

d = 2

n = 25

T_n = a + (n – 1) d

T₂₅ = 5 + (25 – 1) 2

= 5 + 24 x 2

= 5 + 48

T₂₅ = 53

CLASS ACTIVITY

(1) The sum of the first 9 terms of an A.P is 171 and the sum of the next 5 terms is 235.Find the (a) Common difference

(b) First term

(c) Sequence

(2) The sum of the first 8 terms of an A.P is 172 and the sum of the next three terms is 15. Find the

(a) Common difference

(b) First term

(c) 21^st term of the A.P

PRACTICAL  PROBLEMS INVOLVING REAL LIFE SITUATION

Example 1:

A clerk employed by a private establishment on an initial salary of N5000 per annum. If his annual increment in salary is N300. Find the total salary earned by the clerk in 20years.

Solution:

a = 5000

d = 300

n = 20

S_n = ⁿ/₂ {2a + (n – 1) d}

S₂₀ = ²⁰/₂ {2 x 5000 + (20 – 1) 300}

= 10 {10000 + 19 x 300}

= 10 {10000 + 5700}

= 10 {15700}

\ S₂₀ = 157000

\ The total amount earned in 20years is N157000.00

EXAMPLE 2

A sum of money is shared among nine people so that the first gets N75, the next N150, the next N225, and so on.

1. How much money does the ninth person get?
2. How much money is shared altogether?

Solution:

 a=75   d=150-75=75   n=9

T_n=a + (n-1)d

T₉=75 + (9-1)75

     =75+600

=N675

S_n

S₉

=4.5 × 750

=N3375

CLASS ACTIVITY

• The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1^stJanuary 1980 was N10,250.00, what was its value in January 1989 when it was 9years old? Give your answer correct to three significant figures.

                                             (WAEC) 1989

(2) The houses on one side of a particular street are assigned odd numbers,      starting from 11. If the sum of the numbers is 551, how many houses are there? (SSCE 1999)

PRACTICE EXERCISE

(1) The 6th term of an A.P is 26 and the 11th   term is 46. Find the

(i) Common difference

(ii) First term

(iii) 25th term of the A.P

• The 5th term of an A.P is 11 and the 9th term is 19. Find the

(i)

common difference

(ii) First term

(iii) 21^st term of the A.P

(3) The fourth term of an A.P is 37 and the     6th term is 12 more than the

fourth term. Find the first and seventh terms.

SSCE, June 1994, No 11a (WAEC)

(4) The first three terms of an A.P are (x+2) ,(2x-5) and (4x+1). Find the

(i) Value of x

(ii) 7th term.

(5) The first three terms of an A.P are x,    (2x-5) and (x+6). Find the

(i) Value of x

(ii) 21^st term.

ASSIGNMENT

1. If the first three terms of an A.P are  (4x+1), (2x-5) and (x+3). Find the

(i) Value of x

(ii) Sequence

(iii) 11th term of the sequence

2. Given that 9, x, y, 24 are in A.P, find the values of x and y.

(10) If –5, a, b, 16 are in A.P, find the values    of a and b.

3. The 8th term of an arithmetic progression (A.P) is 5 times the third term while the 7th term is 9 times greater than the 4th term. Write the first five terms of the A.P. (SSCE 2009)

4. If 3, x, y, 18 are the arithmetic progression (A.P).  Find the values of x and y. (SSCE 2008)

5. If are successive terms of an arithmetic progression (A.P), show that (SSCE 2007)
6. The 3rd and 8th terms of an arithmetic progression (A.P)are -9 and 26 respectively. Find the:
7. Common difference
8. First term (SSCE 2007)
9. The 2nd, 3rd and 4th terms of an A.P. are Calculate the value of

(SSCE 2006)

8. An arithmetic progression (A.P) has 3 as its first term and 4 as the common difference.
9. Write an expression, in its simplest form for the nth term.
10. Find the least term of the A.P that is greater than 100 (SSCE 2003)

9. The first term of an arithmetic progression (A.P)is 3 and the common difference is 4. Find the sum of the first 28 terms. (SSCE 2002)

10. The first term of an arithmetic progression (A.P) is -8. The ratio of the 7th term to the 9th term is 5:8. Calculate the common difference of the progression. (SSCE 2000)

11. The 6th term of an A.P. is 35 and the 13th term is 77. Find the 20th term. (SSCE 1997)

KEYWORDS: sequence, series, first term, common difference, last term, arithmetic progression, finite sequence etc.

WEEK 4:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC:  Sequence and Series II

Content:

• Examples of geometric progression
• Calculation of; (i) First term (ii) Common ratio (iii) nth term, (iv) Geometric progression (v) sum of terms of geometric progression. (vi) Sum to infinity
• Practical problems involving real life situation.

GEOMETRIC PROGRESSION (G.P) OR EXPONENTIAL SEQUENCE

Given any sequence of terms T₁, T₂, T₃, T₄, … T_n-1, T_n. If the ratio between any term and the one preceding it is constant then the sequence is said to be in geometric progression (G.P). The ratio is called the common ratio denoted by r.

i.e.

r =

where n = 1, 2, 3, 4, …

Example1: find out which of the sequence is a GP

(i) 1, 2, 4, 8, 16, …

Þ     = 2

1

= 2

= 2

The ratio is common, hence the sequence is a G.P  \ r = 2

(ii)   16, 8, 4, 2, …

Þ    =

=

=

The ratio is common, hence the sequence is a G.P        \   r  =

Example 2: find out which of the sequence is a GP

(i)   3, 7, 9, 12, …

Þ   =

3

=

The ratio is NOT common, hence not a G.P.

(ii)    2, -10, +50, -250, …

T_n  Þ   = -5

T_n-1

= -5

= -5

The ratio is common, hence the sequence is a G.P         \ r = -5

CLASS ACTIVITY

 Which of the following sequences is G .P.?

• 3,6,12,24…
• 2,4,6,8,…
• 5,15,45,…
• 1,4,16,…

THE NTH TERM OF A G.P

Let 5 be the first term of a G.P whose common ratio is 2. Then

The 2nd term is 5×2              = 5(2)¹

The 3rd term is 5x2x2           = 5(2)²

The 4th term is 5x2x2x2       = 5(2)³

The 5th term is 5x2x2x2x2   = 5(2)⁴

The nth term is 5x2x2 … 2   = 5(2)^{n – 1}

In general, the nth term of a G.P denoted by T_n, whose first term is “a” and whose common ratio is “r” is ar^n-1.    i.e.

T_n = ar^{n – 1}   where n = 1, 2, 3, 4, …

NOTE

The four examples below show how the formula can be used to find the nth term, n, r and a.

Example 1:      

1. Find the 8th term of the G.P

3, 6, 12, 24, …

Solution

a = 3

r = 2

n = 8

T_n = ar^{n – 1}

T₂₁ = 3(2)^{8 – 1}

T₂₁ = 3×2⁷

T₂₁ = 3×128

\ T₂₁ = 384

1. Find the value of n given that the nth term of a G.P is 2916 and the first term and common ratio are 4and 3

Solution:

T_n = 2916

r = 3

a = 4

T_n = ar^{n – 1}

2916 = 4(3) ^{n – 1}

 2916 = 3^{n – 1}

4

729 = 3^{n – 1}

3⁶ = 3^{n – 1}

n-1 = 6

n = 6+1

\ n = 7.

EXAMPLE 2:

1. Find the common ratio of an exponential sequence whose 10th term is –512and the first term is

Solution

a = 1

T₁₀ = -512

n = 10

T_n = ar^{n – 1}

-512 = 1(r)^{10 – 1}

-512 = r⁹

(-512)^1/9 = r

r = (-2⁹)^1/9

\ r = -2

2. Find the first term of an exponential sequencewhose 7th term is 4096 and the common ratio is

Solution

T₇ = 4096

n = 7

r = 4

T_n = ar^{n – 1}

4096 = a(4)^7-1

4096 = a4⁶

4096 = 4096a

a = 4096

4096

\a = 1

CLASS  ACTIVITY

(1) The 3rd and 6th term of a geometric progression (G.P) are 48 and 14²/₉ respectively. Write down the first four terms of the G.P.

SSCE, June. 1993, No9 (WAEC)

(2) The first and third terms of a G.P are 5 and 80 respectively. What is the 4th term?

SSCE, Nov. 1993, No 11b (WAEC)

GEOMETRIC SERIES

The general expression for a geometric series is given as

S_n = a + ar + ar² + ar³ + … + ar^n-1 —– (1)

Where S_n represents the sum of n terms of the series

Multiply both sides of equation (1) by r to have

rS_n = ar + ar² + ar³ + ar⁴ + — + arⁿ —- (2)

Subtract (2) from (1) to have

S_n –  rS_n = a – arⁿ

S_n (1 – r) = a(1 –  rⁿ)

S_n=   —————— (3)

If the numerator and denominator of equation (3) is multiplied by –1, we have

S_n=       ——————– (4)

If r < 1, formula (3) is more convenient

If r >1, formula (4) is more convenient

Example 1:

Find the sum of the first 8 terms of the G.P 3, 6, 12, 24, …

Solution:

a = 3

n = 8

r = 2

S_n = a(rⁿ – 1)           r > 1

r – 1

S₈ = 3(2⁸ – 1)

2 – 1

= 3(256 – 1)

1

= 3 (255)

S₈ = 765.

Example 2:

Find the sum of the first 10 terms of the G.P  2, -6, 18, -54, …

Solution:

a = 2

n = 10

r = -3

S_n=      r < 1     

S₁₀ = 2(1-(-3)¹⁰)

1-(-3)

= 2(1-59049)

1+3

= 2(-59048)

4

= -59048

2

S₁₀ = -29524

CLASS ACTIVITY

(1) Find the sum of the first 9 terms of the sequence 84, 42, 21, 10½, …

(2) Find the sum of the first 10 terms of the G.P 4, 8, 16, 32, …

SUM TO INFINITY

In general, if the common ratio, r, is a fraction such that -1n approaches zero as n increases towards infinity. It then follows that the sum

S=     becomes

This formula gives the sum to infinity of a geometric progression.

EXAMPLE 1:

Find the sum to infinity of the series S_n=36+24+6+4+….

Solution

a=36  r=

EXAMPLE 2:

The sum to infinity of a GP series is 15/7 and its second term is -6/5. Find the common ratio.

Solution

Now, ar=-6/5

a=-6/5r

15 (1-r)=7 (-6/5r)

15- 15r =-42/5r

5r(15-15r)=-42

75r-75r²= -42

25r – 25r² = -14

25r²-25r-14=0

25r²-35r+10r-14=0

5r(5r-7)+2(5r-7)=0

(5r+2)(5r-7)=0

5r+2=0 or 5r-7=0

r=-2/5 or 7/5

Note that r=7/5 is not a true value since (7/5)ⁿ will not approach zero(0) as n increases towards infinity. Hence the valid common ratio of the series is -2/5.

CLASS ACTIVITY

1. Find the sum to infinity of  S_n=200+120+72+43 +……..
2. the first term of a GP series is 4/9 and the sum to infinity is . find the common ratio.

Practical problems involving real life situation

Example 1:

2. A ball was dropped from a height 80m above a concrete floor. It rebounded to the height of ½ of its previous height at each rebound. After how many bounces is the ball 2.5m high?           NECO 2001

Solution

The rebounds forms a sequence of the order

80, 40, 20, …,2.5.

a = 80

r = ½

T_n = 2.5      T_n   =  ar^n-1

2.5 = 80(½)^n-1

2.5 =  (½)^n-1

80

=  (½)^n-1

  5     = (½)^n-1

2×80

¹/₃₂  = (½)^n-1

                                    (½)⁵  = (½)^n-1

                             n – 1 = 5

n = 5 + 1

\  n  =  6

486. The 3rd term of a G.P is 54 and the 5th term is 486. Find the

• Common ratio
• First term
• 7th term of the G.P

Solution

(a) T₃ = 54          T_n = ar^n-1

n = 3            54 = ar^3-1

54 = ar² ————— (1)

T₅ = 486       486 = ar^5-1

n = 5           486 = ar⁴  ————– (2)

Equation (2) ¸ equation (1)

486  =  ar⁴

54       ar²

9 = r²

r = ±Ö9

r = ±3

(b) Substitute in equation (1)

54 = a(±3)²

54 = 9a

a = ⁵⁴/₉

\ a = 6

(c)   a = 6

r = ±3

n = 7

T_n = ar^{n – 1}

T₇ = 6(±3)^{7 – 1}

= 6×3⁶

= 6×729

= 4373

Example 2: 

1. If 2, x, y, 54 are in G.P, find x and y.

Solution:

a = 2

n = 4

T₄ = 54

T_n = ar^{n – 1}

54 = 2(r)^{4 -1}

54 = 2r³

⁵⁴/₂ = r³

r³ = 27

r ³ = 3³

\r = 3

a = 2

r = 3

T₂ = x 

n = 2

T_n = ar^{n – 1}

x  = 2(3)^{2 – 1}

x = 2×3

\ x = 6

a = 2

r = 3

T₃ = y

n = 3

y = 2(3)^{3 – 1}

y = 2(3)²

y = 2×9

\ y = 18

ii.Given that x, (x-2), (2x-1) are in G.P. Find the

• Value(s) of x
• Sequences
• Possible value(s) of the 8th term of the G.P

Solution:

Since x, (x-2), (2x-1) are in G.P, the ratio between any of the terms and the one preceding must be common.

i.e (x-2)  =  (2x-1)

x          (x-2) 

Cross-multiplying

(x – 2)(x – 2)  =  x(2x – 1)

x² – 2x –2x + 4 = 2x² – x

0 = 2x² – x – x² + 2x + 2x – 4

0 = x² + 3x – 4

i.e x² + 3x – 4 = 0 (Factorize the equation)

-4x²

x² + 4x – x – 4 = 0

x(x + 4) – 1(x + 4) = 0

(x + 4)(x – 1) = 0

x + 4 = 0     or  x – 1 = 0

x = -4   or      x  = 1

(b) Since x = -4 and 1, we shall have two sequences

For x = -4

x, (x-2), (2x-1)

= -4, (-4 – 2), (2(-4)-1)

= -4, -6, -9

For x = 1

x, (x-2), (2x-1)

= 1, (1-2), (2×1-1)

= 1, -1, 1

(c) For x = -4 the sequence = -4, -6, -9

a = -4

n = 8

r = 3

2

T₈ = -4(³/₂) ^{8 – 1}

= -4(³/₂)⁷

= -4 x 2187

128

\T₈  = – 2187

32

For x =1, the sequence = 1, -1, 1

a = 1

n = 8

r = -1

T₈ = -1(-1)^{8 – 1}

                = 1(-1) ⁷

= -1

Class activity

1. Three consecutive terms of a geometric progression are given as n-2,n and n+3. Find the common ratio.
2. The 3^rdand 6^th terms of a geometric progression are 48 and 14 Write down the first four terms of the GP.

PRACTICE EXERCISE

(1) Find the sum of the first 7 terms of the G.P 3, 9, 27, 81, …

(2) Find the sum of the first 11 terms of the G.P 5, 10, 20, 40, …

(3) If the 3rd and 7th terms of a G.P are 12 and 192 respectively, find the sum of the first 6 terms of the sequence.

(4) The third and fifth terms of a geometric progression are ⁹/₂  and  ⁸¹/₈  respectively.

Find the  (i) Common ratio

(ii) First term.

(5) If 2, x, y, -250, … is a geometric progression, find x and y.

                                                (JAMB)

ASSIGNMENT

1. Given that 2, a, b, 686 are in G.P, find the value of a and b.

2. The first three terms of a G.P are x+1, (x+4) and 2x. Find the

• Value (s) of x

(ii) Sequence(s)

(iii) 6th term of sequences

3. What is the 25^thterm of 5, 9, 13?  The 3^rd and 6^th terms of a Geometric progression are 48 and 14 .Write down the first four terms of the G.P. (SSCE 1993)

4. The third term of a Geometric Progression is 360 and the sixth term is 1215.Find the

(a) common ratio (b) first term (c) sum of the first four terms.     (SSCE 1998)

5. The 1^stand 3^rd terms of a geometric Progression are 2 and Find

(i) the common difference (ii) the 5^th term. (SSCE 1999)

6. Write down the 15^thterm of the sequence  , , , …, (SSCE 2003)

7. The sum of the second and third terms of a geometric progression is six times the fourth term. Find the two possible values of the common ratio.(ii) If the second term is 8 and the common ratio is positive, find the first six terms      (SSCE 2008)

8. The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4.

Find its first term. (SSCE 2010)

9. Given the Geometric Progression 6, 12, 24, 48, …

Find (i) its common ratio (ii) the 24^th and 40^th term of the G.P.

10. The sum of the Geometric Progression 2 + 6 + 18 + 54 +…+ 1458 is 2186.

Find the nth term of the G.P.

11. Find the sum of the G.P 3 + 9 + 27 + …. up to the 7^th

12. Given the Geometric Progression (G.P) 4, 16, 64…

Find the 6^th and 7^th term of the G.P respectively.

13. The sum to infinity of a geometric progression is -1/10 and the first term is -1/8. Find the common ratio of the progression.   JAMB 2012
14. Find the sum to infinity of the series ½ ,1/6, 1/18,…. JAMB 2004

KEYWORDS:

sequence, series, first term, common ratio,last term, geometric progression, infinite series etc.

WEEK 5:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC: Quadratic Equations

Content:  

• Revision of factorization of perfect squares
• Making quadratic expression perfect squares by adding a constant K.
• Solution of quadratic equation by the method of completing the square 
• Deducing the quadratic formula from completing the square.
• Construction of quadratic equation from sum and product of roots.
• Word problems leading to quadratic equations.

Revision of factorization of perfect squares

A quadratic expression is a perfect square if it can be expressed as the product of two linear factors that are identical. For example,x²+4x+4=(x+2)(x+2) ,x²+6x+9=(x+3)(x+3)  are perfect squares.

Example 1: Factorize

The expression is a perfect square if the first term and the constant terms are both perfect squares. The sign of the middle term of the quadratic expression can be put between the terms of the linear factors.

For , the first term is the square of  and the constant term is 121 which is the square of 11.

Notice that the middle term is twice the product of the terms of the linear factor.  In this case

Example 2: Factorize

The leading term (term containing the highest power of the variable) is a perfect square. Also the constant term, 400 is a perfect square.

So,

Example 3: Factorize

Solution;

Notice that the middle term of the quadratic expression is twice the product of the terms of the linear factors.

CLASS ACTIVITY

Factorize the following quadratic expressions

MAKING QUADRATIC EXPRESSION PERFECT SQUARES BY ADDING A CONSTANT K

In this section, we shall consider the constant K to be added to a quadratic expression to make it a perfect square.

Example 1: What must be added to  to make the expression a perfect square?

Solution:

Let the constant term be k,  is a perfect square if

i.e.

equating the coefficients of m, we have

Equating the constant terms,

Therefore 16 must be added to the expression to make it a perfect square.

In general, the quantity to be added is the square of half of the coefficient of m (or whatever letter is involved)

Example 2: What must be added to  to make it a perfect square?

Solution:

What is to be added is the square of half of the coefficient of w

The coefficient of w is

Half of the coefficient is

The square of half of the coefficient is  , so  must be added to  to make it a perfect square.               i.e

CLASS ACTIVITY

What must be added to …

1. X²+4x
2. M²-3m

SOLUTION OF QUADRATIC EQUATION BY THE METHOD OF COMPLETING THE SQUARE

In this section, we shall solve quadratic equations using method of completing the square

Example 1:

Solve the quadratic equation using completing the square method:

                                                    Solution

the L.H.S of the equation does not factorize. So, one can rearrange the equation to make the L.H.S a perfect square

We then find half of the coefficient of , square it and add it to both sides of the equation. The reason for that is to make the L.H.S a perfect square.

i.e

Take square root of both sides,

EXAMPLE 2:

Solve  using completing the square method.

                                           Solution

L.H.S of  cannot be factorized,

Make the coefficient of  one by dividing through by 3 to get

CLASS ACTIVITY

Solve the following using completing the square method

DEDUCING THE QUADRATIC FORMULA FROM COMPLETING THE SQUARE

The general form of a quadratic equation is .

We shall now derive the formula for solving the equation by method of completing the square

Given that,

Since  we can divide through by  to get

We shall now make the left hand side a perfect square by adding  to both sides i.e half of the coefficient of  all squared.

Take square root of both sides;

This is known as quadratic formula

Example 1:  Use quadratic formula to solve,

We shall write the equation in the form

So,

Use the formula,

EXAMPLE 2: i. Solve using the quadratic formula,

Solution

Considering the coefficients of

1.  Use quadratic formula to solve;

Solution; first is to clear the fractions by multiplying each term of both sides of the equation by the LCM of the denominators which

On rearranging, we have;

CLASS ACTIVITY

Use quadratic formula to solve the following equations:

1.  i.

CONSTRUCTION OF QUADRATIC EQUATION FROM SUM AND PRODUCT OF ROOTS

We shall now consider how to construct quadratic equations from sum and product of roots. The root on an equation is the value that satisfies the equation.

Let  be roots of a quadratic equation then

The general form of a quadratic equation is

Divide through by

Comparing equations (i) and (ii); equating coefficients of x, we have,

Equating their constant terms; we have

Consequently, a quadratic equation can be expressed or written as;

Example 1: Find the quadratic equation whose roots are

Solution

The equation is

Clear the fraction by multiplying through by the LCM of the denominator i.e. 6

⇒

EXAMPLE 2:

Construct the quadratic equation whose sum and product of roots are respectively  .

Solution

⇒

CLASS  ACTIVITY

1. Find the sum and product of the roots of the following quadratic equations:
4. Find the quadratic equation whose roots are;

WORD PROBLEMS LEADING TO QUADRATIC EQUATIONS

In this section, we shall consider some word problems leading to quadratic equation.

Example 1:  If 63 is subtracted from twice the square of a number, the result is the same as five times of the number. Find the number

Solution

Let the number,

Twice the square of the number is

The equation required is

On rearranging, we have

This is then solved using any of the known methods (completing the square method or quadratic formula or factorization method)

Example 2:   Find two consecutive odd numbers whose product is 399

Solution: 

Let the smaller of the odd number be the other one will be  (note that this approach also goes for consecutive even numbers)

Solving this equation, we obtain,

If then the other will be

If  then the other will be

Therefore, the numbers are -19 & -21 or 19 & 21.

CLASS ACTIVITY

1. A mother is 36 years old and her son is 6 years. When will the product of their ages be 451
2. The hypotenuse of a right angled triangle is one unit more than twice the shortest side. The third side is one unit less than twice the shortest side. Find the; (a) shortest side   (b) hypotenus
3. A man is 3 times as old as his son, 8 years ago, the product of their ages was 112. Find their present ages.

PRACTICE EXERCISE

1. What value of k makes the given expression a perfect square?²
2. 2 4        C. 8        D. 16        E. 64(SSCE 1988)

2. Factorize: ²²
3. (5y-a)(y+3a)       (5y+a)(y-3)       C. (5y²+a)(2y-3a)
4. (y-a)(5y+3a)    E. (y+a)(5y-3a)                              (SSCE 1988)

3.  Factorize the following expression: 2x²+x – 15.
4. (2x+5)(x-3) B.(2x-5)(x-3)   C.  (2x-5)(x-3)   D.(2x-3)(x+5)
5. (2x+5)(x+3)         (SSCE 1989)

4.  If 4x²-12x+c is a perfect square, find the value of c.
5. 36 B. 9       C. 9/4      D. -9/4      E. 24  (SSCE 1990)

5.  Factorize completely 4a³-a
6. a(4a²-1)   B. (2a-1)(2a+1) C. a(2a-1) D. 2a²(a-1) E. (2a+1)(2a-1)

(SSCE 1990)

6.  Factorize 3a²-11a+6
7. (3a-2)(a – 3) B.  (2a-2)(a-3) C. (3a – 2)(a – 3) D. (3a + 2)(a -3)
8.   (2a-3)(a+2)    (SSCE 1991)

7.  Factorize a²– 3a – 10
8. (a+5)(a+2) B. (a – 5)(a – 2) C. (a + 5)(a – 2) D.(a – 5)(a + 2)

8.  Factorize the expressions: ²²
9. 2()-3² B.(2)(2)   C.  ()()
10. E.²

9.  What must be added to the expression ²-18 to make it a perfect square?
10. 3      B. 9      C.36      D.72      E. 81       (SSCE 1992)

10.  Factorize the expression 2s²-3st-2t²
11. (2s-t)(s+2t)    B. (2s+t)(s-2t)   C. (s+t)(2s-t) D. (2s+t)(s-t)
12. (2s+t)(s+2t)               (SSCE 1993)

11. Factorize ²
12. B. C.
13. E. (SSCE 1993)

12.  Factorize ²²
13. B. C.
14. E. (SSCE 1994)

ASSIGNMENT

1. Factorize ² (SSCE 1990)

2. Factorize ² (SSCE 1990)

3. Find the value of m which makes ²a perfect square.                 (SSCE 1990)

4. Given that ²factor            (SSCE 2000)
5. Derive an equation whose coefficients are integers and which has roots of ¹/₂ and -7.
6. Three years ago a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate present ages of the daughter and the father.(SSCE 1989)
7. Solve the equation correct to two decimal places: ².       (SSCE 1991)
8. Using the method of completing the square, find the roots of the equation ²

Correct to 1 decimal place.     (SSCE 1995)

9. The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving  it, find the two numbers.                                       (SSCE 1995)
10. The area of a rectangular floor is ². One side is  longer than the other.  Calculate

(a) The dimensions of the floor

(b) If it costs  per square metre to carpet the floor and only  is

available, what area of the floor  can be covered with the carpet?    (SSCE 1998)

11. A rectangular lawn of length  metres is  metres wide. If the diagonal is

metres, find;

(i) the value of

(ii) the area of lawn.

12. The sum of the ages of a woman and her daughter is 46 years. In 4 years’ time, the ratio of their ages will be 7:2. Find their present ages.             (SSCE 2001)

13. The sides of a rectangular floor are  The diagonal is .

Calculate in metres;

(a) the value of

(b) the area of the floor.                                          (SSCE 2001)

14. Solve correct to two decimal places the equation ² (SSCE2008)

15. The lengths, in cm of the sides of a right angled-triangle are where  Find, in cm, the length of its hypotenuse.

KEYWORDS:

Quadratic equation, Factorize, perfect square, coefficient, roots, quadratic formula, etc.

WEEK 6:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC: Simultaneous Linear and Quadratic Equation 

Content:  

• Simultaneous linear equations (Revision).
• Solution to linear and quadratic equations.
• Graphical solution of linear and quadratic equations.
• Word problems leading to simultaneous equations.
• Gradient of curve.

SIMULTANEOUS LINEAR EQUATIONS (REVISION)

Simultaneous linear equations (revision)

Recall: Simultaneous means happening or done at the same time i.e. following each other. It can be solved either graphically or algebraically. Algebraically involves using either substitution or elimination methods

Example 1: Solve the following pairs of simultaneous equation

Solution:

ELIMINATION METHOD

(a)  ,

On adding both equations, we have

Substituting for  in equation (1)

⇒

Using substitution method;

• ………….(1)

,  …………(2)

In (2),

Substituting in equation (1), we have

In (3),

EXAMPLE 2: Solve the pairs of equations;

and

Solution;

Recall;

⇒

Comparing the powers of 3, we have,

Solving equations (i) and (ii) we obtain values for

i.e

CLASS ACTIVITY

1. Solve the following pairs of simultaneous equations

,

• ,

2. Solve the questions below; (WAEC)

,

SOLUTION OF LINEAR AND QUADRATIC EQUATIONS

When solving a simultaneous equation involving one linear and one quadratic such as  (or ) graphical or substitution not elimination method is frequently used.

EXAMPLE 1: Solve the pair of equation;      ……………….  (i)

…………………(ii)

Solution

from equation (ii),

Substituting for  in (i),

Collecting like terms and rearranging the equation; we have

Using any of the methods learnt previously for solving quadratic equations, we have that

When                                            When

EXAMPLE 2: Solve completely;

Solution:

Note:

……….(ii)

Substituting for m+n in (ii),

Multiply equation (ii) by 7

Add (iii) & (iv)

Substitute for m in (ii),

CLASS ACTIVITY

1. Solve: (a)

(b.)

2. (a)A man is years old while his son is  years old. The sum of their ages is equal to twice the difference of their ages. The product of their ages is 675. Write down the equations connecting their ages and solve the equations in order to find the ages of the man and his son.    (WAEC)

(b)Solve:

GRAPHICAL SOLUTION OF LINEAR AND QUADRATIC EQUATIONS

Note: To use graphical method, plot the graphs of each equation on the same axes. Then read off the x & y coordinates of the points where both lines cross to obtain the required solution.

Example 1:

Solve the following simultaneous equations graphically:

Solution    

Tables of values ()

From the graph, x≃1.6 and y≃1.5  or x≃ -1.6 and y=5.6

EXAMPLE 2:

         Draw the graphs of

• Use your graph to solve the following equations;

• Find the minimum value of

Solution

Tables of values for

• From the graph,
• For we have to make it look like the first graphical equation given. So we add 12 to both sides;

The line  is drawn on the graph

• For this, we add to both sides ;

Then subtract 2 from both sides as well,

The line has tables of values as follows

From the graph above,

(b)  The minimum value of  is where the turning point of the curve is, i.e at and it occurs at

CLASS ACTIVITY

1. Solve the following simultaneous equations graphically;

2. (a)   Copy and complete the following table of values for the relation;

.(b)  Draw the graph of the relation using a scale of 2cm to 1 unit on each axis.

• Using the same axes, draw the graph of
• From your graphs determine the roots of the equation (WAEC)

WORD PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS

Note: Read the question carefully so as to understand what you need to find. Choose letters for the unknowns. Solve the equations formed from the information.

Example 1:

The diagram below is a rectangle. Find the perimeter and the area of the rectangle

Opposite sides of a rectangle are equal, so we equate each of them and we obtain a simultaneous equation;

With the previous knowledge on simultaneous equation, solving equations (i) & (ii)

We obtain

Perimeter = 2L + 2B or 2(L+B)

= 2(25) + 2(17)

= 50 + 34

= 84cm

Area = LB

= 25cm x 17cm

= 425

EXAMPLE 2:

A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.        (WAEC)

Solution:

Let the digits be , where p is the tens digit and q is the unit part, so the number is 10p+q. when the digits are interchanged, the tens digit becomes q and the unit digit becomes p. hence the number is 10q+p

But ,

Divide through by 9

We were told that the sum of the digits 11, i.e

Adding (i) & (ii), 2q = 12

q = 6

substituting for q in (ii)

p + 6 = 11

p = 5

the number =

    CLASS ACTIVITY

1. The sum of two numbers is 110 and their difference is 20. Find the two numbers
2. The perimeter of a rectangular lawn is 24m. if the area of the lawn is how wide is the lawn?      (JAMB)

LINEAR GRAPH (REVISION)

Recall that any equation whose highest power of the unknown is one is a linear equation. The expression y = ax + b where a & b are constants, represents a general linear function. Its graph is a straight line. To plot the graph of a linear function, two distinct points are sufficient to draw the straight line graph.

Example 1:

Draw the graph of y = 3x – 2

Solution: A table of three points is prepared below

Scale: 1cm to 1 unit on x-axis

1cm to 5 units on y-axis

GRADIENT OF A STRAIGHT LINE X & Y INTERCEPTS

The y-intercept of a line is the point () where the line intersects the y-axis. To find b, substitute for x in the equation of the line and solve for y.

The x-intercept of a line is the point (a,) where the line intersects the x-axis. To find a, substitute  for y in the equation of the line and solve for x.

Example 1:

Solve the equation using the intercept method

Solution: To find the x-intercept, we let y be  and solve for x,

4x – 2(0) = 8

4x = 8

X = 2,   the intercept is (2,0)

To find the y-intercept, we let x be 0 and solve for y

4(0) – 2y = 8

Y = -4,   the intercept is (0,-4)

The graph of the given equation should then be plotted.

Gradient of a straight line: The gradient of a non-vertical line is a number that measures the line’s steepness. The gradient of a straight line is the rate of change of y compared with x. For example, if the gradient is 3, then for any increase in x,  y increases three times as much. We can calculate gradient by picking two points on the line and writing the ratio of vertical change (change in y) to the corresponding horizontal change (change in x).

The gradient formula: The gradient of a straight line is generally represented by m.

gradient of any line (m) =

=

 =  

Consider moving from point P to another point Q on a straight line PQ, as in the diagram below

Y-axis

Increase in y

Increase in x

X-axis

The gradient of line PQ passing through points  is;

Example 2:

Find the gradient of the lines joining the following pairs of points(-3,2) , (4,4)

Solution:  from the points

Gradient of AB =

=

=   =

=

CLASS ACTIVITY

Find the gradients of the lines joining the following pairs of points

1. (9,7) , (2,5)
2. (-4,-4) , (-1,5)
3. (7,-2) , (-1,2)

GRADIENT  OF A CURVE

The gradient of any point on a curve is the gradient of the tangent to the curve at that point. The tangent must be produced at equidistant to the point.

Note: there are two major types of quadratic curves from the general form

                Where  is +ve

i.e

where  is –ve

i.e

There are two types of gradients namely: positive and negative gradients

Positive Gradient               Negative Gradient

()                ()

()                                                                                        ()

Note: The gradient of a straight line is the same at any point on the line, but the gradient of a curve changes from point to point.

Example 1:

Copy and complete the following tables of values for

Using a scale of 1cm to 5units on y-axis and 1cm to 1unit on x-axis, draw the graph of

Use the graph to find the gradient of the curve  at

Solution

(b)   At point x = 3 traced to the curve, we draw an equidistant line as shown, then take the slope.

Slope =  =  = 2.78

Example 2:  The table below is for the curve

• Find the value of the constant m
• Using a scale of 1cm to represent 1unit on the x-axis and 1cm to represent 2units on the y-axis, draw the graph of for
• Use the graph to obtain the gradient of the curve at point

Solution

• Using any two values of and  we can solve for ‘m’, let us make use of (2, -3)                                               i.e

⇒

(c.)   Gradient =  =  =  =  =

CLASS ACTIVITY

1. (a) Copy and complete the following table of values for the relation.

(b) Using 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis,

draw the graph of  for -2

2. (a)from your graph, find the:
3. minimum value of y
4. gradient of the curve at x=1

(b) By drawing a suitable straight line, find the values of x for which

(SSCE 2004)

 PRACTICE  EXERCISE

1. Find a two digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits.(JAMB)
2. Find the perimeter and the area of the equilateral triangle in the diagram below.

3. Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110 years, how old are they?
4. Solve for (x,y) in the equations;

5. Draw the graph of for values of x from -4 to 4. Using your graph, find

• The roots of (i) (ii)  , giving your answer correct to one decimal place
• The least value of and the corresponding value of       (WAEC)

ASSIGNMENT

OBJECTIVE QUESTIONS

The graph above represents the relation. Use it to answer questions 1 and 2.

1. Find the values of for which
2. -1.55, 4.55 B. 1.55, -4.55 C. -1.55, -4.55 D. 1.55, 4.55 (SSCE 2011)

2. What is the equation of line of symmetry of the graph?
3. C. D. (SSCE 2011)

The following is a graph of a quadratic function. Use it to answer question 3 and 4.

3. Find the co-ordinates of point
4. (0, 4) (1, 4) C. (0, -4) D. (-4, 0) (SSCE2009)
5. Find the values of when
6. 1, 3 B. 1, 4 C. 2, 3 D.1, 6 (SSCE 2009)

Use the graph below to answer question below

5. What is the equation of the curve?
6. C. D. (SSCE 2006)

THEORY QUESTIONS

2. Draw the graph of 25 and for.

Use the graph(s) to:

1. find the roots of the equation 25 and
2. determine the line of symmetry of the curve25.

3. (a) Copy and complete the table of values for the relation for.

(b)Using scales of 2cm to 1 unit on the x-axis and 2cm to 2 units on the y-axis.

Draw a graph of the relation

(c) From the graph find the:

1. minimum value of y
2. roots of the equation

iii. gradient of the curve at (SSCE 2010)

4. (a) Copy and complete the table

.

(b) Using a scale of 2cm of 1 unit on the x-axis and 2cm to 5 units on the y-axis,

draw the graph of

(c) Use your graph to find:

1. the roots of the equation
2. the values of x for which

iii. the equation of the line of symmetry of the curve. (SSCE 2005)

5. (a) Copy and complete the following table of values for the relation.

(b) Using 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis,

draw the graph of  for -2

(c)from your graph, find the:

1. minimum value of y
2. gradient of the curve at x=1

(d) By drawing a suitable straight line, find the values of x for which

(SSCE 2004)

6. The table shows the values of the relationfor -4

(a) Copy and complete the table

(b) Using 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis,

draw the graph of

(c) Use your graph to find:

1. the roots of the equation
2. the value for which

iii. the gradient of the curve at x=1. (SSCE 2003)

6. a. Draw the table of values for the relation for the interval
7. Using a scale of 2cm to 1 on the x-axis and 2cm to 2 units on the y-axis, draw the graphs of:
9. for
10. Use your graph to find:
11. The roots of the equation
12. The gradient of at
13. a. Draw the table of values for the relation y=for the interval
14. Using a scale of 2cm to 1 on the x-axis and 2cm to 2 units on the y-axis, draw the graphs of:
16. for
17. Use your graph to find:
18. The roots of the equation
19. The gradient of at (SSCE 2001)

8. a. Copy and complete and the following table of values for the relation .

1. Draw the graph of the relation ; using a scale of 2cm to 1 unit on the x-axis,

and 2cm to 2 units on the y-axis

1. Using the same axes, draw the graph of
2. Obtain in the form where a, b and c are integers, the equation which is

satisfied by the x-coordinate of the points of intersection of the two graphs.

1. from your graphs, determine the roots of the equation obtained in d. above.(SSCE 2000)

9. (a) Copy and complete the values for the relation for

(b) Using a scale of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the:

1. Graph of
2. line y=3 on the same axis

(c) Use your graph to find the

1. roots of the equation
2. Maximum value of (SSCE 1998)
3. a. Copy and complete the following table of values for

1. Using a scale of 2cm to represent 1 unit on the x axis and 2cm to represent 5 units on

the y-axis, draw the graph of .

1. Use your graph to find the:
2. Roots of the equation , correct to one decimal place
3. Gradient of the curve at x=3 (SSCE 1994)

KEYWORDS: gradient, line of symmetry, linear graph, quadratic graph, coordinator, x-axis, y-axis, table of values etc.

WEEK 7:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC:  

MID TERM BREAK

WEEK 8:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC: Coordinates geometry of straight lines

Content:  

• Distance between two points.
• Midpoint of line joining two points.
• Gradients and intercept of a straight line.
• Determination of equation of a straight line.
• Angle between two intersecting straight lines.
• Application of linear graphs to real life situation.

DISTANCE BETWEEN TWO POINTS

Let  be two distinct points. The distance  between them can thus be calculated.

Applying Pythagoras theorem to right- angled triangle in the graph above,

Example 1:

Calculated the distance between the points, (4,1) and (3,-2)

Solution:

The distance between the points (4,1) and (3, -2) is

EXAMPLE 2:

Find the length of the line segment with end points (2,8) and (6,5).

Solution

X₁=2

Y₁=8

X₂=6

Y₂=5

 CLASS  ACTIVITY

Find the distances between the given points:

MID-POINT OF A LINE SEGMENT

In the Cartesian plane above, let  be the mid-point of the line segment  with the coordinate  and  .

As triangles  are similar;

Since R is the mid-point,

:.

Similarly,

Hence, the co-ordinates of the mid-point of the line joining  and  are:

Example 1: Find the mid-point  ‘R’ of the line segment  where  and .

Solution:

and

GRADIENT OF A STRAIGHT LINE

The gradient of a line is defined as the ratio, increase in  in going from one point to another on a line.

Is the change in x as the variable x increases or decreases from x₁ to x₂ and  is the change in y with respect to y₁ and y₂ .

The slope (gradient) m of a straight line L is defined as

If  is the angle of inclination to the slope of L, then  ; is called the angle of slope of the line.

Example 2:  Find the slope  and the angle of inclination  of the L through points  and

Solution:

The slope m of points P₁ and P₂ on L is

The slope  of the points P1 and P3 on L is

Therefore,  implies that the slope of the line L is 3.

Since

Therefore, the angle of inclination  is 71.57

It can therefore be concluded from the example above that any given line has one and only one slope.

CLASS  ACTIVITY:  

• Find the angle between  lines L₁, with slope -7 and L2 which passes through (2,-1) and (5,3)
• Find the gradients of the lines joining the following pairs of points :

DETERMINATION OF EQUATION OF A STRAIGHT LINE

                      ONE POINT FORM OF A LINE

The equation of a line passing through any point(x₁,y₁) and with gradient m is given by

LINE THROUGH TWO POINTS

The equation of a line joining the two-points (x₁,y₁) and (x₂,y₂) is given by

EXAMPLE 1:

Find the equation of the straight line which passes through the following pairs of points: (3, -4) and (5,-3)

SOLUTION

X₁=3    y₁= -4   x₂=5   y₂= -3

or 2y-x+11=0

EXAMPLE 2:

Find the equation of the straight line with gradient 5 and passing through the points(3,-5)

SOLUTION

X₁=3   and y₁=-5    m=5

PERPENDICULAR AND PARALLEL LINES

Two lines are said to be perpendicular to each other if the product of their gradient is equal to -1. If m₁ and m₂ are gradients of two perpendicular lines then m₁m₂=-1.

Two lines are said to be parallel to each other if their gradients are equal. If m₁ and m₂ are gradients of two parallel lines, then m₁=m₂

EXAMPLE 1:

Find the equation of the straight line which is perpendicular to the line 5x-2y=3 and passing through (3, -5).

SOLUTION

m₁= 5/2

m₁ × m₂ = -1

=

Using

EXAMPLE 2:

Find the equation of the straight line which is parallel to 4x-5=12-y and is passing through the point (4,7).

SOLUTION

4x-5=12-y

4x+y=17

Y=-4x+17

M₁=-4  and m₂= -4  since they are parallel

CLASS ACTIVITY

1. Find the equation of the line which is perpendicular to the line y=-4x, passing through the points(7,1)
2. Find the equation of the line parallel to 3x+5y=1 and passing through  (4,-2)

ANGLE OF SLOPE

                                      Y                        B

A                           C              X

From the diagram above, line AB makes an angle  with positive x-axis.  is called the angle of slope of the line. The gradient of the line AB =  = , therefore the gradient of the line is equal to tangent of the angle the line makes with the positive x-axis.

Example 1: Find the gradient of the line joining (2,4) and (1,3); find also the angle of slope of the line.

Solution:  Let    and

But, ;

⇒

Example 2: Find the gradient of the line joining (6,-2) and (-3, 2), also the angle of slope.

Solution:

Let    and

Also,

or

CLASS ACTIVITY

1. Find the gradient of the line joining Find the angle of slope of the line.
2. Find the gradient of the line joining Find the angle of slope of the line.

                         ANGLE BETWEEN TWO LINES

The angle between the lines

Y=xtanα+c₁  and

Y=xtan c₂

is given by .

We can also use the formula

Tanθ=

where M₂ and m₁ are the gradients of the straight lines.

EXAMPLE 1:

Find the angle between the lines

Y=3x-4 and 2x-y+1=0

SOLUTION

y=3x-4

tan

Y=2x+1

tan

|

=71.6⁰-63.5⁰

=8.1⁰

2^ND  METHOD

Using tanθ=

tanθ=

tanθ=

tanθ= 0.1429

θ=

θ=8.1⁰

EXAMPLE 2:

Find the acute angle between the following pair of lines

2x-3y+5=0  and

4y-x+2=0

SOLUTION

Tanθ=|

2x-3y+5=0

3y=2x+5

Y= x +  ,m₁=2/3

4y-x+2=0

4y=x-2

Y=    ,m₂= ¼

Tanθ=| |

Tanθ=

Tanθ=

Tanθ= 0.3571

Θ≃ 19.7⁰

CLASS  ACTIVITY

1. Find the acute angle between the pair of lines

3x+4y=2  and

2x+y=5

2. Find the acute angle between the following pairs of lines:

Y=5x-1  and y-3x+2=0

APPLICATION OF LINEAR GRAPHS TO REAL LIFE SITUATION

EXAMPLE 1:

P(-6,1) and Q(6,6) are the two ends of the diameter of a given circle. Calculate the radius

SOLUTION

X₁=-6   Y₁ =1   X₂=6    Y₂=6

units

EXAMPLE  2:

SOLUTION

X₁=1   y₁=α    x₂=   y₂=1

9=

9 -2 +

7+2(

7+2 ×2=

CLASS  ACTIVITY

1. What is the value of r if the distance between the points (4,2) and (1,r) is 3 units.
2. Find the distance between the point Q(4,3) and the point common to the lines . JAMB

PRACTICE EXERCISE

1. Find the acute angle between the lines 3x+2y=1 and 7x+4y=5.
2. Find the coordinates of the mid-point of the x and y intercepts on the line 2y=4x-8
3. Find the equation of the line which is parallel to the line 5x+4y=18 and makes an intercept of 2 units on the x-axis.  WAEC
4. Find the equation of the line which passes through the point P(4,-3) and is perpendicular to the line 2x+5y+1=0   WAEC
5. Find the coordinates of the mid-point of the x and y intercepts on the line 2y=4x-8   JAMB

ASSIGNMENT

1. What is the value of p if the gradient of the line joining (-1,P) and (p,4) is ?   JAMB
2. Find the value of p if the line that passes through (-1,-p) and            (-2p,2) is parallel to the line 8x+2y-17=0   JAMB
3. If the lines 3y=4x-1  and qy=x+3 are parallel to each other, find the value of q.
4. PQ  and RS are two parallel lines. If the coordinates of P,Q,R and S are (1,q), (2,3),(3,4) and (5,2q) respectively, find the value of q. JAMB
5. Find the mid-points of the line segments with the following end points:
6. (3,0) and (4,-4)
7. (-4,-5) and (3,9)
8. (-1/4 ,0) and (1,- ½ )
9. Find the value of α if the lines 2y-αx+4=0 is perpendicular to the line y+ ¼ x-7=0     JAMB

7. Find the point of intersection of the two lines 3x-2y+5=0 and y-4x+3=0

KEYWORDS:

Coordinates, parallel, perpendicular, intercept, gradient, equation, mid-point, intersect etc.

WEEK 9:                                                                                       

DATE……………………….

Subject: Mathematics

Class: SS 2

TOPIC:  Approximations

Content:

• Revision of approximation.
• Accuracy of results using logarithm table and calculators.
• Percentage error.
• Application of approximation to every day life.

REVISION OF APPROXIMATION

An approximation helps us to see and understand easily the size of a number. It is a number taken as close as possible to the actual value of the number.

In order to come close to the actual value, the number must be rounded off. This implies digits 1 to 4 are rounded down while those from 5 to 9 are rounded up. Other methods could be approximating to decimal places or significant figures.

Decimal places mean the number of places after the decimal point while significant figure is the first non-zero digit from the left.

Example 1:  The distance between the earth and the sun is 148729440km. Round this number to the nearest (a) million   ( b) 2s.f   (c) 3s.f

SOLUTION

1. 148729440km = 149000000km  (i.e round 7 up)
2. 148729440km = 150000000km   (i.e round 8 up)
3. 148729440km = 149000000km  (i.e round 7 up)

EXAMPLE 2: (a) Write 78.45831kg to (i) 2d.p, (ii) 3d.p

SOLUTION  

(i) 78.45831kg = 78.46kg (i.e round 8 up)

(ii) 78.45831kg = 78.458kg   (round down 3)

(b) Round 0.0004996 to (i) 1s.f  (ii) 2s.f

SOLUTION

(i) 0.0004996 = 0.0005  (round the first 9 up)

(ii) 0.0004996 = 0.00050  (round the second 9 up)

CLASS  ACTIVITY

• Round off 586.5764 to (i) 1d.p   (ii) 3d.p
• Approximate 964572183665 to   (i) 3s.f   (ii)  6s.f

ACCURACY OF RESULTS USING LOGARITHM TABLE AND CALCULATORS

All measurements are approximations, so they cannot be exact. Any stated measurement has been rounded off to some degree of accuracy.

Example 1:

A plot of land measuring 4532m by 431m.Calculate the area of the plot, using the calculator and the Logarithms table.

Solution: (i) 4532m x 431m = 1953292

(ii) 4532m x 431m

Antilog of 6.2908 = 1954000

EXAMPLE 2: Calculate the difference due to using two different methods of calculation by evaluating

(a) 2321 x 4122

(b) 12204 x 2123

SOLUTION

(a) i.  2321 x 4122 = 9567162

1.  2321 x 4122

Antilog of 6.9808= 9568000

Difference = 9568000 – 9567000

=  838

.(b)  (i)  12204 x 2123 = 25909092

(ii)  12204 x 2123

Antilog 0f 7.4133= 25900000

Difference = 25909092 – 25900000

= 9092

It can be seen that the result from the calculator is more accurate. The logarithm table results are higher because of premature approximation. The last example was different because the table deals with four-figures only while one is 5-figures

CLASS  ACTIVITY

Solve this problems, using the calculator and the logarithm table. Calculate the difference from your answers and state why?

• 153 x 5234
• 12536 x 2413
• 213564 x 1215
• 1211015 x 123143

PERCENTAGE ERROR

Every measurement no matter how carefully carried out is an approximation, not exact. If the height of a wall is 25m, then  the true or actual height is between 24.5m and 25.5m i.e  (range difference). The error involved cannot exceed , the maximum absolute error.

Note: The maximum absolute error is an allowance within which the actual measurement falls.

Relative error is the ratio of the maximum absolute error (precision) and the true measurement value.

In measurements,

Percentage error =   

If the true value is known,

 Percentage error =

Example1:

A sales girl gave a balance of #1.15 to a customer instead of #1.25. Calculate her percentage error.

SOLUTION

Balance given = #1.15

Real or actual bal. = #1.25

Error = #1.25 – #1.15 = #0.10

Percentage error =

= 8

EXAMPLE 2:

A student who was asked to correct 0.02539 to two significant figures gave its value to two decimal places. His percentage error is ………..

SOLUTION:

The correct value in significant figure is 0.025

His answer is 0.03

% error = %

=

CLASS ACTIVITY

• A Man made a table with a rectangular top of dimension 36cm by 44cm instead of 37cm by 41cm. what is the percentage error in the perimeter of the table correct to 1 decimal place?
• A boy measures the length and breadth of a rectangular lawn as 59.6m amd 40.3m respectively instead of 60m and 40m. what is the percentage error in his calculation of the perimeter of the lawn?

APPLICATION OF APPROXIMATION TO EVERYDAY LIFE

This aspect deals with real life situations.

Example 1:

A car travels a distance of 100km for 1h 39mins. Calculate the speed of the car. If the time is rounded up to the nearest 1hr.Find the difference between the actual speed and when the time was rounded off.

SOLUTION

Speed =  =

                                                =

                                                = 66.67km/hr

1hr 39mins to the nearest one hour is 2hrs, speed =  = 50km/hr

Difference = 66.67km/hr – 50km/hr

= 16.67km/hr

EXAMPLE 2:

Find the sum of 34.25 (to 4s.f), 26 (to 3s.f) and 10 (to 2s.f) and leave your answer to a reasonable degree of accuracy.

Solution: the actual values

34.25 0.005

26 0.5

10   0.5

Sum = 70.25 1.105

the actual sum lies between 69.145 and 71.355

Hence, the sum of the values to a reasonable degree of accuracy is 69 to 2s.f

Note: Maximum error can be derived thus;

Whole numbers =  = 0.5

1 d.p numbers =  0.05

2 d.p numbers =  0.005     etc

PRACTICE EXERCISE

• If it takes a proton to move in seconds, find the speed of the proton in metre per second correct to a suitable degree of accuracy
• Instead of writing as a decimal correct to 3s.f, a student wrote it correct to 3d.p, find his error in standard form.
• Find the value of each of the following and the degree of accuracy
• 3 + 200.5 + 670.3 + 504.4
• 56 – 9.062 – 4.147 – 10.20 – 5.108
• A string is 4.8m . a boy measured it to be 4.95m. Find the percentage error
• A sales boy gave a change of N68 instead of N72.calculate his percentage error.

ASSIGNMENT

• The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error.  NECO 2011
• A man estimated his transport fare a journey as N210.00 instead of N220.00. Find the percentage error in his estimate, correct to 3 s.f. NECO 2012
• A boy when rounding up a number wrote 98 instead of 980(to 2 significant figures).What is the percentage error?       NECO 2007
• Approximate 0.0033780 to 3 significant figures
• Express 302.10495 correct to 5 s. f.
• Express the product of 0.007 and 0.057 to 2 s. f.           NECO 2004
• Evaluate (0.13)³correct to 3 s. f.                                     NECO 2006
• A rectangular room has sides 5m by 4m, measured to the nearest metre.
• Write down the limits of accuracy for each length
• Find the greatest area the room could have
• Find the smallest perimeter the room could have
• Evaluate:  , correct to 2 s. f.              NECO 2008

KEYWORDS: error, percentage error, approximate, limits of accuracy, actual value,etc.

WEEK 10    REVISION

WEEK 11    EXAMINATION

Hope you got what you visited this page for? The above is the lesson note for Mathematics for SS2 class. However, you can download the free PDF file for record purposes.

If you have any questions as regards Mathematics lesson note For SS2 class, kindly send them to us via the comment section below and we shall respond accordingly as usual.