SS 2 MATHEMATICS NOTE FOR THIRD TERM 2021/2022 SESSION
WEEK 1
TOPIC:
GRADIENT OF A LINE AND CURVE
OBJECTIVES
At the end of the lesson, the students should be able to:
Find the gradient and equation of a straight line.
Sketch graphs of lines from given data
Draw tangents to a curve at a given point
Determine value(s) that make an algebraic fraction undefined.
PREVIOUS KNOWLEDGE
The students had been introduced to linear and quadratic functions
1
Gradient (Slope) of a Straight Line
The Gradient (also called Slope) of a straight line shows how steep a straight line is.
Calculate
To calculate the Gradient:
Divide the change in height by the change in horizontal distance
Gradient = Change in Y
Change in X
Examples:
The Gradient = 3/3 = 1
So the Gradient is equal to 1
1
The Gradient = 4/2 = 2
The line is steeper, and so the Gradient is larger.
The Gradient = 3/5 = 0.6
The line is less steep, and so the Gradient is smaller.
Applications of Gradients
Gradients are an important part of life. The roof of a house is built with a gradient to enable rain water to run down
the roof. An aeroplane ascends at a particular gradient after take off, flies at a different gradient and descends at
another gradient to safely land. Tennis courts, roads, football and cricket grounds are made with a gradient to assist
drainage.
Example 5
A horse gallops for 20 minutes and covers a distance of 15 km, as shown in the diagram.
1
Find the gradient of the line and describe its meaning.
Solution:
1
In the above example, we notice that the gradient of the distance-time graph gives the speed (in kilometres per
minute); and the distance covered by the horse can be represented by the equation:
Example 6
The cost of transporting documents by courier is given by the line segment drawn in the diagram. Find the gradient
of the line segment; and describe its meaning.
Solution:
1
So, the gradient of the line is 3. This means that the cost of transporting documents is $3 per km plus a fixed charge
of $5, i.e. it costs $5 for the courier to arrive and $3 for every kilometre travelled to deliver the documents.
Equations of straight lines
In this unit we find the equation of a straight line, when we are given some information about the line. The information
could be the value of its gradient, together with the co-ordinates of a point on the line. Alternatively, the information
might be the co-ordinates of two different points on the line. There are several different ways of expressing the final
equation, and some are more general than others.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• find the equation of a straight line, given its gradient and its intercept on the y-axis;
• find the equation of a straight line, given its gradient and one point lying on it;
• find the equation of a straight line given two points lying on it;
• give the equation of a straight line in either of the forms y = mx + c or ax + by + c = 0. Introduction
This unit is about the equations of straight lines. These equations can take various forms depending on the facts we
know about the lines. So to start, suppose we have a straight line containing the points in the following list.
1
y
x
0
1
2
3
y
2
3
4
5
x
There are many more points on the line, but we have enough now to see a pattern. If we take any x value and add 2,
we get the corresponding y value: 0+2 = 2, 1+2 = 3, 2+2 = 4, and so on. There is a fixed relationship between the x
and y co-ordinates of any point on the line, and the equation y = x +2 is always true for points on the line. We can
label the line using this equation.
The equation of a line through the origin with a given gradient
Suppose we have a line with equation y = x. Then for every point on the line, the y co-ordinate must be equal to the x
co-ordinate. So the line will contain points in the following list.
x
0
1
2
3
y
0
1
2
3
1
y
y = x
x
y = x:
We can find the gradient of the line using the formula for gradients,
,
and substituting in the first two sets of values from the table. We get
so that the gradient of this line is 1.
What about the equation y = 2x? This also represents a straight line, and for all the points on the line each y value is
twice the corresponding x value. So the line will contain points in the following list.
x
0
1
2
y
0
2
4
1
y
y = 2x
y = x
x
y = 2x:
If we calculate the gradient of the line y = 2x using the first two sets of values in the table, we obtain
so that the gradient of this line is 2.
Now take the equation y = 3x. This also represents a straight line, and for all the points on the line each y value is three
times the corresponding x value. So the line will contain points in the following list.
y
x
0
1
2
y
0
3
6
y = 3x
y = 2x
y = x
x
y = 3x:
1
If we calculate the gradient of the line y = 3x using the first two sets of values in the table, we obtain
so that the gradient of this line is 3.
We can start to see a pattern here. All these lines have equations where y equals some number times x. And in each
case the line passes through the origin, and the gradient of the line is given by the number multiplying x. So if we had
a line with equation y = 13x then we would expect the gradient of the line to be 13. Similarly, if we had a line with
equation y = −2x then the gradient would be −2. In general, therefore, the equation y = mx represents a straight line
passing through the origin with gradient m.
Key Point
The equation of a straight line with gradient m passing through the origin is given by
y = mx.
The y-intercept of a line
Consider the straight line with equation y = 2x +1. This equation is in a slightly different form from those we have
seen earlier. To draw a sketch of the line, we must calculate some values.
x y
0 1
1
y
1 3
2 5
y = 2x + 1
x
y = 2x +1:
Notice that when x = 0 the value of y is 1. So this line cuts the y-axis at y = 1.
What about the line y = 2x +4? Again we can calculate some values.
y
x
1
0
1
y = 2x + 4
y
2
4
4
6
x
y = 2x +4:
This line cuts the y-axis at y = 4.
What about the line y = 2x −1? Again we can calculate some values.
1
y
x y
- 1 3
0 1
1 1
y = 2x - 1
x
-1
y = 2x −1:
This line cuts the y-axis at y = −1.
The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts
the y-axis. This number c is called the intercept on the y-axis.
Key Point
The equation of a straight line with gradient m and intercept c on the y-axis is
y = mx + c.
We are sometimes given the equation of a straight line in a different form. Suppose we have the equation 3y−2x = 6.
How can we show that this represents a straight line, and find its gradient and its intercept value on the y-axis?
We can use algebraic rearrangement to obtain an equation in the form y = mx + c:
1
3y −2x = 6,
3y = 2x +6,
y = 2/3x + 2
So now the equation is in its standard form, and we can see that the gradient is and the intercept value on the y-axis
is 2.
We can also work backwards. Suppose we know that a line has a gradient of and has a vertical intercept at y = 1.
What would its equation be?
To find the equation we just substitute the correct values into the general formula y = mx + c. Here,
and c is 1,
so the equation is
. If we want to remove the fraction, we can also give the equation in the form 5y = x +5,
or 5y − x −5 = 0.
EVALUATION
1. Determine the gradient and y-intercept for each of the straight lines in the table below.
Equation
Gradient yintercept
y = 3x +2
y = 5x −2
y = −2x +4
y = 12x
2y −10x = 8
1
x + y +1 = 0
2. Find the equation of the lines described below (give the equation in the form y = mx + c):
(a) gradient 5, y-intercept 3; (b) gradient −2, y-intercept −1; (c) gradient 3, passing through the origin; (d)
gradient passing through (0,1); (e) gradient
-intercept .
The equation of a straight line with given gradient, passing through a given point
Example
Suppose that we want to find the equation of a line which has a gradient of and passes through the point (1,2). Here,
whilst we know the gradient, we do not know the value of the y-intercept c.
We start with the general equation of a straight line y = mx + c.
We know the gradient is and so we can substitute this value for m straightaway. This gives
.
We now use the fact that the line passes through (1,2). This means that when x = 1, y must be 2. Substituting these
values we find
so that
So the equation of the line is y = 31x + 53.
We can work out a general formula for problems of this type by using the same method. We shall take a general line
with gradient m, passing through the fixed point A(x1,y1).
We start with the general equation of a straight line y = mx + c.
We now use the fact that the line passes through A(x1,y1). This means that when x = x1, y must be y1. Substituting
these values we find
1
y1 = mx1 + c
so that
c = y1 − mx1 So the equation of the line is y
= mx + y1 − mx1. We can write this in the alternative form
y − y1 = m(x − x1)
This then represents a straight line with gradient m, passing through the point (x1,y1). So this general form is useful
if you know the gradient and one point on the line.
Key Point
The equation of a straight line with gradient m, passing through the point (x1,y1), is
y - y1 = m(x - x1).
For example, suppose we know that a line has gradient −2 and passes through the point (−3,2). We can use the formula
y − y1 = m(x − x1) and substitute in the values straight away:
−
y −2 =
2(x −(−3))
= −2(x +3)
= −2x −6
y = −2x −4.
EVALUATION
Find the equation of the lines described below (give the equation in the form y = mx + c):
(a) gradient 3, passing through (1,4);
(b) gradient −2, passing through (2,0);
1
(d) gradient 0, passing (−1,2); (e)
(c) gradient , passing through (5,−1);
through (1,−1).
gradient −1, passing
The equation of a straight line through two given points
What should we do if we want to find the equation of a straight line which passes through the two points (−1,2) and
(2,4)?
Here we don’t know the gradient of the line, so it seems as though we cannot use any of the formulæ we have found
so far. But we do know two points on the line, and so we can use them to work out the gradient. We just use the
formula m = (y2 − y1)/(x2 − x1). We get
.
So the gradient of the line is . And we know two points on the line, so we can use one of them in the formula y − y1
= m(x − x1). If we take the point (2,4) we get
y −4 = 2/3(x −2)
3y −12 = 2x −4
3y = 2x +8
y = 3/2x + 3/8 .
As before, it will be useful to find a general formula that can be used for examples of this kind. So suppose the general
line passes through two points A(x1,y1) and B(x2,y2). We shall let a general point on the line be P(x,y).
1
y
B ( x2, y2)
P ( x, y )
A ( x1, y1 )
x
Now we know that the gradient of AP must be the same as the gradient of AB, as all three points are on the same line.
But the gradient of AP is
,
whereas the gradient of AB is
.
Then mAP = mAB, so we must have
.
Now this formula is fairly complicated, but it is easier to remember if all the terms involving y are on one side, and
all the terms involving x are on the other. If we manipulate the formula, we get first
and then
.
It might help you to remember this formula if you notice that the pattern on the left-hand side, involving y, is just the
same as the pattern on the right-hand side, involving x.
1
Key Point
The equation of a straight line passing through the two points (x1,y1) and
(x2,y2) is
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑥 − 𝑥1
𝑥2 − 𝑥1
Now we can use this formula for an example. Suppose that we want to find the equation of the straight line which
passes through the two points (1,−2) and (−3,0). We just substitute into the formula, and rearrange. The various steps
are
𝑦 − (−2)
0 − (−2)
=
𝑥− 1
−3 − 1
𝑦+2
2
=
𝑥− 1
−4
-4(y + 2) = 2(x – 1)
-4y – 8 = 2x – 2
Dividing through by -4
y = -1/2x -3/2
So the line has gradient and its intercept on the y-axis is
obtain 2y = −x −3, or 2y + x +3 = 0.
1
. We can also rearrange the equation a little further to
EVALUATION
Find the equation of the lines described below (give the equation in the form y = mx + c):
(a) passing through (4,6) and (8,26), (b) passing through (1,1) and (4,−8),
(c) passing through (3,4) and (5,4), (d) passing through (0,2) and (4,0), (e)
and (2,−5).
passing through (−2,3)
The most general equation of a straight line
There is one more form of the equation for a straight line that is sometimes needed. This is the equation
ax + by + c = 0.
We have written equations in this form for some of our examples. We can see some special cases of this equation by
setting either a or b equal to zero.
If a = 0 then we obtain lines with general equation by + c = 0, i.e.
parallel to the x-axis.
. These lines are horizontal, so that they are
If b = 0 then we obtain lines with general equation ax + c = 0, i.e.
. These lines are vertical, so that they are
parallel to the y-axis. The equation of a vertical line cannot be written in the form y = mx+c. The equation ax+by +c
= 0 is the most general equation for a straight line, and can be used where other forms of equation are not suitable.
Key Point
The most general equation of a straight line is
ax + by + c = 0.
If a = 0 then the line is horizontal, and if b = 0 then the line is vertical.
1
Finding the gradient of a curve
To find the gradient of a curve, you must draw an accurate sketch of the curve. At the point where you need to know
the gradient, draw a tangent to the curve. A tangent is a straight line which touches the curve at one point only. You
then find the gradient of this tangent.
Example
Find the gradient of the curve y = x² at the point (3, 9).
Gradient of tangent = (change in y)/(change in x)
= (9 - 5)/ (3 - 2.3)
= 5.71
Note: this method only gives an approximate answer. The better your graph is, the closer your answer will be to the
correct answer. If your graph is perfect, you should get an answer of 6 for the above question.
Parallel Lines
Two lines are parallel if they have the same gradent.
Example
1
The lines y = 2x + 1 and y = 2x + 3 are parallel, because both have a gradient of 2.
Perpendicular Lines (HIGHER TIER)
Two lines are perpendicular if one is at right angles to another- in other words, if the two lines cross and the angle
between the lines is 90 degrees.
If two lines are perpendicular, then their gradients will multiply together to give -1.
Example
Find the equation of a line perpendicular to y = 3 - 5x.
This line has gradient -5. A perpendicular line will have to have a gradient of 1/5, because then (-5) × (1/5) = -1. Any
line with gradient 1/5 will be perpendicular to our line, for example, y = (1/5)x.
1
CLASSWORK
New General Mathematics for SS2
Exercise 16a, page 187, No 1 – 5
Exercise 16c, page 189, Numbers 1 and 3
ASSIGNMENT:
New General Mathematics for SS2
Exercise 16d, pages 193 and 194, Numbers 9, 10 and 11
For more examples, check New General Mathematics for SS2, pages 186 – 194.
Video links: https://stoplearn.com/gradients-of-a-straight-line-and-gradient-of-a-curve/
https://www.youtube.com/watch?v=bCYzeVXKKPg
https://stoplearn.com/straight-line-graphs-gradient-of-a-curve-and- drawing-tangents-to-acurve/
1
WEEK 2 & 3
TOPIC:
SEQUENCE AND SERIES
OBJECTIVES:
At the end of the lesson, the students should be able to:
Determine the rule that generates a sequence of terms, extending the sequence as required.
Define and find the nth term of an arithmetic progression in terms of common difference d, and first term, a
Define and find the nth term of an geometric progression in terms of common ratio r, and first term, a
Distinguish between a sequence and series
Calculate the sum of the first n terms of arithmetic and geometric series
Calculate the sum to infinity of a geometric series
Apply sequence and series to numerical and real life problems.
Sequences
What is a sequence? It is a set of numbers which are written in some particular order. For example, take the numbers
1, 3, 5, 7, 9, ... .
Here, we seem to have a rule. We have a sequence of odd numbers. To put this another way, we start with the number
1, which is an odd number, and then each successive number is obtained by adding 2 to give the next odd number.
Here is another sequence:
1
1, 4, 9, 16, 25, ... .
This is the sequence of square numbers. And this sequence,
1, −1, 1, −1, 1, −1, ... ,
is a sequence of numbers alternating between 1 and −1. In each case, the dots written at the end indicate that we must
consider the sequence as an infinite sequence, so that it goes on for ever.
On the other hand, we can also have finite sequences. The numbers
1, 3, 5, 9
form a finite sequence containing just four numbers. The numbers
1, 4, 9, 16
also form a finite sequence. And so do these, the numbers
1, 2, 3, 4, 5, 6, ..., n.
These are the numbers we use for counting, and we have included n of them. Here, the dots indicate that we have not
written all the numbers down explicitly. The n after the dots tells us that this is a finite sequence, and that the last
number is n.
Here is a sequence that you might recognise:
1, 1, 2, 3, 5, 8, ... .
This is an infinite sequence where each term (from the third term onwards) is obtained by adding together the two
previous terms. This is called the Fibonacci sequence.
We often use an algebraic notation for sequences. We might call the first term in a sequence u1, the second term u2,
and so on. With this same notation, we would write un to represent the n-th term in the sequence. So
1
u1, u2, u3, ..., un
would represent a finite sequence containing n terms. As another example, we could use this notation to represent the
rule for the Fibonacci sequence. We would write
un = un−1 + un−2
to say that each term was the sum of the two preceding terms.
Key Point
A sequence is a set of numbers written in a particular order. We sometimes write u1 for
the first term of the sequence, u2 for the second term, and so on. We write the n-th term
as un.
EVALUATION 1
(a) A sequence is given by the formula un = 3n + 5, for n = 1,2,3,.... Write down the first five terms of this sequence.
(b)
A sequence is given by un = 1/n2, for n = 1,2,3,.... Write down the first four terms of this sequence. What is the
10th term?
(c)
Write down the first eight terms of the Fibonacci sequence defined by un = un−1+un−2, when u1 = 1, and u2 = 1.
(d)
Write down the first five terms of the sequence given by un = (−1)n+1/n.
Series
A series is something we obtain from a sequence by adding all the terms together. For example, suppose we
have the sequence
1
u1, u2, u3, ..., un .
The series we obtain from this is
u1 + u2 + u3 + ... + un ,
and we write Sn for the sum of these n terms. So although the ideas of a ‘sequence’ and a ‘series’ are related, there is
an important distinction between them. For example, let us consider the sequence of numbers
1, 2, 3, 4, 5, 6, ..., n.
Then S1 = 1, as it is the sum of just the first term on its own. The sum of the first two terms is S2 = 1 + 2 = 3. Continuing,
we get
S3 = 1 + 2 + 3
=
6,
S4 = 1 + 2 + 3 + 4 =
10,
and so on.
Key Point
A series is a sum of the terms in a sequence. If there are n terms in the sequence and
we evaluate the sum then we often write Sn for the result, so that
Sn = u1 + u2 + u3 + ... + un .
1
Arithmetic progressions
Consider these two common sequences
1, 3, 5, 7, ...
and
0, 10, 20, 30, 40, ... .
It is easy to see how these sequences are formed. They each start with a particular first term, and then to get successive
terms we just add a fixed value to the previous term. In the first sequence we add 2 to get the next term, and in the
second sequence we add 10. So the difference between consecutive terms in each sequence is a constant. We could
also subtract a constant instead, because that is just the same as adding a negative constant. For example, in the
sequence
8, 5, 2, −1, −4, ...
the difference between consecutive terms is −3. Any sequence with this property is called an arithmetic progression,
or AP for short.
We can use algebraic notation to represent an arithmetic progression. We shall let a stand for the first term of the
sequence, and let d stand for the common difference between successive terms. For example, our first sequence could
be written as
1, 3,
5,
7,
9,
...
1, 1 + 2, 1 + 2 × 2, 1 + 3 × 2, 1 + 4 × 2, . . . ,
and this can be written as
a, a + d, a + 2d, a + 3d, a + 4d, ...
1
where a = 1 is the first term, and d = 2 is the common difference. If we wanted to write down the n-th term, we would
have
a + (n − 1)d,
because if there are n terms in the sequence there must be (n−1) common differences between successive terms, so
that we must add on (n − 1)d to the starting value a. We also sometimes write ℓ for the last term of a finite sequence,
and so in this case we would have
ℓ = a + (n − 1)d.
Key Point
An arithmetic progression, or AP, is a sequence where each new term after the first is
obtained by adding a constant d, called the common difference, to the preceding term.
If the first term of the sequence is a then the arithmetic progression is
a, a + d, a + 2d, a + 3d, ...
where the n-th term is a + (n − 1)d.
EVALUATION
(a) Write down the first five terms of the AP with first term 8 and common difference 7.
(b)
Write down the first five terms of the AP with first term 2 and common difference −5. (c) What is the common
difference of the AP 11,−1,−13,−25,... ?
(d)
Find the 17th term of the arithmetic progression with first term 5 and common difference 2.
1
(e)
Write down the 10th and 19th terms of the APs
(i)
8,11,14,..., (ii) 8,5,2....
(f)
An AP is given by k,2k/3,k/3,0,....
(i)
(ii)
(iii)
Find the sixth term.
Find the n th term.
If the 20th term is equal to 15, find k.
The sum of an arithmetic series
Sometimes we want to add the terms of a sequence. What would we get if we wanted to add the first n terms of an
arithmetic progression? We would get
Sn = a + (a + d) + (a + 2d) + ... + (ℓ − 2d) + (ℓ − d) + ℓ.
Now this is now a series, as we have added together the n terms of a sequence. This is an arithmetic series, and we can
find its sum by using a trick. Let us write the series down again, but this time we shall write it down with the terms in
reverse order. We get
Sn = ℓ + (ℓ − d) + (ℓ − 2d) + ... + (a + 2d) + (a + d) + a.
We are now going to add these two series together. On the left-hand side, we just get 2Sn. But on the right-hand side,
we are going to add the terms in the two series so that each term in the first series will be added to the term vertically
below it in the second series. We get
2Sn = (a + ℓ) + (a + ℓ) + (a + ℓ) + ... + (a + ℓ) + (a + ℓ) + (a + ℓ),
and on the right-hand side there are n copies of (a + ℓ) so we get
2Sn = n(a + ℓ).
1
But of course we want Sn rather than 2Sn, and so we divide by 2 to get
.
We have found the sum of an arithmetic progression in terms of its first and last terms, a and ℓ, and the number of
terms n.
We can also find an expression for the sum in terms of the a, n and the common difference d. To do this, we just
substitute our formula for ℓ into our formula for Sn. From
we obtain
Key Point
The sum of the terms of an arithmetic progression gives an arithmetic series. If the
starting value is a and the common difference is d then the sum of the first n terms is
.
If we know the value of the last term ℓ instead of the common difference d then we can
write the sum as
.
Example
Find the sum of the first 50 terms of the sequence
1
1, 3, 5, 7, 9, ... .
Solution
This is an arithmetic progression, and we can write down
a = 1,
d = 2,
n = 50.
We now use the formula, so that
Sn =
S50 =
= 25 × (2 + 49 × 2)
=
25 × (2 + 98)
=
2500.
Example
Find the sum of the series
1 + 3·5 + 6 + 8·5 + ... + 101.
Solution
This is an arithmetic series, because the difference betweenWe also know that the first term is 1, and the last term is
101. But we do not know how manythe terms is a constant value, 2·5. terms are in the series. So we will need to use
the formula for the last term of an arithmetic progression,
ℓ = a + (n − 1)d
to give us
1
101 = 1 + (n − 1) × 2·5.
Now this is just an equation for n, the number of terms in the series, and we can solve it. If we subtract 1 from each
side we get
100 = (n − 1) × 2·5
and then dividing both sides by 2·5 gives us
40 = n − 1
so that n = 41. Now we can use the formula for the sum of an arithmetic progression, in the version using ℓ, to give us
Sn =
S41
211 × 41 × (1 +
= 101)
=
2 × 41 × 102
= 41 × 51
= 2091.
Example
An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms.
Find the common difference.
Solution
We are given that a = 3. We are also given some information about the sums S8 and S5, and we want to find the common
difference. So we shall use the formula
1
for the sum of the first n terms. This tells us that
.
and that
So, using the given fact that S8 = 2S5, we see that
1
1
2 × 8 × (6 + 7d) = 2 × 2 × 5 × (6 +
= 4d)
4 × (6 + 7d)
24 + 28d = 5 × (6 + 4d)
30 + 20d
8d = 6
d = 3
.
4
EVALUATION
(a) Find the sum of the first 23 terms of the AP 4,−3,−10,....
(b)
An arithmetic series has first term 4 and common difference . Find (i) the sum of the first 20 terms, (ii)
the sum of the first 100 terms.
(c)
Find the sum of the arithmetic series with first term 1, common difference 3, and last term 100.
(d)
The sum of the first 20 terms of an arithmetic series is identical to the sum of the first 22 terms. If the common
difference is −2, find the first term.
1
Geometric progressions
We shall now move on to the other type of sequence we want to explore.
Consider the sequence
2, 6, 18, 54, ... .
Here, each term in the sequence is 3 times the previous term. And in the sequence
1, −2, 4, −8, ... ,
each term is −2 times the previous term. Sequences such as these are called geometric progressions, or GPs for short.
Let us write down a general geometric progression, using algebra. We shall take a to be the first term, as we did with
arithmetic progressions. But here, there is no common difference. Instead there is a common ratio, as the ratio of
successive terms is always constant. So we shall let r be this common ratio. With this notation, the general geometric
progression can be expressed as
a, ar, ar2, ar3, ... .
So the n-th can be calculated quite easily. It is arn−1, where the power (n−1) is always one less than the position n of
the term in the sequence. In our first example, we had a = 2 and r = 3, so we could write the first sequence as
In our second example, a = 1 and r = −2, so that we could write it as
1
Key Point
A geometric progression, or GP, is a sequence where each new term after the first is
obtained by multiplying the preceding term by a constant r, called the common ratio.
If the first term of the sequence is a then the geometric progression is
a,ar,ar2,ar3,...
where the n-th term is arn−1.
EVALUATION
(a) Write down the first five terms of the geometric progression which has first term 1 and common ratio .
(b)
Find the 10th and 20th terms of the GP with first term 3 and common ratio 2.
(c)
Find the 7th term of the GP 2,−6,18,...,
The sum of a geometric series
Suppose that we want to find the sum of the first n terms of a geometric progression. What we get is
Sn = a + ar + ar2 + ar3 + ... + arn−1 ,
and this is called a geometric series. Now the trick here to find the sum is to multiply by r and
then subtract:
2
3
n−1
Sn = a + ar + ar + ar + ... + ar
rSn =
ar + ar2 + ar3 + ... + arn−1 + arn
Sn − rSn = a
− arn
so that
1
Sn(1 − r) = a(1 − rn).
Key Point
The sum of the terms of a geometric progression gives a geometric series. If the starting
value is a and the common ratio is r then the sum of the first n terms is
a(1 − rn)
Sn = 1 − r
.
Example
Find the sum of the geometric series
1 + 6 + 18 + 54 + ...
where there are 6 terms in the series.
Solution
For this series, we have a = 2, r = 3 and n = 6. So
1
a(1− rn)
1–r
S6 = 2(1 − 36)
1–3
= 2(1 − 729)
−2
=
=
Sn =
Example
Find the sum of the geometric series
8 − 4 + 2 − 1 + ...
where there are 5 terms in the series.
Solution
For this series, we have
and n = 5. So
= 512 .
EVALUATION
(a) Find the sum of the first five terms of the GP with first term 3 and common ratio 2.
1
(b)
Find the sum of the first 20 terms of the GP with first term 3 and common ratio 1.5.
(c)
The sum of the first 3 terms of a geometric series is . The sum of the first six terms is
and common ratio.
. Find the first term
Convergence of geometric series
Consider the geometric progression
and we want to examine this formula in the case of our particular example where
. Now the formula contains the
term rn and, as −1 < r < 1, this term will get closer and closer to zero as n gets larger and larger. So, if −1 < r < 1, we
can say that the ‘sum to infinity’ of a geometric series is
a
S∞ =
,
1−r
where we have omitted the term rn. We say that this is the limit of the sums Sn as n ‘tends to infinity’. You will find
more details of this concept in another unit.
Example Find the sum to infinity of the geometric progression
Solution For this geometric progression we have a = 1 and
. As −1 < r < 1 we can use the formula, so that
.
1
Key Point
The sum to infinity of a geometric progression with starting value a and common
ratio r is given by
a
S∞ =
1−r
where −1 < r < 1.
EVALUATION
(a) Find the sum to infinity of the GP with first term 3 and common ratio .
(b)
The sum to infinity of a GP is four times the first term. Find the common ratio.
(c)
The sum to infinity of a GP is twice the sum of the first two terms. Find possible values of the common ratio.
ASSIGNMENT:
New General Mathematics for SS2
Exercise 18e, pages 214 and 215, Numbers 3, 4, 5 and 6
For more examples, check New General Mathematics for SS2, pages 204 – 215.
Video links: https://www.youtube.com/watch?v=Tj89FA-d0f8
https://www.youtube.com/watch?v=m5Yn4BdpOV0
https://www.youtube.com/watch?v=m5Yn4BdpOV0
1
WEEK 4
TOPIC:
MEASURES OF DISPERSION
OBJECTIVES:
At the end of the lesson, the students should be able to:
Describe and interpret (numerically and verbally) the dispersion or spread of a values in a data set.
Calculate the range, variance and standard deviation of a set of ungrouped and grouped data.
PREVIOUS KNOWLEDGE
The students had been taught measures of central tendency in previous term.
Median, Quartiles And Percentiles (Ungrouped Data)
In these lessons, we will learn how to find the median, quartiles and percentiles of ungrouped data (discrete data).
We have learned that the median is the middle value when a set of data is arranged in order of increasing magnitude.
We
will
now
consider
lower
quartiles
and
upper
quartiles.
The median divides
the
The lower
quartile is
data
the
into
a
middle
1
lower
value
half
of
and
an
the
upper
lower
half.
half.
The upper
quartile is
the
middle
value
of
the
upper
half.
The following figure shows the median, quartiles and interquartile range. Scroll down the page for examples and
solutions.
How to find Median, Quartiles and Percentiles?
Example :
Find the median, lower quartile and upper quartile of the following numbers.
1
12, 5, 22, 30, 7, 36, 14, 42, 15, 53, 25
Solution:
First, arrange the data in ascending order:
Median (middle value) = 22
Lower quartile (middle value of the lower half) = 12
Upper quartile (middle value of the upper half) = 36
If there is an even number of data items, then we need to get the average of the middle numbers.
Example:
Find the median, lower quartile, upper quartile, interquartile range and range of the following numbers.
12, 5, 22, 30, 7, 36, 14, 42, 15, 53, 25, 65
Solution:
First, arrange the data in ascending order:
1
Lower quartile or first quartile =
Median or second quartile =
Upper quartile or third quartile =
Interquartile range = Upper quartile – lower quartile
= 39 – 13 = 26
Range = largest value – smallest value
= 65 – 5 = 60
When evaluating the quartiles, always remember to first arrange the data in increasing order.
1
Mean Deviation
How far, on average, all values are from the middle.
Calculating It
Find the mean of all values ... use it to work out distances ... then find the mean of those distances!
In three steps:
•
•
•
1. Find the mean of all values
2. Find the distance of each value from that mean (subtract the mean from each value, ignore minus signs)
3. Then find the mean of those distances
Like this:
Example: the Mean Deviation of 3, 6, 6, 7, 8, 11, 15, 16
Step 1: Find the mean:
Mean = (3 + 6 + 6 + 7 + 8 + 11 + 15 + 16)/8 = 72/8 = 9
Step 2: Find the distance of each value from that mean:
1
Value
Distance
from 9
3
6
6
3
6
3
7
2
8
1
11
2
15
6
16
7
Which looks like this:
(No minus signs!)
Step 3. Find the mean of those distances:
Mean Deviation = (6 + 3 + 3 + 2 + 1 + 2 + 6 + 7)8 = 30/8 = 3.75
So, the mean = 9, and the mean deviation = 3.75
It tells us how far, on average, all values are from the middle.
In that example the values are, on average, 3.75 away from the middle.
For deviation just think distance
1
Formula
The formula is:
Mean Deviation = Σ|x − μ|/N
•
•
•
•
•
Σ is Sigma, which means to sum up
|| (the vertical bars) mean Absolute Value, basically to ignore minus signs
x is each value (such as 3 or 16)
μ is the mean (in our example μ = 9)
N is the number of values (in our example N = 8)
Let's look at those in more detail:
Absolute Deviation
Each distance we calculate is called an Absolute Deviation, because it is the Absolute Value of the deviation (how
far from the mean).
To show "Absolute Value" we put "|" marks either side like this:
|-3| = 3
For any value x:
Absolute Deviation = |x - μ|
From our example, the value 16 has Absolute Deviation = |x - μ| = |16 - 9| = |7| = 7
1
And now let's add them all up ...
Sigma
The symbol for "Sum Up" is Σ (called Sigma Notation), so we have:
Sum of Absolute Deviations = Σ|x - μ|
Divide by how many values N and we have:
Mean Deviation = Σ|x − μ|/N
Let's do our example again, using the proper symbols:
Example: the Mean Deviation of 3, 6, 6, 7, 8, 11, 15, 16
Step 1: Find the mean:
μ = (3 + 6 + 6 + 7 + 8 + 11 + 15 + 16)/8 = 72/8 = 9
Step 2: Find the Absolute Deviations:
1
x
|x - μ|
3
6
6
3
6
3
7
2
8
1
11
2
15
6
16
7
Σ|x - μ| = 30
Step 3. Find the Mean Deviation:
Mean Deviation = Σ|x - μ|/N = 30/8 = 3.75
Note: the mean deviation is sometimes called the Mean Absolute Deviation (MAD) because it is the mean of the
absolute deviations.
Standard Deviation
It is defined as the positive square-root of the arithmetic mean of the Square of the deviations of the given
observation from their arithmetic mean.
The standard deviation is denoted by s in case of sample and Greek letter σ (sigma) in case of population.
The formula for calculating standard deviation is as follows
for raw data
1
And for grouped data the formulas are
for discrete data
Where d =
C = class interval
for continuous data
Example Raw Data
The weights of 5 ear-heads of sorghum are 100, 102,118,124,126 gms. Find the standard deviation.
Solution
x
100
102
118
124
126
570
x2
10000
10404
13924
15376
15876
65580
1
Standard deviation
Example Discrete distribution
The frequency distributions of seed yield of 50 seasamum plants are given below. Find the standard deviation.
Seed yield in gms
(x)
Frequency (f)
3
4
5
6
7
4
6
15
15
10
Seed
yield in
gms (x)
f
fx
fx2
3
4
5
6
7
Total
4
6
15
15
10
50
12
24
75
90
70
271
36
96
375
540
490
1537
Solution
Here n = 50
1
Standard
deviation
= 1.1677 gms
Example Continuous distribution
The Frequency distributions of seed yield of 50 seasamum plants are given below. Find the standard deviation.
Seed yield in gms
(x)
No. of plants (f)
Seed
yield in
gms (x)
2.5-3.5
3.5-4.5
4.5-5.5
5.5-6.5
6.5-7.5
Total
2.5-35 3.54.5
4
6
4.55.5
15
No. of Mid
Plants x d=
f
4
6
15
15
10
50
3
4
5
6
7
25
-2
-1
0
1
2
0
1
5.56.5
15
6.5-7.5
10
df
d2 f
-8
-6
0
15
20
21
16
6
0
15
40
77
A=Assumed mean = 5
n=50, C=1
=1.1677
Variance
The square of the standard deviation is called variance
(i.e.) variance = (SD) 2.
Coefficient of Variation
The Standard deviation is an absolute measure of dispersion. It is expressed in terms of units in which the original
figures are collected and stated. The standard deviation of heights of plants cannot be compared with the standard
deviation of weights of the grains, as both are expressed in different units, i.e heights in centimeter and weights in
kilograms. Therefore the standard deviation must be converted into a relative measure of dispersion for the purpose of
comparison. The relative measure is known as the coefficient of variation. The coefficient of variation is obtained by
dividing the standard deviation by the mean and expressed in percentage. Symbolically, Coefficient of
1
variation (C.V) =
If we want to compare the variability of two or more series, we can use C.V. The series or groups of data for
which the C.V. is greater indicate that the group is more variable, less stable, less uniform, less consistent or less
homogeneous. If the C.V. is less, it indicates that the group is less variable or more stable or more uniform or more
consistent or more homogeneous.
Example 6
Consider the measurement on yield and plant height of a paddy variety. The mean and standard deviation for yield are
50 kg and 10 kg respectively. The mean and standard deviation for plant height are 55 am and 5 cm respectively.
Here the measurements for yield and plant height are in different units. Hence the variabilities can be compared only
by using coefficient of variation.
For
yield,
For plant
= 20% CV=
height, CV=
= 9.1%
The yield is subject to more variation than the plant height.
1
CLASSWORK
New General Mathematics for SS2
Exercise 19a, page 220, Numbers 3, 6 and 7
ASSIGNMENT:
New General Mathematics for SS2
Exercise 19a, page 220, Numbers 8, 9 and 10
For more examples, check New General Mathematics for SS2, pages 216 – 220.
Video links: https://www.toppr.com/guides/business-mathematics-and-statistics/measures-of-central-tendencyand-dispersion/measure-of-dispersion/
https://www.youtube.com/watch?v=64ELhoTvzk0
https://www.youtube.com/watch?v=zSDAE1Bltrc
1
WEEK 6
TOPIC:
SURDS
OBJECTIVES:.
Surds, and other roots
Roots and powers are closely related, but only some roots can be written as whole numbers. Surds are roots which
cannot be written in this way. Nevertheless, it is possible to manipulate surds, and to simplify formulæ involving
them.
In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that
they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• understand the relationship between negative powers and positive powers;
• understand the relationship between fractional powers and whole-number powers;
• replace formulæ involving roots with formulæ involving fractional powers;
• understand the difference between surds and whole-number roots;
• simplify expressions involving surds;
• rationalise fractions with surds in the denominator.
1
Introduction
In this unit we are going to explore numbers written as powers, and perform some calculations involving them.
In particular, we are going to look at square roots of whole numbers which produce irrational numbers — that
is, numbers which cannot be written as fractions. These are called surds.
1.
Powers and roots
We know that 2 cubed is 2×2×2, and we say that we have 2 raised to the power 3, or to the index 3. An easy way of
writing this repeated multiplication is by using a ‘superscript’, so that we would write 23:
23 = 2×2×2 = 8.
Similarly, 4 cubed is 4×4×4, and equals 64. So we write
43 = 4×4×4 = 64.
But what if we have negative powers? What would be the value of 4−3?
To find out, we shall look at what we know already:
43 = 4×4×4
= 64,
42 =
4×4
= 16,
41 =
4
= 4,
0
and so 4 =
4÷4
=1
4-1 = 1÷4
= 1/4
So a negative power gives the reciprocal of the number that is, -1 over the number. Thus
. Similarly,
1
, and
and
.
A common misconception is that since the power is negative, the result must be negative: as you can see, this is not
so.
Now we know that 40 = 1 and 41 = 4, but what is 41/2?
Using the rules of indices, we know that1 2 41/2 ×41/2 = 41 = 4 because
/
is the square root of 4. It is written as
. So 41/2 equals 2, as 2×2 = 4. Therefore 4
and equals 2:
.
Similarly,
.
And in general, any number a raised to the power equals the square root of a:
So the power, or index, associated with square roots is . Also, in the same way that the index represents the square
root, other fractions can be used to represent other roots. The cube root of the number 4 is written as
where is the index representing cube root. Similarly, the fourth root of 5 may be written as
, and so on. The n-th root is represented by the index 1/n, and the n-th root of a is written as
So, for example, if we have
then this equals 64 to the power ; and then
1
3
1
√64 = 643
= (4×4×4)1/3 = 4.
Key Point
The square root of a is written as a1/2 and is equal to
. The n-th root of a is written
as a1/n and is equal to √n a. The formula
can be used to simplify expressions involving square roots.
Not all roots are unique, for example the square root of 4 is 2 or −2, sometimes written
as ±2. But when written without a sign in front, the square root represents the positive
root. You cannot find the square root of a negative number.
Surds and irrational numbers
We shall now look at some square roots in more detail. Take, for example,
: its value is 5. And the value of
, or 1 . So some square roots can be evaluated as whole numbers or as fractions, in other words as rational numbers.
But what about
? The roots to these are not whole numbers or fractions, and so they have irrational values.
They are usually written as decimals to a given approximation. For example
√2 = 1.414
1
to 3 decimal places,
√3 = 1.732
to 3 decimal places.
When we have square roots which give irrational numbers we call them surds. So
are surds. Other surds are
and so on.
Surds are often found when using Pythagoras’ Theorem, and in trigonometry. So, where possible, it is useful to be able
to simplify expressions involving surds. Take, for example, . This can be written as
, which we can rewrite as
, in other words as
:
√8 =
2√2
In general, the square root of a product is the product of the square roots, and vice versa. This is useful to know when
simplifying surd expressions.
Now suppose we have been given
. At first glance this cannot be simplified. But we can rewrite the expression
as the square root of 5 times 15, so it is the square root of 75, and 75 can be written as 25 times 3. But 25 is a perfect
square, we can use this to simplify the expression.
= 3√5
But watch out if you are given
, which is the square root of 13.This does not equal
, which is 2+3 = 5. Now 5 cannot be the answer as that is the square root 25, not the square root of 13.
1
Key Point
If a positive whole number is not a perfect square, then its square root is called a surd.
A surd cannot be written as a fraction, and is an example of an irrational number.
Simplifying expressions involving surds
Knowing the common square numbers like 4, 9 16, 25, 36 and so on up to 100 is very helpful when simplifying surd
expressions, because you know their square roots straight away, and you can use them to simplify more complicated
expressions. Suppose we were asked to simplify the expression √400×√90:
√400×√90 = √36000 = √3600 ×√10 = 60√10
which cannot be simplified any further.
Key Point
The following formulæ may be used to simplify expressions
involving surds:
√ab = √a ×√b
𝑎
√𝑏 =
√𝑎
√𝑏
But what if we have this
If we have a product of
brackets involving surds,
for example), we can
expand out the brackets in
the usual way:
expression,? If we expand this out and simplify the answer, we get
1
.
So the product
does not involve surds at all. This is an example of a general result known as the
difference of two squares. This general result may be written as
a2 − b2 = (a + b)(a − b)
for any numbers a and b. In our example a = 1 and b = √3, so
.
The expansion of the difference of two squares is another useful fact to know and remember.
Key Point
The formula for the difference of two squares is
a2 − b2 = (a + b)(a − b).
Rationalising expressions containing surds
Sometimes in calculations we obtain surds as denominators, for example
. It is best not to give surd answers in
this way. Instead, we use a technique called rationalisation. This changes the surd denominator, which is irrational,
into a whole number.
To see how to do this, take our example
. To rationalise this, we multiply by the fraction
which is
equal to 1. When we multiply by this fraction we do not change the value of our original expression. We obtain
1
1
√13
=
1 𝑥 √13
√13 𝑥 √13
We can also use the expansion of the
complicated expressions involving surds.
=
√13
13
difference of two squares to rationalise more
Example
Rationalise the expression
1
1+√2 .
Solution
If we multiply this expression by this fraction
we do not change its value, as the new fraction is equal
to 1. When we do this, we use the formula for the difference of two squares to work out
,
and that gives us
, which is 1 − 2 = −1. So now we have a whole number in the
denominator of our fraction, and we can divide through. We get
1
1 − √2
1 − √2
𝑥
=
−1
1 + √2 1 − √2
EVALUATION
1. Simplify the following expressions:
(b) √90×√60
1
(c) √1000/√40
(d) (1−√5)(1+√5)
(e) (√7+3)(√7−3)
2. Rationalise the following expressions:
ASSIGNMENT:
New General Mathematics for SS3
Exercise 1d, page 2, Numbers 5, 6, 7, 8 and 10
Exercise 1e, page 3, Numbers 12, 13 and 14
Exercise 1h, page 5, Numbers 5, 6 and 7
For more examples, check New General Mathematics for SS2, pages 83 – 88.
Video links: https://www.youtube.com/watch?v=568dGLFTom8
https://www.youtube.com/watch?v=OqPbF3nkeyQ
https://www.youtube.com/watch?v=oNA0kOdpb3M
https://www.youtube.com/watch?v=wzAotwNPhm8
1
WEEK 8
TOPIC:
THEORY OF LOGARITHM
OBJECTIVES:
At the end of the lesson, the students should be able to:
Evaluate expressions given in logarithmic form
Recall and use the laws of logarithms to simplify and/or evaluate given expressions without the use of
logarithm tables
PREVIOUS KNOWLEDGE
The students had been taught the use of logarithm tables in calculations
The laws of logarithms
The three main laws are stated here:
First Law logA + logB = logAB
This law tells us how to add two logarithms together. Adding logA and logB results in the logarithm of the product of
A and B, that is logAB. For example, we can write
log10 6 + log10 2 = log10(6 × 2) = log10 12
1
The same base, in this case 10, is used throughout the calculation. You should verify this by evaluating both sides
separately on your calculator.
Second Law logAn = nlogA
So, for example log10 64 = 4log10 6
You should verify this by evaluating both sides separately on your calculator.
Third Law
log A − log B = log
A
B
So, subtracting logB from logA results in log .
For example, we can write
The same base, in this case e, is used throughout the calculation. You should verify this by evaluating both sides
separately on your calculator.
Four other useful results are
log1=0 ,
logm m =1
log 10n = n
log en = n
10
e
The logarithm of 1 to any base is always 0.
The logarithm of a number to the same base is always 1. In particular,
1
log10 10 = 1,
and
loge e = 1
Evaluation
1. Use the first law to simplify the following.
(a) log10 8 + log10 5,
(b) logx + logy, (c) log5x + log3x,
(d) loga + logb2 + logc3.
2. Use the third law to simplify the following.
(a) log10 12 − log10 4,
(b) logx − logy,
(c) log4x − logx.
3. Use the second law to write each of the following in an alternative form.
(a) 3log10 5,
(b) 2logx,
(c) log(4x)2,
(d) 5lnx4,
(e) ln1000.
4. Simplify 7logx − logx5.
1
ASSIGNMENT:
New General Mathematics for SS3
Exercise 2b, pages 12 and 13, Numbers 1, 3, 4, 6, 8, and 9
For more examples, check New General Mathematics for SS2, pages 83 – 88.
Video links: https://stoplearn.com/theory-of-logarithms-and-laws-of-logarithms/
https://www.youtube.com/watch?v=x4feQXMXk5k
1
WEEK 9
TOPIC:
SPHERES
OBJECTIVES
At the end of the lesson, the students should be able to:
Calculate the length of a circular arc, area of sectors and segment of a circle
Calculate the area and volume of a sphere and hemisphere.
PREVIOUS KNOWLEDGE
The students had been introduced to arc and sector of circle in previous classes.
Sphere
A sphere is a three-dimensional circle or we can also say that it is a circle in space. A ring and a ball have something in
common. Both are round but are they the same? Obviously not. One is a circle the other is a sphere. Space below shall
help you delve into the measurements of surface area and volume of the sphere.
Sphere
Circle and sphere, both are round, both are measured using radius. Then what is the difference between the two? A
circle can be drawn on a plane, but can a sphere be drawn on a paper? The answer is no! This is because a sphere is a 3dimensional circle.
1
A circle is a closed figure which can be drawn using a constant length from a fixed point center. This fixed point is
called the center of the circle and the distance with which a circle is drawn is called the radius. The line passing through
the center from one end to the other is called the diameter.
If a circular ring is tied to a string and rotated then we see a change in the shape. The changed shape is called the
Sphere. When a circle changes into a sphere, nothing in it changes, neither the radius nor the diameter. It just becomes a
three-dimensional version of a planar circle. So a spherical shape is a three-dimensional counterpart of a circle, with all
its points lying in space at a constant distance from the fixed point or the center, called the radius of that sphere.
Surface Area
For finding the surface area of any spherical item we need to perform the following activity: A ball of any game is the
best example of a sphere. Take a ball and drive a nail into it. Tie a string from the nail and wind it across the ball. Wind
the string in such a manner that the two layers of string do not overlap each other.
1
We need to cover the ball with just one layer of string. On reaching the center use pins to keep the string intact. Cover
the whole ball in a similar fashion. Now mark the starting and end points of the string wrapped around the ball. Try
measuring the diameter of the ball with the help of a scale.
The diameter gives us the radius of the ball. With the same radius, draw four circles on plain paper. With the help of the
string that was used to cover the ball, fill the circles on paper sequentially.
You will notice that the string used to cover the ball, covers the four circles on paper. This brings us to the conclusion
that the surface area of the ball is equal to the area of four circles. Therefore, the Surface area of a ball = 4 × Area of a
circle = 4 × πr2. Therefore,
The Surface Area of a Sphere = 4πr2
Hemisphere
A hemisphere is the half of a sphere. When a sphere is cut into two halves, then the shape we get is called the
hemisphere. A hemisphere has a curved surface and a flat base. The curved surface area of the hemisphere is half of the
surface area of the sphere. Therefore,
The Curved Surface Area of Hemisphere =1/2 × 4 × πr2
Curved surface area of a hemisphere = 2πr2. Since a sphere is a combination of a curved surface and a flat base, to find
the total surface area we need to sum up both the areas. The flat base being a plane circle has an area πr2. Total surface
area of a hemisphere is 2πr2+πr2. So,
The Total surface area of Hemisphere = 3πr2
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Volume
For finding the volume of a spherical body we use the Archimedes principle. What is Archimedes principle? According
to the Archimedes principle, when a solid figure is immersed in a container filled with water, then the volume of water
that overflows from the container is equal to the volume of that solid figure.
Similarly when we place a spherical figure into a container filled with water, then the amount of water that overflows is
equal to the volume of that figure. Archimedes principle practically gives you the volume, but following this principle is
not possible every time. For finding the formula for the volume of the sphere, we insert the spherical body in a
cylindrical container.
The figure above shows a spherical body inside a cylindrical container. We notice that the radius of the circular bases of
the cylinder is equal to the radius of the spherical body. And since the spherical body touches the top and bottom of the
container, its diameter is equal to the height of the container.
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The volume of a spherical body is assumed to be 2/3 of the cylindrical container, this gives us: Vsphere = 2/3 Vcylinder or 2 /
3 π r2h. Thus, we have: h = 2r = 2 / 3 πr2 (2r) = 4 / 3 πr3. Thus,
The Volume of the Sphere = 4 / 3 πr3
The Volume of Hemisphere
As already said a hemisphere is half the sphere, hence its volume will also be half the volume of the sphere. The volume
of the hemisphere = 1 / 2 × 4 / 3 πr3.
The Volume of the Hemisphere = 2 / 3 π r3
Solved Examples for You
Question: The diameter of the moon is approximately 1/4th of the diameter of the earth. What fraction of the volume of
the earth is equal to the volume of the earth?
Solution: Let the diameter of Earth be x, and the radius of the earth be = x/2
Volume of Earth = 4 / 3π (x / 2)3
The diameter of moon = 1 / 4x = x / 4
Radius of moon = 1 / 2 × 1 / 4x = x / 8
Volume of Moon = 4 / 3 π (x / 8)3
Volume of moon/Volume of Earth = { 4 / 3 π (x / 8)3 } / { 4 / 3 π (x / 2)3 }
= (x / 8)3 / (x / 2)3 = x3 / 83 × 23 /x3 = 23 / 83 = 8 / 8 × 8 × 8 = 1/64.
Therefore, the volume of the moon is 1/64th of Earth.
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Ques. Is sphere a circle?
Ans. Any circle of a sphere is basically a circle that lies on a sphere. Circle-like this can be formed as the meeting of a
sphere and a plane, or of 2 spheres. On a sphere a circle’s plane passes through the middle of the sphere is known as a
great circle.
Ques. How many circles make a sphere?
Ans. A diagram can show us this; where the intersection of any sphere and a cylinder contains 2 circles. Would the
radius of the cylinder be equal to the radius of the sphere? The intersection would be a circle, where both the surfaces
will be tangent.
Ques. What is the equation for the sphere?
Ans. Equation of a Sphere is given in mathematics as follows:
The general equation of a sphere is:
(x – a)² + (y – b)² + (z – c)² = r²
Here, a, b, and c symbolizes the middle of the sphere.
Here, r represents the radius and x, y, and z are respectively the coordinates of the points on the surface of the sphere.
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For more examples, check New General Mathematics for SS 3, pages 22 to 34.
Video links: https://stoplearn.com/surface-area-and-volume-of-a-sphere-hemisphere-and-composite-shape/
https://www.youtube.com/watch?v=sfoB1e2ss6k
https://www.youtube.com/watch?v=klkBxP3CZiE
CLASSWORK
New General Mathematics for SS3,
Exercise 4b, page 26, Numbers 1 and 2
ASSIGNMENT
New General Mathematics for SS3,
Exercise 4b, page 26, Numbers 3, 4 and 5.
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WEEK 10
TOPIC: COMMERCIAL ARITHMETIC
OBJECTIVES
At the end of the lesson, the students should be able to:
Calculate the simple and compound interest on a given principal at a given rate over a given period of time.
Calculate the profit and loss percent on given items.
PREVIOUS KNOWLEDGE
Students had been introduced to simple interest, profit and loss percent at the previous classes
Simple and Compound Interest
If you save (invest) money in a bank or building society, they will pay you interest on the money you deposit
with them. Likewise, if you borrow money, you pay interest to the lender. Ususally the interest rates for borrowers
are higher than for savers, that’s how the bank pays the salaries for all their employers, the cost of buying and
maintaining their premises, and the dividends paid to shareholders. The interest rates can vary a lot, it’s good to
shop around.
Interest on an investment or loan can be calculated in many different ways, it is important to understand the
details. The first important distinction is between Simple and Compound interest. With Simple interest, the
interest is paid (usually every year), but the interest is no reinvested so it doesn’t attract extra interest if you leave
it in the bank. With Compound interest, the interest is reinvested in the same account and earns interest in later
years.
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With Compound interest it is important to know how often the interest is calculated and added to your account
(this is called compounding), for example monthly, yearly, etc., and whether the interest rates quoted are annual
rates or rates per compounding period (for example, monthly).
Simple interest. Example: You invest £200 in a new account which pays 5% interest annually, calculated using
simple interst. How much do you have after one year, after two years? How long does it take to reach £300 in
total?
Ans: After one year you have
210.After two year you have
After each year we add 5% of 200 = 200×5/100 = 10. After n years we have in total
£200 + n × 10 so we have
200 + 10n = 300
with the solution n = 10 .
Compound interest.
Example: £200 invested at 5% compounded annually. How much do you have after one year, after two years? how
long does it take to reach £300 in total?
Ans: After one year you have
210, the same as the Simple Interest case. But in the
second year all the £210 earns interest, so the interest in the 2nd year is
So your total investment
after two years is worth £220.50. We could also calculate this directly by mustiplying the formula for one year by
again.
After each year we multiply the amount invested by (1+5%) = (1+5/100) = 1.05.
After n years we have in total £200 × (1.05)n so we have to solve
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200(1.05)n = 300
Solution is first simplify, then take logs on both sides
200(1.05)n = 300 ⇒ (1.05)n = 1.5
using logs to the base 10 we find n = 8.31
Since n in this case must be an integer, we will need n = 9 to make sure of exceeding the £300
figure. In fact after 8 and 9 years respectively we will have
200 × 1.058 = 295.49 ,200 × 1.059 = 310.27
Before the next example we write out the general formulas which will be useful for any problems involving
compound interest. First consider the case of an amount (the principal) P invested for n years at an annual rate
of interest of r%, compounded annually. The final value of the investment is called the future value and is
denoted by S.
After one year we have
after two years
etc., so after n years the future value is given by
S
P
r
n
We have already seen various examples of how to use this formula to determine one of the variables S,P,r,n if
the other three are given.
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Now consider the case of a problem where the interest is compounded more frequently. It is important to realise
that if the interest rate is quoted at an annual rate but the compounding is done, for example quarterly, the interest
rate per quarter is not the annual rate but a quarter of the annual rate. Let’s do the previous example with the 5%
annual rate compounded quarterly. The quarterly interest rate is then 1.25%. After one quarter the original
principal of £200 is now worth
after 2 quarters the value is now
so after n quarters the future value is
If we want this to be £300, we need to solve the equation
for n. We use logs as in our previous examples, first dividing through by 200 to
simplify
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So we need 33 quarters, or 8 yrs and one quarter, to exceed £300 (in fact after
33 quarters we will have
35, whereas after 32 quarters we would have
Note that when the interest was compounded annually, after 8 years or 32
quarters we had only £295.49, so compounding quarterly gives us slightly more.
The general formula for compounding at intervals which are not annual is just the same as the annual one
S
P
r
n
except now we must interpret r% as the interest rate per period and n as the number of periods.
Example: A credit card debt of £400 at a monthly interest rate of 2% - if you pay back nothing each month, how
long before you hit your credit limit of £500?
Ans: After each month your total debt is multiplied by (1 + 2/100), so after n months your debt is £400(1 +
0.02)n. We could also get this from our general formula. We need to find n such that
400(1 + 0.02)n = 500
First divide by 400 on both sides to get (1 + 0.02)n = (1.02)n = 1.25. Then take logs to the base 10 on both sides
to get
nlog10 1.02 = log10 1.25
So n = log10 1.25log10 1.02 = 11.27
Since n must be an integer, you will be still under your credit limit after month 11
, but over your limit after month
1
.
Continuous compounding Recall the formula for compound interest of a principal P invested at r% per period
for n periods:
S
P
r
n
Now if the annual rate is a% and the interested is compounded every 1/m of a year, the rate per period is a/m.
So the future value is
S
P
a
n
m
Suppose we chose n such that n periods is one year, i.e. we take n = m. So our future value after one year is
S
P
a
m
m
Suppose we take bigger and bigger values of m, i.e. we compound more and more freqently. Does the future
value increase without limit? Let’s look at an example, take a = 5%,P = 100 and calculate the result for a range
of different m values.
You can see that the final result is getting closer and closer to a fixed number, in fact the number £105.13 to the
nearest pence..
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Profit And Loss
The cost price of an item is the price at which a merchant buys an item. The selling price is the price at which the
merchant sells the item. When the selling price is greater than the cost price, the merchant makes a gain, which is
commonly known as a profit. It is given by:
Profit = Selling price – Cost price.
When the selling price is less than the cost price, the merchant incurs a loss, which is given by:
Loss = Cost price – Selling price.
For example, if a calculator costing $40 is sold for $56,
the profit = $56 – $40
= $16;
if a calendar costing $7 is sold for $4,
the loss = $7 – $4
= $3.
To compare profits gained (or losses incurred) from buying and selling of different items, we usually
express the profit (or loss) on an item as a percentage of the cost price of the item.
In the above example, the profit as a percentage of the cost price of the calculator =
100%
= 40%;
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3
the loss as a percentage of the cost price of the calendar = 3 100%
= 42.9%
(correct to 3 sig. fig.).
Example 1
The cost price of a monitor is $480. It is sold at a loss of 12.5% on the
cost price. Find its selling price.
Solution Loss = 12.5% $480
= $60
\ selling price of the monitor = $480 – $60
= $420
For more examples, check New General Mathematics for SS3, pages 35 to 46.
Video links: https://www.youtube.com/watch?v=-45Ft7i7pS
https://www.investopedia.com/articles/investing/020614/learn-simple-and-compound-interest.asp
https://insideiim.com/percentages-profit-loss-and-simple-compound-interest-insideiim-virtualclass-with-cplcs-parag-chitale
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CLASSWORK
New General Mathematics for SS3,
Exercise 5a, page 37 and 38, Numbers 2, 3, and 4
ASSIGNMENT
New General Mathematics for SS3,
Exercise 5a, page 38, Numbers 5, 7, 8 and 11.
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