MATHEMATICS
Learner’s Study and
Revision Guide for
Grade 12
SEQUENCES & SERIES
Revision Notes, Exercises and Solution Hints
by
Roseinnes Phahle
Examination Questions by the Department of Basic Education
Preparation for the Mathematics examination brought to you by Kagiso Trust
Contents
Unit 4
Arithmetic sequence 3
Geometric sequence 4
Quadratic sequence 5
Summing up sequences 6
Convergence and sum to infinity of a geometric series 8
Sigma notation 9
Answers 10
Examination questions with solution hints and answers 11
More questions from past examination papers 17
Answers 25
How to use this revision and study guide
1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you
are required to make a response based on your prior learning of the topic from your teacher in class or
from a textbook. Furthermore, the notes cover all the Mathematics from Grade 10 to Grade 12.
2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do
not read the answer or hints until you have tried to work out a question and are having difficulty.
3. The notes and exercises are followed by questions from past examination papers.
4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out
of the question inside these spaces or boxes.
5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not
read the hints until you have tried to work out the question and are having difficulty.
6. What follows next are more questions taken from past examination papers.
7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and
notes to enrich your knowledge.
8. Finally, don’t be a loner. Work through this guide in a team with your classmates.
Sequences & Series
REVISION UNIT 4: SEQUENCES & SERIES
ARITHMETIC SEQUENCES
Let the first term of a sequence be a .
Let the difference between successive terms be d .
Then the following is what is called an arithmetic sequence:
a; a + d ; a + 2d ; a + 3d ; a + 4d ; .......... .......... .......
The terms of the sequence can be denoted by
T1 ; T2 ; T3 ; T4 ; T5 ; ................; Tn ; ................;
where Tn is the n th term.
This means that The n th term is also called the general term.
T1 = a In order to write an expression for the n th term:
T2 = a + d To determine if the sequence is arithmetic. This
you do by verifying there is a common difference
T3 = a + 2d d between successive terms:
T − T1 = T3 − T2 = T4 − T3 = ............ = d
T4 = a + 3d 2
and so on.
EXERCISE 4.1
We can deduce the n th term as follows:
1. Find an expression for the n th term of the
sequence ‐5; 3; 11; 19; ……………
T1 = a = (1 − 1)a
2. Find the number of terms in the following
T2 = a + d = a + (2 − 1)d arithmetic sequence:
T3 = a + 2d = a + (3 − 1)d 1; 6; 11; 16; ……………..; 486
T4 = a + 3d = a + (4 − 1)d 3. Find the 18th term of a series that has an n th
term given by (5 + 2n )
so that (you complete statement)
4. In an arithmetic sequence T4 = 93 and
Tn = T11 = 44 ; find a and d .
3
Preparation for the Mathematics examination brought to you by Kagiso Trust
GEOMETRIC SEQUENCES
Let the first term of a sequence be a .
Let the ratio between successive terms be r .
Then the following is what is called an geometric sequence:
a; ar; ar 2 ; ar 3 ; ar 4 ; ...........................
The terms of the sequence can be denoted by
T1 ; T2 ; T3 ; T4 ; T5 ; ................; Tn ; ................;
where Tn is the n th term.
This means that
The n th term is also called the general term.
T1 = a
In order to write an expression for the n th term:
T2 = ar
To determine if the sequence is geometric. This you
do by verifying there is a common ratio r between
T3 = ar 2
successive terms:
T4 = ar 3 T ÷ T1 = T3 ÷ T2 = T4 ÷ T3 = ............ = r
2
T5 = ar 4
EXERCISE 4.2
and so on. We can deduce the n th term as
follows: 1. Find an expression for the n th term of the
sequence 5/64; 5/32; 5/1`6; 5/8; ……………
T1 = a = ar 1−1
2. Find the number of terms in the following
geometric sequence:
T2 = ar = ar 2−1
24; 12; 6; 3; ……………..; 3 128
T3 = ar 2 = ar 3−1
3. Find the 4th term of a series that has an n th term
T4 = ar 3 = ar 4−1 given by 27(4 3)
n −1
.
so that (you complete the statement)
4. In a geometric sequence T5 = 3 / 4 and
Tn = T12 = 96 ; find a and r .
Sequences & Series
QUADRATIC SEQUENCES
We want to find an expression that would represent the following sequence of numbers:
3; 11; 23; 39; 59; 83; … … … … … ..
Is the sequence arithmetic?
Is the sequence geometric?
It is neither. What we now do is to find the first and second differences as follows:
Given sequence 3 11 23 39 59 83 . . . . . .
First differences 8 12 16 20 24 . . . . . . .
Second differences 4 4 4 4 . . . . .
A sequence with the same values for the second differences is called a QUADRATIC SEQUENCE. Because
it is quadratic its general term or what is called the n th term and symbolized by Tn must take the form
of a quadratic such as ax 2 + bx + c . Instead of x we use n . Thus the expression or equation
representing a quadratic sequence is
Tn = an 2 + bn + c
Our task is to find the values of a, b and c. We first use this quadratic expression to write down T1 , T2 ,
T3 and so on; and then we find their first and second differences as follows:
Putting n = 1, 2, 3, . . . . . . . we obtain:
Sequence a+b+c 4a+2b+c 9a+3b+c 16a+4b+c . . . . . .
First differences 3a+b 5a+b 7a+b . . . .
Second differences 2a 2a 2a . . . . .
Comparing the second differences, T2 =4(2)+b(2)+c =8+2b+c =11 or 2b+c =3 (2)
we see that 2a = 4 Subtract (1) from (2): b = 2
a =2 Substitute value of b in either (1) or (2):
Thus T1 = 2+b+c = 3 or b+c=1 (1) 2+c=1
and So c = ‐1
Thus putting the values of a=2, b=2 and c=‐1 in Tn = an 2 + bn + c we get the equation of the n th term:
Tn = 2n 2 + 2n − 1
5
Preparation for the Mathematics examination brought to you by Kagiso Trust
EXERCISE 4.3
Consider the sequence 5; 16; 29; 44; 61; . . . . . .
4.3.1 Write down the next three terms of the sequence assuming that the pattern continues.
4.3.2 Determine the n th term of the sequence.
4.3.3 Find the position of the term 236 in the sequence.
SERIES: SUMMING UP SEQUENCES
The sum of the first n terms of a series is denoted by S n .
Sum of the first n terms of an arithmetic progression
Write down the first n terms of the an arithmetic sequence and add them up as shown below :
S n = T1 + T2 + T3 + . . . . . . . . . . . . . + Tn −2 + Tn −1 + Tn
Rewrite the sum of the first n in reverse order:
S n = Tn + Tn −1 + Tn −2 + . . . . . . . . . . . . . + T2 + T1
Now write each term in terms of a, d and n . You need to work this out on a clean sheet of paper or the
space below:
in order to prove the results shown in the boxes below.
Show that by adding the two series and simplifying noting that the (a + d ) terms will cancel out the
result will be
Firstly
Sn =
n
[2a + (n − 1)d ]
2
Sequences & Series
and secondly
Sn =
n
(a + l ) where l = a + (n − 1)d is the last term of the series
2
There are thus two formulae for evaluating the sum of the first n terms of an arithmetic series.
EXERCISE 4.4
4.4.1 Given the series 1,25+3,50+5,75+8,00+ . . . . . , write down the sum of the first n terms and
hence find the value of S16 .
4.4.2 In an arithmetic series S n = n(2n + 3) , find the 10th term.
4.4.3 Calculate the sum of 5+9+13+ . . . . . . . +57+61.
Sum of the first n terms of a geometric progression
Here too you must follow the method above and rewrite the T terms in terms of a , r and n .
Complete the following statements (again, you will need a clean sheet of paper to do all this):
Sn =
rS n =
Now by carrying out subtraction show that
Sn =
(
a 1− rn
)
1− r
If the common ration r is greater than 1, the formula is usually written more conveniently in the form
Sn =
(
a r n −1
)
r −1
7
Preparation for the Mathematics examination brought to you by Kagiso Trust
CONVERGENCE AND THE SUM TO INFINITY OF A GEOMETRIC PROGRESSION
Any series whose sum does not approach some finite value as n takes on larger and larger integer
values is said to be divergent. In mathematical language, n taking on larger and larger values is usually
expressed as n approaching or tending to infinity; in symbols, this is written as n → ∞ .
Without going into any details, a geometric series is divergent in the following cases: r ≤ −1 and r ≥ 1 .
But what happens in the case − 1 < r < 1 and when n → ∞ ?
To answer the question, let’s focus on r n .
Take any value of r that lies between ‐1 and +1. For example, r = 0,5 .
Then, for n = 1 , r n = 0,51 = 0,5
n = 2 , r n = 0,5 2 = 0,25
n = 3 , r n = 0,5 3 = 0,125
Can you see that r n is becoming smaller as n becomes larger? This can only happen if − 1 < r < 1 .
Take now more larger values of n :
When n = 10 : r n = 0,510 = 9,765625 × 10 −4 or 0,0009765625
When n = 100 : r n = 0,5100 = 7,888609052 × 10 −31
You need only look at the negative exponents of 10 to realize that r n → 0 as n → ∞ for r = 0,5 which
is a value of r lying between ‐1 and +1.
Thus if we write Sn =
(
a 1− rn
)
1− r
a a
in the form Sn = − ⋅rn
1− r 1− r
then if − 1 < r < 1 and when n → ∞ , we will have
a a
S∞ = − ⋅0
1− r 1− r
a
S∞ =
1− r
Sequences & Series
which is known as the sum to infinity to which the sum of a geometric series converges if and only if
− 1 < r < 1 , sometimes written as r < 1 .
EXERCISE 4.5
4.5.1 Evaluate S10 for 150; 30; 6; 6/5; . . . . . .
4.5.2 Determine the 9th term in the above series and leave your answer as a fraction?
4.5.3 Determine S ∞ for the above series.
4.5.4 For what values of x does the series 150; 30 x ; 6 x 2 ; 6 x 3 /5; . . . . converge?
SIGMA NOTATION
The sum to n terms and the sum to infinity can be denoted in terms of sigma as follows:
n
S n = ∑ Ti = T1 + T2 + T3 + . . . . . . . + Tn
i =1
∞
S ∞ = ∑ Ti = T1 + T2 + T3 + . . . . . . . .
i =1
EXERCISE 4.6
4.6.1 Write 2+5+8+ . . . . . . up to the 10th term in sigma notation.
15
4.6.2 Calculate ∑ (4i − 3)
i =1
30
4.6.3 Calculate ∑ (2r + 3)
r =15
n
4.6.4 For what value of n does ∑ (5r + 3) first exceed 500?
r =1
9
Preparation for the Mathematics examination brought to you by Kagiso Trust
ANSWERS
EXERCISE 4.1 EXERCISE 4.5
1. Tn = 8n − 13 4.5.1 S10 = 187,49998 ≅ 187,5
2. n = 98
3. T18 = 41 6
4.5.2
4. a = 114; d = −7 78125
EXERCISE 4.2 4.5.3 S ∞ = 187,5
1. Tn = 5 ⋅ 2 n −7
4.5.4 − 5 < x < 5
2. n = 12
3. T4 = 64 EXERCISE 4.6
3
4. a = ;r = 2 10
∑ (3i − 1)
64 4.6.1
i =1
EXERCISE 4.3
4.6.2 435
4.3.1 80; 101; 124; . . .
4.3.2 Tn = n 2 + 8n − 4 4.6.3 HINT:
4.3.3 12th position (reject ‐20 because the position
30 30 14
must be given by a positive integer)
∑ (2r + 3) = ∑ (2r + 3) − ∑ (2r + 3)
r =15 r =1 r =1
EXERCISE 4.4
4.4.1 S n =
n
(2,25n + 0,25) ; S16 = 290 Answer is 762
2
4.4.2 T10 = 41 (HINT: S10 − S 9 = T10 . Can you 4.6.4 HINT: If the expression does not factorise
reason this out?) then use an approximate or graphical
4.4.3 You must first find n =15. S15 = 495 method to find an n such that S n > 500 .
Answer is n = 14
Sequences & Series
PAPER 1 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008
PAPER 1 QUESTION 3 DoE/ADDITIONAL EXEMPLAR 2008
11
Preparation for the Mathematics examination brought to you by Kagiso Trust
PAPER 1 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008
Number Hints and answers Work out the solutions in the boxes below
2.1 First, determine if the sequence is
geometric or arithmetic.
If arithmetic, d = what?
If geometric, r = what?
Then use the appropriate formula
from the formula sheet for the n th
term to find n .
Answer:
n =112
2.2.1 You should be able to determine the
common ratio r.
Answer:
p
r=
3
2.2.2 The formula for convergence is given
on the formula sheet.
Answer:
− 3 < p < 3, p ≠ 0
2.2.3 The formula for S ∞ is also on the
formula sheet.
Answer:
S ∞ = 486
Sequences & Series
PAPER 1 QUESTION 3 DoE/ADDITIONAL EXEMPLAR 2008
Number Hints and answers Work out the solutions in the boxes below
3.1 Test if the sequence is arithmetic or
geometric.
Obviously it is not arithmetic.
So it must be either geometric or
quadratic or both.
Assume it is geometric in order to
show that Tebogo is right or wrong. If
geometric, see if the fourth term is 54.
Assume it is quadratic in order to
show that Thembe is right or wrong. If
quadratic, work out the fourth term to
see if it is 38.
3.2 Use the appropriate formula for each
case above.
NOTE: Formula for the n th term of a
quadratic sequence does not appear
on the formula sheet given to you in
the exam. So memorise the formula
and how it is used to find the n th
term.
Answer:
Thembe’s sequence:
Tn = 4n 2 − 8n + 6
Tebogo’s sequence: 2.3 n −1
3.3 The 11th term is easily obtained by
using the n th term found above.
Answer:
T11 = 402
3.4 Use the appropriate formula to set up
an equation to solve for n .
Answer:
n = 12
13
Preparation for the Mathematics examination brought to you by Kagiso Trust
PAPER 1 QUESTION 2 DoE/NOVEMBER 2008
PAPER 1 QUESTION 3 DoE/NOVEMBER 2008
Sequences & Series
PAPER 1 QUESTION 2 DoE/NOVEMBER 2008
Number Hints and answers Work out the solutions in the boxes below
2.1.1 Simply by inspection you should tell if
the sequence is arithmetic or
geometric.
If neither, see if alternate terms are.
Clearly, the alternate terms are simple
cases of recognizable patterns.
2.1.2 How many terms will there be in each
sequence of alternate terms?
Answer that and then apply the right
formula to work out their sums.
Add the sums to get the answer.
Answer:
S 50 = 1001,00
2.2.1 Test by inspection if the sequence is
arithmetic or geometric.
If neither, test if it is quadratic.
If so, work out the next two terms.
Answer: You write it down.
2.2.2 Use the appropriate formula to work
out the n th term.
Answer:
Tn = n 2 + 7n = n(n + 7 )
2.2.3 Put Tn = 330 and solve.
Answer:
15th term is 330.
15
Preparation for the Mathematics examination brought to you by Kagiso Trust
PAPER 1 QUESTION 3 DoE/NOVEMBER 2008
Number Hints and answers Work out the solutions in the boxes below
3.1 What is a = ?
What is r = ?
Use the appropriate formula to find
the n th term.
Answer:
Tn = 2 4−n x n +1
This answer can be written in several
ways. If yours is not in this form, see if
you can manipulate it to be like this.
3.2 What is the condition for the series to
converge?
Substitute for r in this condition and
solve.
Answer:
− 2 < x < 2
3.3 What is the formula for the sum to
infinity?
Substitute for r in this formula and
solve.
Answer:
S ∞ = 72
Sequences & Series
MORE QUESTIONS FROM PAST EXAMINATION PAPERS
Exemplar 2008
17
Preparation for the Mathematics examination brought to you by Kagiso Trust
Preparatory Examination 2008
Sequences & Series
Feb – March 2009
19
Preparation for the Mathematics examination brought to you by Kagiso Trust
Sequences & Series
November 2009 (Unused paper)
21
Preparation for the Mathematics examination brought to you by Kagiso Trust
November 2009 (1)
Sequences & Series
Feb – March 2010
23
Preparation for the Mathematics examination brought to you by Kagiso Trust
Sequences & Series
ANSWERS
Exemplar 2008 Feb/March 2009
3.1 2; 21 2
3.2 5250 2.1.1 S 2 =
3
4.1 130; 173 3
4.2 Tn = 3n 2 + 4n − 2 2.1.2 S 3
4
4.3 n = 20
4
1 3 9 2.1.3 S 4 =
5.1 Pattern 3: + + 5
4 16 64
n
2.2 S n =
n +1
1 3 9 27
Pattern 4: + + +
4 16 64 256 2008
2.3 S n =
2009
n
3k −1
5.2 ∑
k =1 4
k
3.1 p = 3
5.3 S = 1
3.2.1 T1 = −2
Preparatory Examination 2008 3.2.2 d = 5
3.1 14
3.2 S = 975 3.3 After the first term ‐2, all the other terms end
in either 3 or an 8. Perfect squares never end
4.1 T6 = 38 in a 3 or an 8.
T7 = 51
4.1 ‐34
4.2 T p = p + 2
2
4.2 Tn = −2n 2 + 6n + 2
4.3 p = 5
4.3 T60 = −6838
5.1.1 Total amount = R511,50
5.1.2 Yes; he will have enough money to but the 15
5.1 Area of unshaded squares =
boots ( you must prove that this is the case). 16
5.2 Sum of unshaded squares of 1st seven
5.2.1 0 < x < 4 squares = 567
8 2
5.2.2 S ∞ = =2
3 3
25
Preparation for the Mathematics examination brought to you by Kagiso Trust
November 2009 (Unused papers) Feb/March 2010
2.1 100km on the 21st day. 2.1 Tn = n 2 − 42n + 440
2.2 S14 = 644km 2.2 n = 20 or n = 22
2.3 Not possible. For example, T1000 = 4016km
which cannot be covered in a single day by 2.3 The lowest value is the 21st term.
cycling.
a(r n − 1)
3.1 45 3.1 S n =
r −1
3.2 Tn = n 2 + 4n
9
3.2 S ∞ =
4.1 S n =
(
a 1− r n
) 2
1− r
4.1 30000 + 27000 + 24000 + . . . . . + 0
4.2.1 Series is convergent because − 1 < r < 1 4.2 After 11 years.
45 4.3 x = 13250
4.2.2 S ∞ = = 22,5
2
4.3.1 S 24 = 67108860
4.3.2 T24 = 2 25 or 33554432
4.3.3 Tn = 2 n +1
November 2009(1)
2.1.1 Tn = 4n + 1
2.1.2 Tn = 5(25)
n −1
2.2 Nomsa is correct.
3.1 ‐1; 2; 5
3.2 S100 = 14750
4.1 Tn = −2n + 3
4.2 35th difference = ‐67
4.3 Pn = − n 2 + 4n − 6
4.4 The function has a maximum value of ‐2 and
so the pattern will never have positive
values.
5.1 Growth in 17th year is 3,08cm
5.2 Height = 255,88cm
5.3 The tree will never attain a height greater
than 312cm.
