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
3
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:
........................... ;4 ;3 ;2 ; ; dadadadaa ++++
The terms of the sequence can be denoted by
......;.......... ;T ......;.......... ;T ;T ;T ;T ;T 54321 n
where nT is the thn term.
This means that
a=1T
da +=2T
da 2T3 +=
da 3T4 += and so on. We can deduce the thn term as follows:
( )aa 11T1 −==
( )dada 12T2 −+=+=
( )dada 132T3 −+=+=
( )dada 143T4 −+=+=
so that (you complete statement) =nT
The thn term is also called the general term.
In order to write an expression for the thn term: To determine if the sequence is arithmetic. This you do by verifying there is a common difference d between successive terms:
d==−=−=− ............ TTTTTT 342312
EXERCISE 4.1
1. Find an expression for the thn term of the
sequence ‐5; 3; 11; 19; ……………
2. Find the number of terms in the following arithmetic sequence:
1; 6; 11; 16; ……………..; 486
3. Find the 18th term of a series that has an thnterm given by ( )n25+
4. In an arithmetic sequence 934 =T and
4411 =T ; find a and d .
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:
........................... ; ; ; ; ; 432 arararara
The terms of the sequence can be denoted by
......;.......... ;T ......;.......... ;T ;T ;T ;T ;T 54321 n
where nT is the thn term.
This means that
a=1T
ar=2T
23T ar=
3
4T ar=
45T ar=
and so on. We can deduce the thn term as follows:
111T −== ara
12
2T −== arar
1323T −== arar
143
4T −== arar
so that (you complete the statement) =nT
The thn term is also called the general term.
In order to write an expression for the thn term: To determine if the sequence is geometric. This you do by verifying there is a common ratio r between successive terms:
r==÷=÷=÷ ............ TTTTTT 342312
EXERCISE 4.2
1. Find an expression for the thn term of the sequence 5/64; 5/32; 5/1`6; 5/8; ……………
2. Find the number of terms in the following
geometric sequence: 24; 12; 6; 3; ……………..; 1283
3. Find the 4th term of a series that has an thn term
given by ( ) 13427 −n.
4. In a geometric sequence 4/35 =T and
9612 =T ; find a and r .
Sequences & Series
5
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 First differences Second differences
3 11 23 39 59 83 . . . . . . 8 12 16 20 24 . . . . . . . 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 thn term and symbolized by nT must take the form
of a quadratic such as cba 2 ++ xx . Instead of x we use n . Thus the expression or equation representing a quadratic sequence is
cbaT 2 ++= nnn
Our task is to find the values of a, b and c. We first use this quadratic expression to write down 1T , 2T ,
3T and so on; and then we find their first and second differences as follows:
Putting . . . . . . . ,3 ,2 ,1=n we obtain:
Sequence First differences Second differences
a+b+c 4a+2b+c 9a+3b+c 16a+4b+c . . . . . . 3a+b 5a+b 7a+b . . . . 2a 2a 2a . . . . .
Comparing the second differences, we see that 2a = 4 a =2 Thus 1T = 2+b+c = 3 or b+c=1 (1) and
2T =4(2)+b(2)+c =8+2b+c =11 or 2b+c =3 (2) Subtract (1) from (2): b = 2 Substitute value of b in either (1) or (2): 2+c=1 So c = ‐1
Thus putting the values of a=2, b=2 and c=‐1 in cbaT 2 ++= nnn we get the equation of the thn term:
122T 2 −+= nnn
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 thn 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 nS .
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 :
nnnn TTTTTTS ++++++= −− 12321 . . . . . . . . . . . . .
Rewrite the sum of the first n in reverse order:
1221 . . . . . . . . . . . . . TTTTTS nnnn +++++= −−
Now write each term in terms of nda and , . 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 ( )da + terms will cancel out the result will be Firstly
( )[ ]dnanSn 122
−+=
Sequences & Series
7
and secondly
( )lanSn +=2
where ( )dnal 1−+= is the last term of the series
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 16S .
4.4.2 In an arithmetic series ( )32 += nnSn , 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 nra and , . Complete the following statements (again, you will need a clean sheet of paper to do all this):
=nS
=nrS
Now by carrying out subtraction show that
( )rraS
n
n −−
=11
If the common ration r is greater than 1, the formula is usually written more conveniently in the form
( )1
1−−
=rraS
n
n
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: 1−≤r and 1≥r .
But what happens in the case 11 <<− r and when ∞→n ?
To answer the question, let’s focus on nr .
Take any value of r that lies between ‐1 and +1. For example, 5,0=r .
Then, for 1=n , 5,05,0 1 ==nr
2=n , 25,05,0 2 ==nr
3=n , 125,05,0 3 ==nr
Can you see that nr is becoming smaller as n becomes larger? This can only happen if 11 <<− r .
Take now more larger values of n :
When 10=n : 250,00097656or 10765625,95,0 410 −×==nr
When 100=n : 31100 10888609052,75,0 −×==nr
You need only look at the negative exponents of 10 to realize that 0→nr as ∞→n for 5,0=r which
is a value of r lying between ‐1 and +1.
Thus if we write ( )
rraS
n
n −−
=11
in the form nn r
ra
raS ⋅
−−
−=
11
then if 11 <<− r and when ∞→n , we will have
011
⋅−
−−
=∞ ra
raS
r
aS−
=∞ 1
Sequences & Series
9
which is known as the sum to infinity to which the sum of a geometric series converges if and only if
11 <<− r , sometimes written as 1<r .
EXERCISE 4.5
4.5.1 Evaluate 10S 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 2x ; 6 3x /5; . . . . converge?
SIGMA NOTATION
The sum to n terms and the sum to infinity can be denoted in terms of sigma as follows:
n
n
iin TTTTTS ++++== ∑
=
. . . . . . . 3211
. . . . . . . . 3211
+++== ∑∞
=∞ TTTTS
ii
EXERCISE 4.6 4.6.1 Write 2+5+8+ . . . . . . up to the 10th term in sigma notation.
4.6.2 Calculate ( )∑=
−15
1
34i
i
4.6.3 Calculate ( )∑=
+30
1532
rr
4.6.4 For what value of n does ( )∑=
+n
rr
135 first exceed 500?
Preparation for the Mathematics examination brought to you by Kagiso Trust
ANSWERS
EXERCISE 4.1 1. 138 −= nTn
2. 98=n 3. 4118 =T
4. 7;114 −== da EXERCISE 4.2 1. 725 −⋅= n
nT
2. 12=n 3. 644 =T
4. 2;643
== ra
EXERCISE 4.3 4.3.1 80; 101; 124; . . .
4.3.2 482 −+= nnTn
4.3.3 12th position (reject ‐20 because the position must be given by a positive integer) EXERCISE 4.4
4.4.1 ( )25,025,22
+= nnSn ; 29016 =S
4.4.2 4110 =T (HINT: 10910 TSS =− . Can you
reason this out?) 4.4.3 You must first find n =15. 49515 =S
EXERCISE 4.5 4.5.1 5,18749998,18710 ≅=S
4.5.2 78125
6
4.5.3 5,187=∞S 4.5.4 55 <<− x EXERCISE 4.6
4.6.1 ( )∑=
−10
113
ii
4.6.2 435 4.6.3 HINT:
( ) ( ) ( )∑∑∑===
+−+=+14
1
30
1
30
15323232
rrrrrr
Answer is 762 4.6.4 HINT: If the expression does not factorise then use an approximate or graphical method to find an n such that 500>nS . Answer is 14=n
Sequences & Series
11
PAPER 1 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008
PAPER 1 QUESTION 3 DoE/ADDITIONAL EXEMPLAR 2008
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 thn term to find n . Answer: n =112
2.2.1 You should be able to determine the common ratio r. Answer:
3pr =
2.2.2 The formula for convergence is given on the formula sheet. Answer:
0,33 ≠<<− pp
2.2.3 The formula for ∞S is also on the formula sheet. Answer:
486=∞S
Sequences & Series
13
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 thn 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 thn term. Answer: Thembe’s sequence:
684 2 +−= nnTn
Tebogo’s sequence: 13.2 −n
3.3 The 11th term is easily obtained by
using the thn term found above. Answer:
40211 =T
3.4 Use the appropriate formula to set up an equation to solve for n . Answer:
12=n
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
15
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:
00,100150 =S
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 thn term. Answer:
( )772 +=+= nnnnTn
2.2.3 Put 330=nT and solve.
Answer: 15th term is 330.
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 thn term. Answer:
142 +−= nnn xT
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:
22 <<− x
3.3 What is the formula for the sum to infinity? Substitute for r in this formula and solve. Answer:
72=∞S
Sequences & Series
17
MORE QUESTIONS FROM PAST EXAMINATION PAPERS
Exemplar 2008
Preparation for the Mathematics examination brought to you by Kagiso Trust
Preparatory Examination 2008
Sequences & Series
19
Feb – March 2009
Preparation for the Mathematics examination brought to you by Kagiso Trust
Sequences & Series
21
November 2009 (Unused paper)
Preparation for the Mathematics examination brought to you by Kagiso Trust
November 2009 (1)
Sequences & Series
23
Feb – March 2010
Preparation for the Mathematics examination brought to you by Kagiso Trust
Sequences & Series
25
ANSWERS Exemplar 2008 3.1 2; 21 3.2 5250 4.1 130; 173
4.2 243 2 −+= nnTn
4.3 20=n
5.1 Pattern 3: 649
163
41
++
Pattern 4: 25627
649
163
41
+++
5.2 ∑=
−n
kk
k
1
1
43
5.3 1=S Preparatory Examination 2008 3.1 14 3.2 975=S 4.1 386 =T
517 =T
4.2 22 += pTp
4.3 5=p 5.1.1 Total amount = R511,50 5.1.2 Yes; he will have enough money to but the boots ( you must prove that this is the case). 5.2.1 40 << x
5.2.2 322
38==∞S
Feb/March 2009
2.1.1 32
2 =S
2.1.2 43
3S
2.1.3 54
4 =S
2.2 1+
=n
nSn
2.3 20092008
=nS
3.1 3=p 3.2.1 21 −=T
3.2.2 5=d 3.3 After the first term ‐2, all the other terms end in either 3 or an 8. Perfect squares never end in a 3 or an 8. 4.1 ‐34
4.2 262 2 ++−= nnTn
4.3 683860 −=T
5.1 Area of unshaded squares = 1615
5.2 Sum of unshaded squares of 1st seven squares = 567
Preparation for the Mathematics examination brought to you by Kagiso Trust
November 2009 (Unused papers) 2.1 100km on the 21st day. 2.2 kmS 64414 =
2.3 Not possible. For example, kmT 40161000 =
which cannot be covered in a single day by cycling. 3.1 45
3.2 nnTn 42 +=
4.1 ( )
rraS
n
n −−
=11
4.2.1 Series is convergent because 11 <<− r
4.2.2 5,22245
==∞S
4.3.1 6710886024 =S
4.3.2 33554432or 22524 =T
4.3.3 12 += nnT
November 2009(1) 2.1.1 14 += nTn
2.1.2 ( ) 1255 −= nnT
2.2 Nomsa is correct. 3.1 ‐1; 2; 5 3.2 14750100 =S
4.1 32 +−= nTn
4.2 35th difference = ‐67
4.3 642 −+−= nnPn
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.
Feb/March 2010
2.1 440422 +−= nnTn
2.2 22or 20 == nn 2.3 The lowest value is the 21st term.
3.1 ( )
11
−−
=rraS
n
n
3.2 29
=∞S
4.1 30000 + 27000 + 24000 + . . . . . + 0 4.2 After 11 years. 4.3 13250=x
Top Related