Unit 4: Sequences and Series

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.

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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 .


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 .

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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


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

−+=

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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


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

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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?


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

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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

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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


PAPER 1 QUESTION 2 DoE/NOVEMBER 2008


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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.



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

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MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Exemplar 2008


Preparatory Examination 2008

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Feb – March 2009

Sequences & Series

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November 2009 (Unused paper)


November 2009 (1)

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Feb – March 2010

Sequences & Series

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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


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

Unit 4: Sequences and Series

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Transcript of Unit 4: Sequences and Series