KT Classroom 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.

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

T1 = a

T2 = a + d

T3 = a + 2d

T4 = a + 3d

and so on.

We can deduce the n th term as follows:

T1 = a = (1 − 1)a

T2 = a + d = a + (2 − 1)d

T3 = a + 2d = a + (3 − 1)d

T4 = a + 3d = a + (4 − 1)d

so that (you complete statement)

Tn =

The n th term is also called the general term.

In order to write an expression for the n th term:

To determine if the sequence is arithmetic. This you do by verifying there is a common difference d between successive terms:

T − T1 = T3 − T2 = T4 − T3 = ............ = d 2

EXERCISE 4.1

1. Find an expression for the n th 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 n th term given by (5 + 2n )

4. In an arithmetic sequence T4 = 93 and

T11 = 44 ; find a and d .

3

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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

T1 = a

T2 = ar

T3 = ar 2

T4 = ar 3

T5 = ar 4

and so on. We can deduce the n th term as follows:

T1 = a = ar 1−1

T2 = ar = ar 2−1

T3 = ar 2 = ar 3−1

T4 = ar 3 = ar 4−1

so that (you complete the statement)

Tn =

The n th term is also called the general term.

In order to write an expression for the n th term:

To determine if the sequence is geometric. This you do by verifying there is a common ratio r between successive terms:

T ÷ T1 = T3 ÷ T2 = T4 ÷ T3 = ............ = r 2

EXERCISE 4.2

1. Find an expression for the n th 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; ……………..; 3 128

3. Find the 4th term of a series that has an n th term given by 27(4 3)

n −1

.

4. In a geometric sequence T5 = 3 / 4 and

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: 3 11 23 39 59 83 . . . . . .

Given sequence 8 12 16 20 24 . . . . . . .

First differences 4 4 4 4 . . . . .

Second differences

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:

a+b+c 4a+2b+c 9a+3b+c 16a+4b+c . . . . . .

Sequence 3a+b 5a+b 7a+b . . . .

First differences 2a 2a 2a . . . . .

Second differences

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

So c = ‐1 and

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

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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

(

)

a 1− rn Sn =

1− r

If the common ration r is greater than 1, the formula is usually written more conveniently in the form

(

)

a r n −1 Sn =

r −1

7

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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

a 1− rn Sn =

1− r

in the form

Sn =

a a − ⋅rn 1− r 1− r

then if − 1 < r < 1 and when n → ∞ , we will have

S∞ =

a a − ⋅0 1− r 1− r

S∞ =

a

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.

4.6.2

Calculate

15

∑ (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

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ANSWERS EXERCISE 4.5

4.5.1 S10 = 187,49998 ≅ 187,5

EXERCISE 4.1

1. Tn = 8n − 13 2. n = 98 3. T18 = 41

4.5.2

4. a = 114; d = −7

4.5.3 S ∞ = 187,5

4.5.4 − 5 < x < 5

EXERCISE 4.6

EXERCISE 4.2 1. Tn = 5 ⋅ 2 n −7 2. n = 12 3. T4 = 64 3 4. a = ;r = 2 64

10

EXERCISE 4.3

4.3.1 80; 101; 124; . . .

4.3.2 Tn = n 2 + 8n − 4 4.3.3 12th position (reject ‐20 because the position must be given by a positive integer)

EXERCISE 4.4

n (2,25n + 0,25) ; S16 = 290 2 4.4.2 T10 = 41 (HINT: S10 − S 9 = T10 . Can you

4.4.1 S n =

reason this out?)

4.4.3 You must first find n =15. S15 = 495

6

78125

4.6.1

∑ (3i − 1) i =1

4.6.2 435

4.6.3 HINT:

30

30

14

r =15

r =1

r =1

∑ (2r + 3) = ∑ (2r + 3) − ∑ (2r + 3)

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 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

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PAPER 1 QUESTION 2

Number Hints and answers 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:

r=

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

DoE/ADDITIONAL EXEMPLAR 2008

Work out the solutions in the boxes below

p

3

Sequences & Series

PAPER 1 QUESTION 3

Number Hints and answers

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 3.3

3.4

Tebogo’s sequence: 2.3 n −1 The 11th term is easily obtained by using the n th term found above.

Answer: T11 = 402 Use the appropriate formula to set up an equation to solve for n .

Answer: n = 12

13

DoE/ADDITIONAL EXEMPLAR 2008

Work out the solutions in the boxes below

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PAPER 1 QUESTION 2

DoE/NOVEMBER 2008

PAPER 1 QUESTION 3

DoE/NOVEMBER 2008

Sequences & Series

PAPER 1 QUESTION 2

Number Hints and answers 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

2.2.2

2.2.3

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.

Use the appropriate formula to work out the n th term.

Answer: Tn = n 2 + 7n = n(n + 7 )

Put Tn = 330 and solve.

Answer: 15th term is 330.

15

DoE/NOVEMBER 2008

Work out the solutions in the boxes below

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PAPER 1 QUESTION 3

Number Hints and answers 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

DoE/NOVEMBER 2008

Work out the solutions in the boxes below

Sequences & Series

MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008

17

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Preparatory Examination 2008

Sequences & Series

Feb – March 2009

19

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Sequences & Series

November 2009 (Unused paper)

21

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November 2009 (1)

Sequences & Series

Feb – March 2010

23

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Sequences & Series

ANSWERS Exemplar 2008 3.1 2; 21 3.2 5250

4.1 130; 173 4.2 Tn = 3n 2 + 4n − 2 4.3 n = 20

2.1.1 S 2 =

2.1.2 S 3

Pattern 4:

1 3 9 27

+ + + 4 16 64 256

3k −1

∑ k k =1 4 n

5.3 S = 1

Preparatory Examination 2008 3.1 14 3.2 S = 975

4.1 T6 = 38 T7 = 51

4.2 T p = p + 2 2

4.3 p = 5

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 0 < x < 4

2.1.3 S 4 =

8 2 =2 3 3

4

5

n

n +1

2.3 S n =

2008

2009

3.1 p = 3

3.2.1 T1 = −2 3.2.2 d = 5

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 Tn = −2n 2 + 6n + 2

4.3 T60 = −6838

5.1 Area of unshaded squares =

15

16

5.2 Sum of unshaded squares of 1st seven squares = 567

25

3

4

2.2 S n =

5.2.2 S ∞ =

2

3

1 3 9 5.1 Pattern 3: +

+ 4 16 64

5.2

Feb/March 2009

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November 2009 (Unused papers) 2.1 100km on the 21st day.

2.2 S14 = 644km 2.3 Not possible. For example, T1000 = 4016km which cannot be covered in a single day by cycling.

3.1 45 3.2 Tn = n 2 + 4n

4.1 S n =

(

a 1− r 1− r

n

)

4.2.1 Series is convergent because − 1 < r < 1 4.2.2 S ∞ =

45 = 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)

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.

n −1

Feb/March 2010 2.1 Tn = n 2 − 42n + 440 2.2 n = 20 or n = 22

2.3 The lowest value is the 21st term.

a(r n − 1) 3.1 S n =

r −1

3.2 S ∞ =

9

2

4.1 30000 + 27000 + 24000 + . . . . . + 0 4.2 After 11 years.

4.3 x = 13250