2 unit ap and gp

7/24/2019 2 Unit AP and GP

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First published by

John Kinny-Lewis in 2012

c John Kinny-Lewis 2012

National Library of Australia

Cataloguing-in-publication data

ISBN: 978-0-9872782-1-0

This book is copyright. Apart from any fair dealing for the purposes

of private study, research, criticism or review as permitted under theCopyright Act 1968, no part may be reproduced, stored in a retrievalsystem, or transmitted, in any form by any means, electronic, mechan-ical, photocopying, recording, or otherwise without prior written per-mission. Enquiries to be made to John Kinny-Lewis.

Copying for educational purposes.

Where copies of part or the whole of the book are made under Sec-tion 53B or Section 53D of the Copyright Act 1968, the law requiresthat records of such copying be kept. In such cases the copyright owner

is entitled to claim payment.

Typeset by Christopher Hines

Edited by Christopher Hines


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Preface

This revision of Mathematics Topics has been writtento follow the syllabus of the NSW Higher School Cer-tificate course of Mathematics.

It is assumed that the student is familiar with the con-tent of the corresponding Mathematics syllabus.


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2 ARITHMETIC AND GEOMETRIC SERIES

Arithmetic and Geometric Series

Example 1

The fifth term of an arithmetic sequence is 44, and the tenth term is 64.

(i) Find the value of the first term and the common difference.

(ii) Find the sum of the first 100 terms.

Solution:

(i)

Tn=a + (n 1)d

T5=a + 4d= 44 (1)

T10=a + 9d= 64 (2)

(2) (1) 5d= 20d = 4

In (1) a + 4

4 = 44

a = 28

Therefore the first term is 28 and the common difference is 4.

(ii)

Sn

= n

2(2a + (n 1)d)

S100=100

2

(2

28 + 99

4)

S100= 22, 600


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ARITHMETIC AND GEOMETRIC SERIES 3

Example 2

A company plans to sell 400 computers in its first month of operation. Theyplan to increase their sales by 40 computers each month. How many com-puters do they plan to sell,

(i) in the last month of the first year of operation?

(ii) over the entire 12 months?

Solution:

(i)The sequence is 400, 440, 480 . . .

a= 400 and d= 40

Tn= a + (n 1)d

= 400 + (n 1) 40= 400 + 40n 40= 360 + 40n

T12= 360 + 40 12= 840

Therefore in the last month, 840 computers should have been sold.

(ii)

Sn

= n

2(a + T12)

=

12

2(400 + 840)

= 7440

Therefore in the entire 12-month period 7440 computers should have beensold.


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Example 3

A geometric series has a second term 4.5 and the ratio of the fourth term tothe third term is 1.5.

(i) Find the common ratio r.

(ii) What is the first term a?

(iii) Calculate the sum of the first 9 terms.

Solution:

(i)The common ratio is given as

r= T

n

Tn1

= T4

T3= 1.5

(ii)

Tn=ar

n1

T2=ar = 4.5

a 1.5 = 4.5 the first term is 3

(iii)

Sn

= a(rn 1)

r 1

S9=3((1.5)9 1)

1.5 1

= 6

3

2

9

1

= 6

19683

512 1

= 224169

256


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Example 4

From his sons first birthday, a father makes a series of annual paymentsof $1000 to a financial institution given an interest of 7% pa. He continuesto make these payments until his sons 21st birthday, when the last paymentis made. He then withdraws the whole amount and gives it to his son. Howmuch does the son receive, to the nearest dollar?

Solution:

From the sons first till his 21st birthday there is a period of 20 years.

the first amount earns A1=1000 1.0720

the second amount A2= 1000 1.0719

and so on until A20= 1000 1.07the total amount = 1000 + 1000 1.07 + 1000 1.072 + + 1000 1.0720

(including the final payment made on the 21st birthday)

now 1000 + 1000 1.07+ + 1000 1.0720 is a GP

where a= 1000, r= 1.07 and n= 21

S21= 1000 (1.07)21 1

1.07 1 the son receives $44, 865. (nearest dollar)

Example 5

A loan of $10,000 is to be repaid by equal annual instalments, repaymentscommencing at the end of the first year of the loan. Interest at the rate of 8%

is calculated each year on the balance owing at the beginning of that year,and added to the balance.

If the annual instalment is Mdollars, prove that

(i) the amount owing at the beginning of the second year of the loan is(10, 800 M) dollars.

(ii) the amount owing at the beginning of the third year of the loan is

(11, 664 2.08 M) dollars.(iii) If the loan, including the interest charges, is exactly repaid at the end

ofnyears, then

M= 800(1.08)n

(1.08)n 1


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Solution:

(i)Let A1 be the amount owing at the beginning of the second year.

A1= 10, 000

1 + 8

100

1

M

= 10, 000 1.08 MA1= (10, 800 M) dollars

(ii)

Let A2 be the amount owing at the beginning of the third year.

A2=A1 (1.08)1 M= (10, 000 (1.08)1 M) (1.08)1 M= 10, 000

(1.08)2

M(1.08)1

M

= 10, 000 (1.08)2 M(1.08 + 1)A2= 11, 664 2.08M dollars.

(iii)

A2= 10, 000 (1.08)2 M(1 + 1.08)A3= 10, 000

(1.08)3

M(1 + 1.08 + (1.08)2)

An

= 10, 000 (1.08)n M1 + 1.08 + (1.08)2 + + (1.08)n1

1 + 1.08 + (1.08)2 + + (1.08)n1 is a GPa= 1 and r= 1.08

Sn =

a(rn 1)r 1

=1((1.08)n 1)

1.08 1Sn

=(1.08)n 1

0.08


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When the loan is repaid An

= 0

0 = 10, 000(1.08)n M((1.08)n 1)

0.08

M((1.08)n 1)

0.08 = 10, 000(1.08)n

M=10, 000(1.08)n 0.08

(1.08)n 1

M= 800(1.08)n

(1.08)n

11) The sum of the first 5 terms of an arithmetic series is 45, the sum of the

first 10 terms is 145. Find

(i) the common difference

(ii) the first term

2) The sixth term of an arithmetic progression is 18 and the eleventh termis 24.

(i) Find the common difference(ii) Find the sum of the first 20 terms

3) The sum ton terms of an arithmetic sequence is given by Sn= 2n2 + 3n.Find

(i) the first term.

(ii) the common difference.

(iii) the number of terms required to give a sum of 1325.

4) A truck travels 100 km to load 1 ton of coal and then returns to its base.It then travels 120 km, loads another 1 ton of coal further away fromthe first and returns to base.

If the distances that the truck travels each time form an arithmeticsequence then

(i) show that the total distance travelled, forn trips, is 20n(n +9) km.

(ii) If the price for coal is $600 per ton and the travelling costs are$2 per km, calculate the number of trips possible before the total

distance travelled makes the collection unprofitable.(iii) Using your calculator, determine the number of trips that will

yield a maximum profit.


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5) The sum of the first three terms of a geometric progression is 152 and thesum of the next three terms 513. Find

(i) the common ratio.

(ii) the first term.

6) By considering the limiting sum of a suitable infinite series, express 0.272as a fraction in its simplest form.

7) The first three terms of a geometric series are 2

w,

2

w2 and

2

w3. Given that

|w

|>1.

(i) Find an expression in w for the limiting sum.

(ii) Find the value ofw when the limiting sum is w.

8) A golf ball is dropped from a height of 12 m onto a stone pavement. Aftereach bounce, the maximum height reached by the ball is one third ofthe previous maximum height.

(i) What height does it reach on its second bounce?

(ii) How far has it travelled when it strikes the pavement on the fourth

occasion?9) A sum of $5000 is invested in a superannuation fund at the start of each

year for 25 years at 5% per annum. Find the value of superannuationavailable at the end of 25 years.

10) Tony invests $P in an account which earns 6% interest, compoundedannually. He intends to withdraw $1000 at the end of each year, im-mediately after the interest has been paid. He wishes to be able to dothis for exactly 15 years.

(i) How much does he have in his account, immediately after he has

made his first withdrawal?

(ii) Write an expression in terms of P for the money in the account,immediately after his 2nd withdrawal.

(iii) Calculate the value of P, which leaves his account empty afterthe 15th withdrawal.

(iv) Suppose Tony wished to withdraw $1900 per year from his accountif the initial investment is the same as the one calculated in (iii).For how many years could he withdraw this amount (at the same

interest rate).

END OF CHAPTER


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ARITHMETIC AND GEOMETRIC SERIES - SOLUTIONS 9

Arithmetic and Geometric Series - Solutions1) The sum of the first 5 terms of an arithmetic series is 45, the sum of the first10 terms is 145. Find

(i) the common difference

Sn= n

2(2a + (n 1)d)

S5=5

2(2a + 4d) = 45

2a + 4d= 18 (1)

S10=10

2(2a + 9d) = 145

2a + 9d= 29 (2)

(2) (1) 5d= 11

d = 2.2

Therefore the common difference is 2.2.

(ii) the first term

In (1) 2a + 4 2.2 = 18

2a= 9.2

a = 4.6

Therefore the first term is 4.6.

2) The sixth term of an arithmetic progression is 18 and the eleventh term is

24.

(i) Find the common difference

Tn= a + (n 1)d

T6= a + 5d= 18 (1)

T11= a + 10d= 24 (2)

(2) (1) 5d= 6

d= 1.2

Therefore the common difference is 1.2.


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10 ARITHMETIC AND GEOMETRIC SERIES - SOLUTIONS

(ii) Find the sum of the first 20 terms

Using d= 1.2 (from i)

In (1) a + 5 1.2 = 18

a + 6 = 18

a = 12

Sn= n

2(2a + (n 1)d)

S20

=

20

2(2 12 + 19 1.2) = 468Therefore the sum of the first 20 terms is 468.

3) The sum to n terms of an arithmetic sequence is given by Sn = 2n2 + 3n.Find

(i) the first term.

Sn= 2n2 + 3n

T1= S1= 2 (1)2 + 3 1

T1= 5 a = 5(ii) the common difference.

T1+ T2= S2

5 + T2= 2 (2)2 + 3 2

T2= 8 + 6 5

= 9

d= T2 T1= 9 5 = 4

Therefore the common difference is 4.

(iii) the number of terms required to give a sum of 1325.

1325 = 2n2 + 3n 2n2 + 3n 1325 = 0

n=3 10, 609

4

n = 25 (n >0)

there are 25 terms required.


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4) A truck travels 100 km to load 1 ton of coal and then returns to its base.It then travels 120 km, loads another 1 ton of coal further away from the first andreturns to base.

If the distances that the truck travels each time form an arithmetic sequence then

(i) show that the total distance travelled, for n trips, is 20n(n + 9) km.

The trips form the arithmetic progression 200, 240, 280... Where a = 240 andd= 40.

Let Sn= D (total distance travelled)

D= Sn= n

2(2a + (n 1)d)

D = n

2(2 200 + (n 1) 40)

= n

2(400 + 40n 40)

= n

2(360 + 40n)

D = 20n(n + 9)

(ii) If the price for coal is $600 per ton and the travelling costs are $2 per km,calculate the number of trips possible before the total distance travelled makes thecollection unprofitable.

the total revenue of coal is $600n

the total cost is 20n(n + 9) $2 = $40n(n + 9)

when the profit is 0 ,

600n= 40n(n + 9)

40(n + 9) = 600

n + 9 = 15

n= 6

Therefore there are 6 trips possible.


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(iii) Using your calculator, determine the number of trips that will yield a maxi-mum profit.

let P = profit

P= revenue cost

P= 600n 40n(n + 9)

= 600n 40n2 360n

= 240n 40n2

= 40n(6 n)

n 1 2 3 4 5 6

P $ 200 $320 $360 $320 $200 $0

Therefore the maximum profit is $360.

Therefore 3 trips will yield a maximum profit.

5)The sum of the first three terms of a geometric progression is 152 and the sum

of the next three terms 513. Find

(i) the common ratio.

S3= a + ar+ ar2 = 152

S6= a + ar+ ar2 + ar3 + ar4 + ar5

S6 S3= ar3 + ar4 + ar5 = 513

a + ar+ ar2 = 152 (1)

ar3 + ar4 + ar5 = 513 (2)

(2) (1) ar3 + ar4 + ar5

a + ar+ ar2 =

513

152

ar3(1 + r+ r2)

a(1 + r+ r2) =

27

8

r3 =27

8

r =3

2

Therefore the common ratio is 1.5.


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(ii) the first term.

In (1) a

1 +

3

2+

3

2

2

= 152

a 194

= 152

a = 32

Therefore the first term is 32.

6) By considering the limiting sum of a suitable infinite series, express 0.27

2 as afraction in its simplest form.

express 0.272 as 0.2 + 0.0727272...

0.272 = 2

10+

72

1000+

72

100, 000+ . . .

Note that 72

1000+

72

100, 000+ . . . is a geometric progression, with

a= 721000

and r= 1100

.

S= a

1 r =72

1000

1 1100

= 72

1000 99

100

=

4

55

0.272 = 2

10+

72

1000+

72

100, 000+ . . .

= 2

10+

4

55

= 3

11


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7) The first three terms of a geometric series are 2

w,

2

w2 and

2

w3. Given that

|w| >1.

(i) Find an expression in w for the limiting sum.

2

w,

2

w2,

2

w3, . . .

a = 2

w, r=

1

w

S= a

1

r

=

2

w

1 1w

= 2

w

1 1

w

= 2

w

w 1

w

=

2

w w

w 1

S= 2

w 1(ii) Find the value ofw when the limiting sum is w.

S= w

w = 2

w 1

w(w 1) = 2w2 w= 2

w2 w 2 = 0

(w 2)(w+ 1) = 0

w = 2 or w= 1

Since|w| >1 then w= 2 is the only valid solution.


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8) A golf ball is dropped from a height of 12 m onto a stone pavement. Aftereach bounce, the maximum height reached by the ball is one third of the previousmaximum height.

(i) What height does it reach on its second bounce?

The ball first falls 12 metres, bounces to a height of 4 metres, falls and bounces a

second time reaching 4 3 = 1 13

m high.

(ii) How far has it travelled when it strikes the pavement on the fourth occa-sion?

Striking the pavement for the 4th time includes the initial 12m drop as well as3 symmetrical rise and falls,

D = 12 + 2

4 +4

3+

4

32

= 235

9

9) A sum of $5000 is invested in a superannuation fund at the start of each yearfor 25 years at 5% per annum. Find the value of superannuation available at the

end of 25 years.let A1 be the final value of the first investment ,

let A2 be the final value of the second investment ,

... and so on. Finally let A25 be the final value of the final investment ,

A1= 50001 + 5

10025

= 5000(1.05)25

A25= 5000(1.05)

total = 5000(1.05) + + 5000(1.05)24 + 5000(1.05)25

This is a GP with a= 5000(1.05) and r= 1.05.

Sn= a(rn 1)

r 1

S25= 5000(1.05)(1.0525 1)

1.05

1

= $250, 567.27

Therefore the amount of superannuation is $250,567.27


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10) Tony invests $P in an account which earns 6% interest, compounded annu-ally. He intends to withdraw $1000 at the end of each year, immediately after theinterest has been paid. He wishes to be able to do this for exactly 15 years.

(i) How much does he have in his account, immediately after he has made hisfirst withdrawal?

Let A1 be the amount remaining after the first withdrawal,

A1= P(1.06) 1000(ii) Write an expression in terms ofPfor the money in the account, immediatelyafter his 2ndwithdrawal.

Let A2 be the amount remaining after the second withdrawal,

A2= A1(1.06) 1000

= [P(1.06) 1000](1.06) 1000

=P(1.06)2 1000(1.06 + 1)

=P(1.06)2 2060(iii) Calculate the value of P, which leaves his account empty after the 15th

withdrawal. Let A15 be the amount remaining after the final withdrawal.A15= P(1.06)

15 1000(1 + 1.06 + 1.062 + + 1.0614)

1 + 1.06 + 1.062 + + 1.0614 is a GP

a= 1, r= 1.06

S15=1(1.0615 1)

1.06 1 = 23.27597

A15= P(1.06)15 1000 23.27597

to leave the account empty A15= 0 P(1.06)15

= 1000 23.27597P= $9712.25

(iv) Suppose Tony wished to withdraw $1900 per year from his account if theinitial investment is the same as the one calculated in (iii). For how many yearscould he withdraw this amount (at the same interest rate).

Again let A1 be the amount remaining after the first withdrawal,

A1= 9712.25(1.06) 1900Also,

A2= [9712.25(1.06) 1900](1.06) 1900

= 9712.25(1.06)2 1900(1 + 1.06)


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After n withdrawals,

An= 9712.25(1.06)n 1900(1 + 1.06 + . . . (1.06)n1)

1 + 1.06 + . . . (1.06)n1 is a GP.

Sn=1(1.06n 1)

0.06

Now let An= 0

9712.25(1.06)n = 1900 (1.06n 1)

0.06

582.735(1.06)n = 1900 (1.06)n 1900

1317.265 (1.06)n = 1900

(1.06)n = 1.4424

n= log1.4424

log1.06

= 6.3

Therefore he could withdraw the amount for 6 years.

END OF CHAPTER

2 unit ap and gp

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