Preview Extract

INSTRUCTORโS
SOLUTIONS MANUAL
BRIAN BEAUDRIE
BARBARA BOSCHMANS
Northern Arizona
University
Northern Arizona
University
to accompany
A P ROBLEM S OLVING A PPROACH
TO M ATHEMATICS
F OR E LEMENTARY S CHOOL T EACHERS
THIRTEENTH EDITION
Rick Billstein
University of Montana
Barbara Boschmans
Northern Arizona University
Shlomo Libeskind
University of Oregon
Johnny W. Lott
University of Montana
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author.
Copyright ยฉ 2020, 2016, 2013 by Pearson Education, Inc. 221 River Street, Hoboken, NJ 07030. All rights reserved.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-13-499560-1
ISBN-10: 0-13-499560-0
Contents
Chapter 1 An Introduction to Problem Solving
Chapter 2 Introduction to Logic and Sets
Chapter 3 Numeration Systems and Whole Number Operations
Chapter 4 Number Theory
Chapter 5 Integers
Chapter 6 Rational Numbers and Proportional Reasoning
Chapter 7 Decimals, Percents, and Real Numbers
Chapter 8 Algebraic Thinking
Chapter 9 Probability
Chapter 10 Data Analysis/Statistics: An Introduction
Chapter 11 Introductory Geometry
Chapter 12 Congruence and Similarity with Constructions
Chapter 13 Area, Pythagorean Theorem, and Volume
Chapter 14 Transformations
Copyright ยฉ 2020 Pearson Education, Inc.
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CHAPTER 1
AN INTRODUCTION TO PROBLEM SOLVING
List the numbers:
3 +
6 + ๏ + 297 + 300
300 + 297 + ๏ +
6 +
3
303 + 303 + ๏ + 303 + 303
Assessment 1-1A: Mathematics and
Problem Solving
1. (a)
List the numbers:
1 +
2 + ๏ +
98 +
99
99 +
98 + ๏ +
2 +
1
There are 100 sums of 303. Thus the
total can be found by computing
100โ
303 = 15,150.
2
100 + 100 + ๏ + 100 + 100
There are 99 sums of 100. Thus the total can
be found by computing 99โ
2100 = 4950.
(Another way of looking at this problem is
to realize there are 99
= 49.5 pairs of
2
sums, each of 100; thus 49.5๏ 100 = 4950.)
(b) The number of terms in any sequence of
numbers may be found by subtracting the
first term from the last, dividing the result
by the common difference between terms,
and then adding 1 (because both ends must
-1 + 1 = 501
be accounted for). Thus 1001
2
(d) The number of terms in any sequence of
numbers may be found by subtracting the
first term from the last, dividing the result
by the common difference between terms,
and then adding 1 (because both ends must
be accounted for). Thus 4004-4 + 1 = 100
terms.
List the numbers:
4 +
8 + ๏ + 396 + 400
400 + 396 + ๏ +
8 +
4
404 + 404 + ๏ + 404 + 404
terms.
List the numbers:
3 + ๏ + 999 + 1001
1001 + 999 + ๏ +
3 +
1
1002 + 1002 + ๏ + 1002 + 1002
There are 100 sums of 404. Thus the
total can be found by computing
100โ
404 = 20, 200.
2
1 +
2. (a)
There are 501 sums of 1002. Thus the total
can be found by computing
501โ
1002 = 251, 001.
2
(c) The number of terms in any sequence of
numbers may be found by subtracting the first
term from the last, dividing the result by the
common difference between terms, and then
adding 1 (because both ends must be
accounted for). Thus 3003-3 + 1 = 100
(b)
terms.
Copyright ยฉ 2020 Pearson Education, Inc.
1
2 Chapter 1: An Introduction to Problem Solving
When the stack in (a) and a stack of the
same size is placed differently next to the
original stack in (a), a rectangle containing
100 (101) blocks is created. Since each
block is represented twice, the desired sum
is 100 (101) 2 = 5050.
While the above represents a specific
example, the same thinking can be used for
any natural number n to arrive at a formula
n(n ๏ซ 1) 2.
3. There are 1471-36 + 1 = 112 terms.
List the numbers:
36 + 37 + ๏ + 146 + 147
147 + 146 + ๏ + 37 + 36
183 + 183 + ๏ + 183 + 183
There are 112 sums of 183. Thus the total can
be found by computing 1122โ
183 = 10, 248.
4. (a) Make a table as follows; there are 9 rows
so there are 9 different ways.
6-cookie
packages
2-cookie
packages
single-cookie
packages
1
1
1
0
0
0
0
0
0
2
1
0
5
4
3
2
1
0
0
2
4
0
2
4
6
8
10
(b) Make a table as follows; there are 12 rows
so there are 12 different ways.
6-cookie 2-cookie single-cookie
packages packages
packages
2
0
0
1
3
0
1
2
2
1
1
4
1
0
6
0
0
6
5
0
2
0
4
4
0
3
6
0
2
8
0
1
10
0
0
12
5. If each layer of boxes has 7 more than the
previous layer we can add powers of 7:
70 = 1 (red box)
71 = 7 (blue boxes)
72 = 49 (black boxes)
73 = 343 (yellow boxes)
74 = 2401 (gold boxes)
1 + 7 + 49 + 343 + 2401 = 2801 boxes
altogether.
6. Using strategies from Poylaโs problem solving
list identify subgoals (solve simpler problems)
and make diagrams to solve the original
problem.
1 triangle; name this the โunitโ triangle.
This triangle is made of 4 unit triangles.
Counting the large triangle there are
5 triangles
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-1A: Mathematics and Problem Solving 3
The first row sums to 17 + a + 7 = 24 + a.
So
a = 66 – 24 = 42. The last column sums to
Unit triangles 4 unit triangles 9 unit triangles
9
3
1
13 total triangles
7 + b + 27 = 34 + b. So
b = 66 – 34 = 32. The first column sums to
17 + 12 + c = 29 + c. So
c = 66 – 29 = 37. The second column sums
to 42 + 22 + d = 64 + d . So
d = 66 – 64 = 2.
Unit triangles 4 unit triangles 9 unit triangles 16 unit triangles
16
7
3
1
There are 27 triangles in the original figure.
7. Observe that E = (1 + 1) + (3 + 1) + ๏ +
(97 + 1) = O + 49. Thus, E is 49 more than O.
Alternative strategy:
๏ + 98
O + E = 1 + 2 + 3 + 4 + 5 + 6 + ๏ + 97 ๏ซ
=
11. Debbie and Amy began reading on the same
day, since 72 pages for Debbie ๏ธ 9 pages per
day = 8 days. Thus Amy is on
6 pages per day ยด 8 days = page 48.
12. The last three digits must sum to 20, so the
second to last digit must be 20 – (7 + 4) = 9.
Since the sum of the 11th , 12th, and 13th digits is
also 20, the 11th digit is 20 – (7 + 9) = 4.
98(99)
= 49(99)
2
รฆ 49(50) รถรท
E = 2(1 + 2 + 3 + 4 + ๏ + 49) = 2 รงรง
= 49(50)
รงรจ 2 รทรทรธ
O = O + E – E = 49(99) – 49(50) = 49(49).
So O is 49 less than E.
8. Bubba is last; Cory must be between Alababa
and Dandy; Dandy is faster than Cory. Listing
from fastest to slowest, the finishing order is
then Dandy, Cory, Alababa, and Bubba.
9. Make a table.
$20 bills
$10 bills
$5 bills
2
2
1
1
1
1
0
0
0
0
0
0
1
0
3
2
1
0
5
4
3
2
1
0
0
2
0
2
4
6
0
2
4
6
8
10
There are twelve rows so there are twelve
different ways.
10. The diagonal from the left, top corner to the
right, bottom corner sums to
17 + 22 + 27 = 66.
We can continue in this fashion until we find
that A is 9, or we can observe the repeating
pattern from back to front, 4, 9, 7, 4, 9,
7, โฆ and discover that A is 9.
13. Choose the box labeled Oranges and Apples
(Box B). Retrieve a fruit from Box B. Since
Box B is mislabeled, Box B should be labeled
as having the fruit you retrieved. For example,
if you retrieved an apple, then Box B should be
labeled Apples. Since Box A is mislabeled, the
Oranges and Apples label should be placed on
Box A. These leave only one possibility for Box
C; it should be labeled Oranges. If an orange
was retrieved from Box B, then Box C would be
labeled Oranges and Apples and Box A should
be labeled Apples.
14. The electrician made $1315 for 4 days at $50
per hour. She spent $15 per day on gasoline so 4
โข $15 = $60 on gasoline. The total is then $1315
+ $60 = $ 1375. At $50 per hour, she worked
1375
๏ฝ 27.5 hours.
50
15. Working backward: Top ๏ญ 6 rungs ๏ญ 7 rungs + 5
rungs ๏ญ 3 rungs = top ๏ญ 11 rungs, which is
located at the middle. From the middle rung
travel up 11 to the top or down 11 to the
bottom. Along with the starting rung, then, there
are 11 + 11 + 1 = 23 rungs.
16. There are several different ways to solve this.
One is to use a variable. So, let a be equal to the
number of apple pies that are baked. This means
the number of cherry pies that are baked is
Copyright ยฉ 2020 Pearson Education, Inc.
4 Chapter 1: An Introduction to Problem Solving
represented by 4 โ a. So, using 9 slices for an
apple pie and 7 slices for a cherry pie, we get:
9a ๏ซ 7(4 ๏ญ a) ๏ฝ 34
9a ๏ซ 28 ๏ญ 7a ๏ฝ 34
2a ๏ฝ 6
a๏ฝ3
So the number of apple pies is 3. Therefore, the
number of cherry pies is 1.
Since the number of pies is small, another
solution strategy is to make a table of all
possible cases:
Apple Cherry Apple Cherry
Total
Pies
Pies
slices slices
slices
0
4
0
28
28
1
3
9
21
30
2
2
18
14
32
3
1
27
7
34
4
0
36
0
36
From the table, we can see there are 34 slices
when there are three apple pies and one cherry
pie.
17. Al made $50. Examine it step by step. First, Al
spent $100 on the CD player. At this point, he is
down $100. Once he sold it for $125, he is now
$25 ahead. When he later bought it back for
$150, he was down $125; but after selling it
again for $175, he is now ahead $50.
18. Since the bat is $49 more than the ball, and the
total spent is $50, we can use the guess and
check method to solve. The table below
represents possible guesses:
Cost of bat Cost of ball sum
$49.00
$0.00
$49.00
$49.25
$0.25
$49.50
$49.50
$0.50
$50.00
Another method would be to use a variable. Let
the cost of the ball (in dollars) be the variable b.
The cost of a bat in dollars, therefore, would be
b + 49. Together the two costs must add up to
50, so:
b + (b + 49) = 50
2b + 49 = 50
2b = 1
b = ยฝ. In terms of money, b = $0.50, 50 cents.
Since the ball costs 50 cents, the bat must cost
$49.50, and the sum of the price of bat and ball
does equal $50.
Assessment 1-1B
1. (a) List the numbers:
1 + 2 + ๏ + 48
49 + 48 + ๏ + 2
50 + 50 + ๏ + 50
+
+
+
49
1
50
There are 49 sums of 50. Thus the total can
be found by computing 492โ
50 = 1225.
(Another way of looking at this problem is
= 24.5 pairs of
to realize there are 49
2
sums, each of 50; thus 24.5 โ
50 = 1225.)
(b) The number of terms in any sequence
of numbers may be found by subtracting
the first term from the last, dividing the
result by the common difference between
terms, and then adding 1 (because both
ends must be accounted for). Thus
2009 – 1
+ 1 = 1005 terms.
2
List the numbers:
1 +
3 + ๏ + 2007
2009 + 2007 + ๏ +
3
2010 + 2010 + ๏ + 2010
+
+
+
2009
1
2010
There are 1005 sums of 2010. Thus the
total can be found by computing
1005 โ
2010
= 1, 010, 025.
2
(c) The number of terms in any sequence
of numbers may be found by subtracting
the first term from the last, dividing the
result by the common difference between
terms, and then adding 1 (because both
ends must be accounted for). Thus
600 – 6
+ 1 = 100 terms.
6
List the numbers:
6 + 12 + ๏ + 594
600 + 594 + ๏ + 12
606 + 606 + ๏ + 606
+
+
+
600
6
606
There are 100 sums of 606. Thus the total
can be found by computing
606โ
100 = 30, 300.
2
(d) The number of terms in any sequence
of numbers may be found by subtracting
the first term from the last (or the last from
the first if the first is greater than the last),
dividing the result by the common
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-1B 5
difference between terms, and then adding
1 (because both ends must be accounted
for). Thus
1000 – 5
+ 1 = 200 terms.
5
List the numbers:
1000 + 995 + ๏ +
10
5 +
10 + ๏ + 995
1005 + 1005 + ๏ + 1005
+
5
+ 1000
+ 1005
There are 200 sums of 1005. Thus the
total can be found by computing
1005 โ
200
= 100, 500.
2
2. (a) The diagram illustrates how the numbers
can be paired to form 50 sums of 101. The
sum of the first 100 natural numbers is
50(101) = 5050.
(b) A diagram similar to the one in 2a would
illustrate how the numbers can be paired to
form 100 sums of 202. Because there are
an odd number of terms, the middle term,
101, is left unpaired. So, the sum of the
first 201 natural numbers is 100 โข 202 +
101 = 20,301.
3. There are
203 – 58
+ 1 = 146 terms. terms
1
List the numbers:
58 + 59 + ๏ + 202
203 + 202 + ๏ + 59
261 + 261 + ๏ + 261
+
+
+
203
58
261
1
1
9
2
1
8
3
1
7
4
1
6
5
1
5
1
2
8
5. Since there are two different color socks in the
drawer, drawing only two socks does not
guarantee finding a matching pair; you could
get one sock of each color. However, once you
draw a third sock, you are guaranteed to have a
matching pair, since the third sock must match
one or the other of the previous two socks.
6. There are 13 squares of one unit each; 4 squares
of four units each; and one square 9 units; for a
total of 18 squares.
7.
P = 1 + 3 + 5 + 7 + … + 99
Q =
5 + 7 + … + 99 + 101
Q – P = (5 + … + 99 + 101) (1 + 3 + 5 + … + 99)
= (101) – (1 + 3)
= 97
Q is larger than P by 97.
8. A diagram will help.
There are 146 sums of 261. Thus the total can
be found by computing
146 โ
261
= 19, 053.
2
4. There are many answers to this problem. A
systematic list is a good approach. Using only
two numbers and addition, 5 rows give 5
different ways.
One number
Eleven minus the number
1
10
2
9
3
8
4
7
5
6
We can view 6 + 5 as a different way than 5 + 6
and continue in this manner to find 10 different
ways. Or, we can find 7 more ways
using three numbers as follows.
The next step is 120 miles + 40 miles = 160
miles from Missoula.
9. (a) Marc must have five pennies to make an
even $1.00. The minimum number of coins
would have as many quarters as possible,
or three quarters. The remaining 20/c must
consist of at least one dime and one nickel;
the only possibility is one dime and two
nickels. The minimum is 5 pennies, 2
nickels, 1 dime, and 3 quarters, or 11 coins.
(b) The maximum number of coins is achieved
by having as many pennies as possible. It is
a requirement to have one quarter, one
dime, and one nickel = 40/c, so there may
then be 60 pennies for a total of 63 coins.
Copyright ยฉ 2020 Pearson Education, Inc.
6 Chapter 1: An Introduction to Problem Solving
10. Adding all the numbers gives 99. This means
that each row, diagonal, and column must add to
99 ๏ธ 3 = 33. Write 33 as a sum of the numbers
in all possible ways:
19 + 11 + 3
19 + 9 + 5
17 + 13 + 3
17 + 11 + 5
17 + 9 + 7
15 + 13 + 5
15 + 11 + 7
13 + 11 + 9
Summarizing the pattern:
Number
Nr. sums with number
3
2
5
3
7
2
9
3
11
4
13
3
15
2
17
3
19
2
(ii) If one side is heavier, take two of the
three marbles and weigh them. If they
are the same weight, the remaining
marble is the heavier. If not, the
heavier will be evident on this second
weighing.
12. (a) There are:
1 partridge ยด 12 days = 12 gifts;
2 doves ยด 11 days = 22 gifts;
3 hens ยด 10 days = 30 gifts;
4 birds ยด 9 days = 36 gifts;
5 rings ยด 8 days = 40 gifts;
6 geese ยด 7 days = 42 gifts;
7 swans ยด 6 days = 42 gifts;
8 maids ยด 5 days = 40 gifts;
9 ladies ยด 4 days = 36 gifts;
10 lords ยด 3 days = 30 gifts;
11 pipers ยด 2 days = 22 gifts; and
12 drummers ยด 1 day = 12 gifts.
So the gifts given the most by your true
love was 42 geese and 42 swans.
(b) 12 + 22 + ๏ + 22 + 12 = 364 gifts
total.
Thus 11 must be in the center of the square
and 5, 9, 13, and 17 must be in the corners.
One solution would be:
17
7
9
3
11
19
13 15 5
11. Answers may vary; two solutions might be to:
(a) Put four marbles on each tray of the
balance scale. Take the heavier four and
weigh two on each tray. Take the heavier
two and weigh one on each tray; the
heavier marble will be evident on this third
weighing.
(b) This alternative shows the heavier marble
can be found more efficiently, two steps
rather than three. Put three marbles in each
tray of the balance scale.
(i) If the two trays are the same weight,
the heavier marble is one of the
remaining two. Weigh them to find the
heavier.
13. (a) There must be 1 or 3 quarters for an
amount ending in 5. Then dimes can add to
$1.15 plus 4 pennies to realize $1.19. Thus:
Quarters
Dimes
Pennies
Total
3
4
4
$1.19
1
9
4
$1.19
and in neither case can change for $1.00 be
made.
(b) Two or zero quarters would allow an
amount ending in 0. Then more
combinations of dimes or pennies could
add to $1.00.
14. If the price of 15 sandwiches equals the price
20 = 4
of 20 salads, each sandwich will buy 15
3
( )
salads. Thus 3 sandwiches = 3 43 = 4 salads.
15. Use a variable and a table
12 AM
T
5 AM
9 AM
12 PM
T – 15 2(T – 15) 2(T – 15) + 10
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-2A: Explorations with Patterns 7
2(T – 15) + 10 = 32
So,
2T – 30 + 10 = 32
2T – 20 = 32
2T = 52
T = 26 degrees.
16. One way to solve this problem is to write an
equation using a variable. For example, let p
equal the number of puzzles Seth bought. So the
number of trucks Seth bought would be 5 โ p.
Since each puzzle cost $9 and each truck cost
$5, we get
9 p ๏ซ 5(5 ๏ญ p ) ๏ฝ 33
9 p ๏ซ 25 ๏ญ 5 p ๏ฝ 33
4p ๏ฝ 8
p๏ฝ2
So Seth bought 2 puzzles. Since he bought 5
gifts all together, he also bought 3 trucks.
Another way to solve it is to guess and check,
making a table to keep track of results
Puzzles
Trucks
Cost of
puzzles
Cost of
Trucks
Total
cost
0
5
$0
$25
$25
1
4
$9
$20
$29
2
3
$18
$15
$33
3
2
$27
$10
$37
The table shows that by buying 2 puzzles and 3
trucks, Seth will spend $33.
17. The first line tells us that hamburgers equal $10.
Using that information, we can use the second
line to figure out 4 hot dogs cost $8, so each hot
dog costs $2. Going to the third line, since two
hot dogs cost $4, each drink must cost $1. So on
the fourth line, one drink and one hamburger
cost $11. The value of the question mark is
$11.
18. This is difficult to visualize, so the strategy of
examining a simpler case and looking for a
pattern might be the best method to solve this
problem. It would also help if you are able to
make a model of the shapes and act out the
situation.
Imagine a 3 ยด 3 ยด 3 large cube. It would be
made up of a total of 27 small cubes. Taking off
one layer all around would involve taking off
the top layer, the bottom layer, the front, the
back, the left side, and the right side. Doing this
one step at a time: taking off the top layer takes
away 9 cubes. The same if you take off the
bottom layer, so youโve removed 18 cubes so
far. Taking off the front and the back would
remove 3 more cubes each (the other 6 cubes on
those faces were removed when the top and
bottom were removed). Finally, taking off the
left and right sides removes one more cube
each. In total, 26 cubes are removed, leaving
only 1 cube.
Now, imagine a 4 ยด 4 ยด 4 cube. There would
be a total of 64 small cubes making it.
Removing the top and bottom layers takes a
total of 32 cubes away; removing the front and
the back would take away 16 more cubes;
finally, removing the left and right sides takes
away 8 more cubes, leaving only 8 cubes.
So, when starting with a 3 ยด 3 ยด 3 large cube,
we are left with 1 small cube (1 ยด 1 ยด 1 = 1);
when starting with a 4 ยด 4 ยด 4 large cube, we
are left with 8 small cubes
(2 ยด 2 ยด 2 = 8). Therefore, the pattern seems
to be, when given large cube of
dimension n ยด n ยด n, to find the number of
small cubes that make it after removing one
layer of small cubes all around the larger cube,
you would take
(n – 2) ยด (n – 2) ยด (n – 2). So for a
10 ยด 10 ยด 10 large cube, the number of small
cubes left would be 8 ยด 8 ยด 8 = 512 small
cubes.
Assessment 1-2A: Explorations
with Patterns
1. (a) Each figure in the sequence adds one
box each to the top and bottom rows.
The next would be:
(b) Each figure in the sequence adds one
upright and one inverted triangle. The next
would be:
(c) Each figure in the sequence adds one box
to the base and one row to the overall
triangle. The next would be:
Copyright ยฉ 2020 Pearson Education, Inc.
8 Chapter 1: An Introduction to Problem Solving
2. (a) Terms that continue a pattern are 17, 21,
25, , โฆ . This is an arithmetic sequence
because each successive term is obtained
from the previous term by addition of 4.
(b) Terms that continue a pattern are 220, 270,
320, โฆ . This is arithmetic because each
successive term is obtained from the
previous term by addition of 50.
(c) Terms that continue a pattern are 27, 81,
243, โฆ . This is geometric because each
successive term is obtained from the
previous term by multiplying by 3.
(d) Terms that continue a pattern are
109 ,1011 ,1013 , โฆ . This is geometric
because each successive term is obtained
from the previous term by multiplying by
102.
(e) Terms that continue a pattern are
193 + 10ร 230 ,193 + 11ร 230 ,193 + 12ร 230 ,
โฆ . This is arithmetic because each
successive term is obtained from the
previous term by addition of 230 .
3. In these problems, let an represent the nth term
in a sequence, a1 represent the first term,
d represent the common difference between
terms in an arithmetic sequence, and r represent
the common ratio between terms in a geometric
sequence. In an arithmetic sequence,
an = a1 + (n – 1)d ; in a geometric sequence
(d) Geometric sequence:
a1 = 10 and r = 102:
(i ) a100 = 10 โ
(102 )(100-1) = 10 โ
(102 )99
= 10 โ
10198 = 10199.
(ii ) an = 10 โ
(102 )(n-1)
= 10 โ
10(2n-2) = 102n-1.
(e) Arithmetic sequence:
a1 = 193 + 7 โ
230 and d = 230:
(i )
a100 = 193 + 7 โ
230 + (100 – 1) โ
230
= 193 + 7 โ
230 + 99 โ
230
= 193 + 106 ร 230.
(ii ) an = 193 + 7 โ
230 + (n – 1) โ
230
= 193 + (n + 6) ร 230.
4. 2, 7, 12, โฆ . Each term is the 5th number on
a clock face (clockwise) from, the preceding
term.
5. (a) Make a table.
Number of term
Term
1
2
1โ
1โ
1 = 1
2โ
2โ
2 = 8
3
3 โ
3 โ
3 = 27
4
4 โ
4 โ
4 = 64
5
5 โ
5 โ
5 = 125
6
6 โ
6 โ
6 = 216
7
7 โ
7 โ
7 = 343
an = a1r n-1. Thus:
(a) Arithmetic sequence: a1 = 1 and d = 4:
(i ) a100 = 1 + (100 – 1) โ
4
= 1 + 99 โ
4 = 397.
(ii ) an = 1 + (n – 1) โ
4
= 1 + 4n – 4 = 4n – 3.
(b) Arithmetic sequence:
a1 = 70 and d = 50:
(i ) a100 = 70 + (100 – 1) โ
50
= 70 + 99 โ
50 = 5020.
(ii ) an = 70 + (n – 1) โ
50
= 70 + 50n – 50 or 50n + 20.
(c) Geometric sequence: a1 = 1 and r = 3:
(i ) a100 = 1 โ
3100-1 = 399.
n -1
(ii ) an = 1 โ
3
n-1
= 3
.
8
8 โ
8 โ
8 = 512
9
9 โ
9 โ
9 = 729
10
10 โ
10 โ
10 = 1000
11
11 โ
11 โ
11 = 1331
The 11th term 1331 is the least 4-digit
number greater than 1000.
(b) The 9th term 729 is the greatest 3-digit
number in this pattern.
(c) 104 = 10,000; The greatest number less
than 104 is 21 ๏ 21 ๏ 21 ๏ฝ 9261.
(d) The cell A14 corresponds to the 14th term,
which is 14 โ
14 โ
14 = 2744.
6. (a) The number of matchstick squares in each
windmill form an arithmetic sequence with
a1 = 5 and d = 4. The number of
matchstick squares required to build the
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-2A: Explorations with Patterns 9
10th windmill is thus 5 + (10 – 1) โ
4 =
5 + 9 โ
4 = 41 squares.
(b) The nth windmill would require
5 + (n – 1) โ
4 = 5 + 4n – 4 = 4n + 1
squares.
(c) There are 16 matchsticks in the original
windmill. Each additional windmill adds
12 matchsticks.
This is an arithmetic sequence with
a1 = 16 and d = 12, so an = 16 +
(n – 1) โ
12 = 12n + 4 matchsticks.
7. (a) Each cube adds four squares to the
preceding figure; or 6, 10, 14, โฆ . This is
an arithmetic sequence with a1 = 6 and
d = 4. Thus a15 = 6 + (15 – 1) โ
4 =
62 squares to be painted in the 10th figure.
(b) This is an arithmetic sequence with
a1 = 6 and d = 6. The nth term is thus:
an = 6 + (n – 1) โ
4 = 4n + 2.
8. Since the first year begins with 700 students,
after the first year there would be 760, after
the second there would be 820, โฆ , and after the
twelfth year the number of students would
be the 13th term in the sequence.
This then is an arithmetic sequence with
a1 = 700 and d = 60, so the 13th term
(current enrollment + twelve more years) is:
700 + (13 – 1) โ
60 = 1420 students.
9. Using the general expression for the nth term of
an arithmetic sequence with a1 = 24, 000 and
a9 = 31, 680 yields:
31680 = 24000 + (9 – 1)d
31680 = 24000 + 8d ๏ d = 960,
the amount by which Juanโs income increased
each year.
To find the year in which his income was
$45120:
45120 = 24000 + (n – 1) โ
960
45120 = 23040 + 960n
๏ n = 23.
Juanโs income was $45,120 in his 23rd year.
10. The number that fits into the last triangle is 8.
The numbers inside the triangle are found by
multiplying the number at the top of the triangle
by the number at the bottom left of the triangle;
then subtracting from that the number at the
bottom right of the triangle. So,
2 ยด 5 – 2 = 8.
11. (a) To build an up-down-up staircase with 3 steps
up and 3 steps down, each current step will
have a block placed on it; plus a block will be
added to each end. In total, there will be five
more blocks, for a total of 9 blocks used. To
build an up-down-up staircase with 4 steps up
and 4 steps down, each step from the previous
iteration will have a block placed on it (5
blocks) plus a block at each end (2 blocks),
adding seven total blocks, bringing the total
to 9 + 7 = 16 blocks. The table below
illustrates the pattern:
Steps
up/down
Blocks
added
Total
blocks
1
2
3
4
0
3
5
7
1
4
9
16
Therefore, to build an up-down-up staircase
with 5 steps up and 5 steps down, 9 blocks
need to be added to the previous 16 blocks
to arrive at a total of 25 blocks.
(b) Based on the table and answer above, the
total blocks are always the square of the
number of steps up and down. So, if the
number of steps up and down are n, then
the total number of blocks will be n2.
12. (a) Using the general expression for the nth
term of an arithmetic sequence with
a1 = 51, an = 251, and d = 1 yields:
251 = 51 + (n – 1) โ
1 ๏
251 = 50 + n ๏ n = 201.
There are 201 terms in the sequence.
(b) Using the general expression for the nth
term of a geometric sequence with a1 = 1,
an = 260 , and r = 2 yields:
260 = 1(2) n-1 = 2n-1
๏ n – 1 = 60 ๏ n = 61.
There are 61 terms in the sequence.
(c) Using the general expression for the nth term
of an arithmetic sequence with a1 = 10,
an = 2000, and d = 10 yields:
2000 = 10 + (n – 1) โ
10 ๏
2000 = 10n ๏ n = 200.
There are 200 terms in the sequence.
Copyright ยฉ 2020 Pearson Education, Inc.
10 Chapter 1: An Introduction to Problem Solving
(d) Using the general expression for the nth
term of a geometric sequence with a1 = 1,
an = 1024, and r = 2 yields:
1024 = 1(2)n-1 ๏
210 = 2n-1 ๏ n – 1 = 10
๏ n = 11.
There are 11 terms in the sequence.
13. (a) First term:
(1) 2 + 2 = 3;
Second term:
(2)2 + 2 = 6;
Third term:
(3) 2 + 2 = 11;
Fourth term:
(4)2 + 2 = 18; and
Fifth term:
2
(5) + 2 = 27.
(b) First term:
Second term:
Third term:
Fourth term:
Fifth term:
5(1) + 1 = 6;
5(2) + 1 = 11;
5(3) + 1 = 16;
5(4) + 1 = 21; and
5(5) + 1 = 26.
(c)
10(1) – 1 = 9;
First term:
Second term:
10
(2)
– 1 = 99;
(3)
– 1 = 999;
Third term:
10
Fourth term:
10(4) – 1 = 9999; and
Fifth term:
10(5) – 1 = 99999.
(d) First term:
Second term:
Third term:
Fourth term:
Fifth term:
3(1) – 2 = 1;
3(2) – 2 = 4;
3(3) – 2 = 7;
3(4) – 2 = 10; and
3(5) – 2 = 13.
14. Answers may vary; examples are:
5
(a) If n = 5, then 5+
= 2 ยน 5 + 1 = 6.
5
(b) If n = 2, then (2 + 4)2 = 62 = 36
does not equal 22 + 42 = 20.
15. (a) There are 1, 5, 11, 19, 29 tiles in the five
figures. Each figure adds 2n tiles to the
preceding figure, thus a6 the 6th term has
29 + 12 = 41 tiles.
(b) n 2 = 1, 4,9, 16, 25, โฆ . Adding (n – 1)
to n2 yields 1, 5, 11, 19, 29, โฆ , which is the
proper sequence. Thus the nth term has
n2 + n – 1.
(c) If n 2 + (n – 1) = 1259;
Then n 2 + n – 1260 = 0. This implies
(n – 35)(n + 36) = 0, so n = 35.
There are 1259 tiles in the 35th figure.
16. The nth term of the arithmetic sequence is
200 + n (200). The sequence can also be
generated by adding 200 to the previous term.
The nth term of the geometric sequence is 2n.
The sequence can also be generated by
multiplying the previous term by 2. Make a
table.
Number of
the term
7
8
9
10
11
12
Arithmetic
term
1600
1800
2000
2200
2400
2600
Geometric
term
128
256
512
1024
2048
4096
With the 12th term, the geometric sequence is
greater.
17. (a) Start with one piece of paper. Cutting it
into five pieces gives us 5. Taking each of
the pieces and cutting it into five pieces
again gives 5 โ
5 = 25 pieces. Continuing
this process gives a geometric sequence: 1,
5, 25, 125, โฆ . After the 5th cut there are
55 = 3125 pieces of paper.
(b) The number of pieces after the nth cut
would be 5n.
18. (a) For an arithmetic sequence there is a
common difference between the terms.
Between 39 and 69 there are three
differences so we can find the common
difference by subtracting 39 from 69 and
dividing the answer by three:
69 ๏ญ 39 ๏ฝ 30 and 30 ๏ธ 3 ๏ฝ 10. The common
difference is 10 and we can find the
missing terms: 39 โ 10 = 29 and 39 + 10 =
49 and 49 + 10 = 59.
(b) For an arithmetic sequence there is a
common difference between the terms.
Between 200 and 800 there are three
differences so we can find the common
difference by subtracting 200 from 800 and
dividing the answer by three:
800 ๏ญ 200 ๏ฝ 600 and 600 ๏ธ 3 ๏ฝ 200. The
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-2B 11
common difference is 200 and we can find
the missing terms: 200 โ 200 = 0 and 200 +
200 = 400 and 400 + 200 = 600.
(c) For a geometric sequence there is a
common ration between the terms.
Between 54 and 510 there are three common
ratios used so we can find the common
ratio by dividing 510 by 54 and then taking
the cube root:
๏จ1๏ฉ
510 ๏ธ 54 ๏ฝ 56 and (56 ) 3 ๏ฝ 52. The
common ratio is 52 and we can find the
missing terms:
54 ๏ธ 52 ๏ฝ 52 ,54 ๏ 52 ๏ฝ 56 ,56 ๏ 52 ๏ฝ 58.
19. (a) Letโs call the missing terms a, b, c, d, e and
f, then the sequence becomes:
a, b,1,1, c, d , e, f
b ๏ซ1 ๏ฝ 1 ๏ฎ b ๏ฝ 0
a ๏ซb ๏ฝ1๏ฎ a ๏ซ0 ๏ฝ1๏ฎ a ๏ฝ1
1๏ซ1 ๏ฝ c ๏ฎ c ๏ฝ 2
1๏ซ c ๏ฝ d ๏ฎ 1๏ซ 2 ๏ฝ d ๏ฎ d ๏ฝ 3
c๏ซd ๏ฝ e ๏ฎ 2๏ซ3 ๏ฝ e ๏ฎ e ๏ฝ 5
d ๏ซ e ๏ฝ f ๏ฎ 3 ๏ซ 5 ๏ฝ f ๏ฎ๏ฝ 8.
The missing terms are 1, 0, 2, 3, 5, and 8.
(b) Letโs call the missing terms a, b, c, and d,
then the sequence becomes:
a, b, c,10,13, d ,36,59
c ๏ซ 10 ๏ฝ 13 ๏ฎ c ๏ฝ 3
b ๏ซ c ๏ฝ 10 ๏ฎ b ๏ซ 3 ๏ฝ 10 ๏ฎ b ๏ฝ 7
๏ญ
a ๏ซ b ๏ฝ c ๏ฎ a ๏ซ 7 ๏ฝ 3 ๏ฎ๏ฝ 4
10 ๏ซ 13 ๏ฝ d ๏ฎ d ๏ฝ 23
The missing terms are -4, 7, 3, and 23.
(c) If a Fibonacci-type sequence is a sequence
in which the first two terms are arbitrary
and in which every term starting from the
third is the sum of the previous two terms,
then we can add 0 and 2 to get the third
term and continue the pattern:
0๏ซ2 ๏ฝ 2
2๏ซ2 ๏ฝ 4
2๏ซ4 ๏ฝ 6
4 ๏ซ 6 ๏ฝ 10
6 ๏ซ 10 ๏ฝ 16
10 ๏ซ 16 ๏ฝ 26
The missing terms are 2, 4, 6, 10, 16,
and 26.
20. (a)
Year 1
Year 2
Year 3
Year 4
Year 5
80 ๏ซ .05(80) ๏ฝ 84
84 ๏ซ .05(84) ๏ฝ 88.2
88.2 ๏ซ .05(88.2) ๏ฝ 92.61
92.61 ๏ซ .05(92.61) ๏ฝ 97.2405
97.2405 ๏ซ .05(97.2405) ๏ฝ 102.102525
๏ป $102.10.
(b) This is a geometric sequence with
a1 ๏ฝ 80 and r = 1.05, so the price after n
years is 80 โข 1.05n.
Assessment 1-2B
1. (a) In a clockwise direction, the shaded area
moves to a new position separated from the
original by one open space, then two open
spaces, then by three, etc. The separation in
each successive step increases by one unit;
next would be:
(b) Each figure in the sequence adds one row
of boxes to the base. Next would be:
(c) Each figure in the sequence adds one box
to the top and each leg of the figure. Next
would be:
2. (a) Terms that continue a pattern are 18, 22,
26, โฆ . This is an arithmetic sequence
because each successive term is obtained
from the previous term by addition of 4.
(b) Terms that continue a pattern are 39, 52,
65, โฆ . This is an arithmetic sequence
because each successive term is obtained
from the previous term by addition of 13.
(c) Terms that continue a pattern are 44, 45,
46, โฆ . This is a geometric sequence
because each successive term is obtained
from the previous term by multiplying by
4.
Copyright ยฉ 2020 Pearson Education, Inc.
12 Chapter 1: An Introduction to Problem Solving
(d) Terms that continue a pattern are 214, 218,
222, โฆ . This is a geometric sequence
because each successive term is obtained
from the previous term by multiplying by
24.
(e) Terms that continue a pattern are
100 + 10 ร 250 ,100 + 12 ร 250 ,100 + 14 ร 250 ,….
This is an arithmetic sequence because
each successive term is obtained from the
previous term by adding by 2โข250.
3. In these problems, an represents the nth
term in a sequence, a1 represents the first term,
d represent the common difference between
terms in an arithmetic sequence, and r
represents the common ratio between terms in a
geometric sequence.
In an arithmetic sequence,
an = a1 + (n – 1) d ; in a geometric sequence,
an = a1r n-1. Thus:
(a) Arithmetic sequence: a1 = 2 and d = 4.
(i)
a100 = 2 + (100 – 1) โ
4 = 398.
(ii) an = 2 + (n – 1) โ
4
(i)
a100 = 100 + 4 โ
250 + (100 – 1) โ
251
= 100 + 2 โ
251 + 99 โ
251
= 100 + 101 โ
251
= 100 + 101 โ
2 โ
250
= 100 + 202 ร 250.
(ii) an = 100 + 4 โ
250 + (n – 1) โ
251
= 100 + 2 โ
251 + (n – 1) โ
251
= 100 + (n + 1) โ
251
= 100 + (n + 1) โ
2 โ
250
= 100 + 2(n + 1)250.
4. The hands must move 8 hours to move from 1
to 9 on the clock face. To move from 9 to 5, the
hand must move 8 hours also. To move from 5
to 1, the hand must move another 8 hours. If we
add 8 hours to 1 oโclock, we will land on the 9.
This pattern will continue, so the next three
terms are 9, 5, 1.
= 2 + 4n – 4 = 4n – 2.
(b) Arithmetic sequence: a1 = 0 and
d = 13.
(i)
a100 = 0 + (100 – 1) โ
13 = 1287.
(ii) an = 0 + (n – 1) โ
13
= 13n – 13.
(c) Geometric sequence: a1 = 4 and r = 4.
(i)
5. (a) Answers may vary: two possible answers
are:
(i)
The sum of the first n odd numbers is
a100 = 4 โ
499 = 4100.
n2; e.g., 1 + 3 + 5 + 7 = 42.
(ii) Square the average of the first and
last terms; e.g.,
(ii) an = 4 โ
4n-1 = 4n.
(d) Geometric sequence: a1 = 22 and
1+3+5+ 7 =
r = 24.
(i)
a100 = 22 โ
(24 )99 = 22 โ
2396 = 2398.
4 (n -1)
2
(ii) an = 2 โ
(2 )
a1 = 100 + 4 โ
2
1+ 7
2
2
35 – 1
+ 1 = 18 terms in this
2
sequence.
(i) 1 + 3 + 5 + 7 + ๏ + 35 = 182
= 22 โ
24n-4 = 24n- 2.
(e) Arithmetic sequence:
50
(b) There are
2
( ) =4.
= 324.
50
and d = 2 โ
2
51
= 2 :
(ii)
(
1 + 35
2
2
) = 18 = 324.
2
6. (a) Note that 5 toothpicks are added to form
each succeeding hexagon. This is an
arithmetic sequence a1 = 6 and d = 5,
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 1-2B 13
so a10 = 6 + (10 – 1) โ
5 =
6 + 9 โ
5 = 51 toothpicks.
(b) n hexagons would require
6 + (n – 1) โ
5 = 6 + 5n – 5 = 5n + 1
toothpicks.
7. (a) Looking at the second figure, there are 3 +
1 = 4 triangles. In the third figure, there are
5 + 3 + 1 = 9 triangles. The fourth
figure would then have
7 + 5 + 3 + 1 = 16 triangles. An
alternative to simply adding 7, 5, 3, and 1
together is to note that 7 + 1 = 8 and
5 + 3 = 8. There are 42 = 2 of these
sums, and 2 โ
8 = 16. Then the 100th
figure would have 100 + 99 = 199
triangles in the base, 99 + 98 = 197
triangles in the second row, and so on until
the 100th row where there would be 1
triangle. 199 + 1 = 200;
197 + 3 = 200; etc. and so the sum of
each pair is 200 and there are 100
= 50 of
2
these pairs. 50 โ
200 = 10, 000, or 10,000
triangles in the 100th figure.
(b) The number of triangles in the nth figure is
n (number of triangles in base + 1). The
2
number of triangles in the base is
n + (n – 1), or 2n – 1. (2n – 1) +
1 = 2n. Then n2 (2n) = n 2 , or n2
triangles in the nth figure.
8. This is a geometric sequence with
15,360
and r = 12 . The nth term of a
a1 =
2
geometric sequence is an = a1r n-1; thus the
10
( ) = 15 liters.
10th term would be 15360 12
Note the progression of terms in the following
table:
After
Day
Amount of Water Remaining
1
15,360 โ
12 = 7680 liters
2
7680 โ
12 = 3840 liters
๏
๏
9
60 โ
12 = 30 liters
10
30 โ
12 = 15 liters
9. This is an arithmetic sequence with a1 = 8 16
of an hour)
(i.e., 8 a.m. plus 10 minutes, or 10
60
and d = 56 (or 50
of an hour). Thus
60
a8 = 8 16 + (8 – 1) โ
56 = 14, or 2:00 p.m.
(14 is 2:00 p.m. on a 24-hour clock.)
10. Answers will be a rotation of the following
figure:
11. (a) In the first drawing, there are 6 toothpicks;
in the second drawing, there are 10, and the
third drawing has 14 toothpicks. This it is
an arithmetic sequence with
a1 = 6, d = 4. So, for the tenth figure, we
a10 = 6 + (10 – 1) โ
4
would have a10 = 6 + 9 โ
4
a10 = 42.
(b) The nth term for this arithmetic sequence is
an = 6 + (n – 1) โ
4
an = 6 + 4n – 4
an = 4n + 2.
(c) For a total of 102 toothpicks, to find the
figure, we have
102 = 4n + 2
100 = 4n
25 = n.
12. (a) The nth term for this geometric sequence is
3n-1. Thus 399 = 3n-1.
So 99 = n – 1, and n = 100.
There are 100 terms in the sequence.
(b) The nth term for this arithmetic sequence
is 9 + (n – 1) โ
4. Thus 353 = 9 +
(n – 1) โ
4. Solving for n, n = 87. There
are 87 terms in the sequence.
(c) The nth term for this arithmetic sequence is
38 + (n – 1) โ
1. Thus 238 = 38 +
(n – 1) โ
1. Solving for n, n = 201.
There are 201 terms in the sequence.
Copyright ยฉ 2020 Pearson Education, Inc.
14 Chapter 1: An Introduction to Problem Solving
13. (a)
First term:
5(1) – 1 = 4
Second term:
5(2) – 1 = 9
Third term:
5(3) – 1 = 14
Fourth term:
multiplying the previous term by 3. Make a
table
5(4) – 1 = 19
Number of
the term
Arithmetic
term
Geometric
term
Fifth term:
5(5) – 1 = 24
7
2000
729
(b) First term:
Second term:
Third term:
Fourth term:
Fifth term:
6(1) – 2 = 4
8
2300
2187
6(2) – 2 = 10
9
2600
6561
6(3) – 2 = 16
(c)
First term:
5โ
1+1 = 6
Second term:
5 โ
2 + 1 = 11
Third term:
(d)
6(4) – 2 = 22
6(5) – 2 = 28
th
With the 9 term, the geometric sequence is
greater.
17. Use a table of Fibonacci numbers to find the
pattern, Fn is the nth Fibonacci number:
5 โ
3 + 1 = 16
Generation
Male
Female
Number in
Generation
Total
Fourth term:
5 โ
4 + 1 = 21
1
1
0
1
1
Fifth term:
5 โ
5 + 1 = 26
2
0
1
1
2
First term:
12 – 1 = 0
3
1
1
2
4
Second term:
22 – 1 = 3
4
1
2
3
7
32 – 1 = 8
5
2
3
5
12
Third term:
6
3
5
8
20
Fourth term:
2
4 – 1 = 15
52 – 1 = 24
๏
๏
๏
๏
๏
Fifth term:
n
Fnโ2
Fnโ1
Fn
Fn+ 2 โ1
14. Answers may vary; examples are:
(a) If n = 6, then
3+6
= 3 ยน 6.
3
(b) If n = 4, then
(4 – 2) 2 = 4 ยน 42 – 22 = 12.
15. (a) The first figure has 2 tiles, the second has 5
tiles, the third has 8 tiles, โฆ . This is an
arithmetic sequence where the nth term is
2 + (n – 1) โ
3.
Thus the 7th term has 2 + (7 – 1) โ
3 = 20
tiles.
(b) The nth term is 2 + (n – 1) โ
3 = 2 +
3n – 3 = 3n – 1.
(c) The question can be written as: Is there
an n such that 3n – 1 = 449. Since
3n – 1 = 449 ๏ 3n = 450 ๏ n = 150,
the answer is yes, the 150th figure.
16. The nth term of the arithmetic sequence is
-100 + n(300). The sequence can also be
generated by adding 300 to the previous term.
The nth term of the geometric sequence is 3n-1 .
The sequence can also be generated by
The sum of the first n Fibonacci numbers is
Fn + 2 – 1. F12 = 144, so there are 143 bees in
all 10 generations.
18. (a) For an arithmetic sequence there is a
common difference between the terms.
Between 49 and 64 there are three
differences so we can find the common
difference by subtracting 49 from 64 and
dividing the answer by three:
64 ๏ญ 49 ๏ฝ 15and15 ๏ธ 3 ๏ฝ 5. The common
difference is 5 and we can find the missing
terms: 49 โ 5 = 44 and 49 + 5 = 54 and
54 + 5 = 59.
(b) For a geometric sequence there is a
common ratio between the terms. Between
1 and 625 there are four common ratios
used so we can find the common ratio by
dividing 625 by 1 and then taking the
๏จ1๏ฉ
fourth root: 625 ๏ธ 1 ๏ฝ 625and 625 4 ๏ฝ 5.
The common ratio is 5 and we can find the
missing terms:
1 ๏ 5 ๏ฝ 5,5 ๏ 5 ๏ฝ 25, 25 ๏ 5 ๏ฝ 125.
Copyright ยฉ 2020 Pearson Education, Inc.
Mathematical Connections 1-2: Review Problems
(c) For a geometric sequence there is a
common ratio between the terms. Between
310 and 319 there are three common ratios
used so we can find the common ratio by
dividing 319 by 310 and then taking the
๏จ1๏ฉ
cube root: 319 ๏ธ 310 ๏ฝ 39 and (39 ) 3 ๏ฝ 33.
The common ratio is 33 and we can find
the missing terms:
310 ๏ธ 33 ๏ฝ 37 ,310 ๏ 33 ๏ฝ 313 ,313 ๏ 33 ๏ฝ 316.
(d) For an arithmetic sequence there is a
common difference between the terms.
Between a and 5a there are four
differences so we can find the common
difference by subtracting a from 5a and
dividing the answer by four:
5a ๏ญ a ๏ฝ 4a and 4a ๏ธ 4 ๏ฝ a. The common
difference is a and we can find the missing
terms: a + a = 2a, 2a + a = 3a, 3a + a =
4a.
19. (a) Letโs call the missing terms x and y, then
the sequence becomes 1, x, y, 7, 11 and if it
is a Fibonacci-type sequence then:
1๏ซ x ๏ฝ y
x๏ซ y ๏ฝ7
y ๏ซ 7 ๏ฝ 11 ๏ฎ y ๏ฝ 11 ๏ญ 7 ๏ฝ 4
and x ๏ซ y ๏ฝ 7 ๏ฎ x ๏ซ 4 ๏ฝ 7 ๏ฎ x ๏ฝ 3.
The missing terms are 3 and 4.
(b) Letโs call the missing terms x, y. and z, then
the sequence becomes x, 2, y, 4, z and if it
is a Fibonacci-type sequence then:
x๏ซ2๏ฝ y
2๏ซ y ๏ฝ 4 ๏ฎ y ๏ฝ 2
y๏ซ4 ๏ฝ z ๏ฎ 2๏ซ4 ๏ฝ z ๏ฎ z ๏ฝ 6
and x ๏ซ 2 ๏ฝ 2 ๏ฎ x ๏ฝ 0.
The missing terms are 0, 2, and 6.
(c) Letโs call the missing terms x, y. and z, then
the sequence becomes x, y, 3, 4, z and if it
is a Fibonacci-type sequence then:
x๏ซ y ๏ฝ3
y ๏ซ3 ๏ฝ 4 ๏ฎ y ๏ฝ 4๏ญ3 ๏ฝ1
3๏ซ 4 ๏ฝ 7
and x ๏ซ y ๏ฝ 3 ๏ฎ x ๏ซ 1 ๏ฝ 3 ๏ฎ x ๏ฝ 2.
The missing terms are 2, 1, and 7.
15
Mathematical Connections 1-2:
Review Problems
15. Order the teams from 1 to 10, and consider a
simpler problem of counting how many games
are played if each team plays each other once.
The first team plays nine teams. The second
team also plays nine teams, but one of these
games has already been counted. The third team
also plays 9 teams, but two of these games were
counted in the previous two summands.
Continuing in this manner, the total is 10 + 9 +
8 + โฆ + 3 + 2 + 1 = 9(10) / 2 = 45 games.
Double this amount to obtain 90 games must
be played for each team to play each other
twice.
16. 7 ways. Make a table:
Quarters
Dimes
Nickels
1
1
1
1
0
3
0
4
0
0
3
2
0
2
4
0
1
6
0
0
8
17. If the problem is interpreted to stated that at
least one 12-person tent is used, then there are
10 ways. This can be seen by the table below,
which illustrates the ways 2-,3-,5-, and 6-person
tents can be combined accommodate 14 people.
6-Person
5-Person
3-Person
2-Person
2
0
0
1
1
1
1
0
1
0
2
1
1
0
0
4
0
2
0
2
0
1
3
0
0
1
1
3
0
0
4
1
0
0
2
4
0
0
0
7
Copyright ยฉ 2020 Pearson Education, Inc.
16 Chapter 1: An Introduction to Problem Solving
Chapter 1 Review
1. Make a plan. Every 7 days (every week) the day
will change from Sunday to Sunday. 365 days
per year ยธ 7 days per week ยป 52 weeks per
year + 17 weeks per year. Thus the day of the
week will change from Sunday to Sunday 52
times and then change from Sunday to Monday.
July 4 will be a Monday.
2.
$5.90 ๏ธ 2 ๏ฝ $2.95 more on one of the items.
That is $20 ๏ซ $2.95 ๏ฝ $22.95 for the more
expensive item and $20 ๏ญ $2.95 ๏ฝ $17.05 for
the less expensive item. Check that both items
add up to $40: $22.95 ๏ซ $17.05 ๏ฝ $40.
3. (a) 15, 21, 28. Neither. The successive
differences of terms increases by one; e.g.,
10 + 5, 15 + 6, โฆ .
(b) 32, 27, 22. Arithmetic Subtract 5 from
each term to obtain the subsequent term.
(c) 400, 200, 100. Geometric Each term is half
the previous term.
(d) 21, 34, 55. Neither Each term is the sum of
the previous two termsโthis is the
Fibonacci sequence.
(e) 17, 20, 23. Arithmetic Add 3 to each term
to obtain the subsequent term.
(f) 256, 1024, 4096. Geometric Multiply
each term by 4 to obtain the subsequent
term.
(g) 16, 20, 24. Arithmetic Add 4 to each term
to obtain the subsequent term.
(h) 125, 216, 343. Neither Each term is the 3rd
power of the counting numbers = 13, 23,
33, โฆ .
4. (a) The successive differences are 3. Each term
is 3 more than the previous term. This
suggests that it is an arithmetic sequence of
the form 3n + ?. Since the first term is 5,
3(1) + ? = 5. The nth term would be
3n + 2.
(b) Each term given is 3 times the previous
term. This suggests that the sequence is
geometric. nth term will be 3n.
(c) The only number that changes in
successive terms is the exponent. For the
first term, the exponent is 2; for the second
term, the exponent is 3; for the third term,
the exponent is 4; and so on. So, for the nth
term, the exponent will be n + 1. Therefore,
the nth term will be 2n+1 โ 1.
5. (a) 3(1) – 2 = 1;
3(2) – 2 = 4;
3(3) – 2 = 7;
3(4) – 2 = 10; and
3(5) – 2 = 13.
(b)
12 + 1 = 2;
22 + 2 = 6;
32 + 3 = 12;
42 + 4 = 20; and
52 + 5 = 30.
(c)
4(1) – 1 = 3;
4(2) – 1 = 7;
4(3) – 1 = 11;
4(4) – 1 = 15; and
4(5) – 1 = 19.
6. (a) a1 = 2, d = 2, an = 200.
So 200 = 2 + (n – 1) โ
2 ๏ n = 100.
Sum is
100(2 + 200)
= 10,100.
2
(b) a1 = 51, d = 1, an = 151.
So 151 = 51 + (n – 1) โ
1 ๏ n = 101.
Sum is
101โ
(51+151)
= 10, 201.
2
7. (a) Answers will vary; for example 5 and 3 are
odd numbers, but 5 + 3 = 8, which is not
odd.
(b) 15 is odd; and it does not end in a 1 or a 3.
(c) The sum of any two even numbers is
always even. An even number is one
divisible by 2, so any even number can be
represented by 2 + 2 + 2 + ๏ .๏ซ
Regardless of how many twos are added,
the result is always a multiple of 2, or an
even number.
8. All rows, columns, and diagonals must add to
34; i.e., the sum of the digits in row 1. Complete
rows or columns with one number missing, then
two, etc. to work through the square:
16
3
5
10 11
8
9
6
7
12
4
15 14
1
Copyright ยฉ 2020 Pearson Education, Inc.
2
13
Chapter 1 Review 17
9. The ten middle tables will hold two each and
the two end tables will hold three each, totaling
26 people.
10. (a)
๏ซ 260 ๏ฝ 261
๏ฝ 2 (2 ๏ญ 1)
3m + m + (m – 10) = 90 ๏
60
5m = 100.
Thus
๏ฝ 625
2
๏ฌ + m + s = 90 ๏
So
60
2
and
Then ๏ฌ = 3m and s = m – 10.
๏ฝ 2(260 ) ๏ญ 260
(b)
m = length of the middle-sized piece,
s = lenth of the shortest piece.
๏ฝ 261 ๏ญ 260
๏ฝ2
17. Let ๏ฌ = length of the longest piece,
m = 20cm;
๏ฌ = 3m = 60cm; and
๏ฝ 625
s = m – 10 = 10 cm.
๏ฝ 25
11. 100 ยธ 5 = 20 plus 1 = 21 posts. 1 must be
added because both end posts must be counted.
12. 1 mile = 5280 feet.
18. Make a diagram that demonstrates all the ways
four-digit numbers can be formed from left
(thousands place) to right (ones place). 12 fourdigit numbers can be formed.
5280 ยธ 6 feet = 880 turns per mile.
880 ยด 50000 miles = 44, 000, 000 turns.
13. There are 9 students between 7 and 17 (8
through 16). There must be 9 between them in
both directions, since they are direct opposites.
9 + 9 + 2 = 20 students.
14. Let l be a large box, m be a medium box, and s
be a small box:
3l + (3l ยด 2m each) + [(3 ยด 2)m ยด 5s each]
3l + 6m + 30s = 39 total boxes.
15. Extend the pattern of doubling the number of
ants each day. This is a geometric sequence
with a1 = 1500, an = 100, 000, and r = 2.
100, 000 = 1500 โ
2n-1 ๏
66 23 = 2n-1.
Since 27 ๏ญ1 ๏ผ 66 23 and 28๏ญ1 ๏พ 66 23 , the ant farm
will fill sometime between the 7th and 8th day.
16. The best strategy would be one of guessing and
checking:
(i) Ten 3โs + two 5โs = 40๏ close but
too low.
(ii) Nine 3โs + three 5โs = 42๏ still too
low.
(iii) Eight 3โs + Four 5โs = 44.
19. Answer may vary. Fill the 4-cup container with
water and pour the water into the 7-cup
container. Fill the 4-cup container again and
pour water into the 7-cup container until it is
full. Four minus three (1) cups of water will
remain in the 4-cup container. Empty the 7-cup
container and pour the contents of the 4-cup
container into the 7- cup container. The 7-cup
container now holds 1 cup of water. Refill the 4cup container and pour it into the 7-cup
container. The 7-cup container now contains
exactly 5 cups of water.
They must have answered four 5-point
questions.
Copyright ยฉ 2020 Pearson Education, Inc.
18 Chapter 1: An Introduction to Problem Solving
20. A possible pattern is to increase each rectangle
by one row of dots and one column of dots to
obtain the next term in the sequence. Make a
table.
Number
of the
term
Row
of
dots
Column
of dots
Term
(row ร
column)
1
1
2
2
2
2
3
6
3
3
4
12
4
4
5
20
5
5
6
30
6
6
7
42
7
7
8
56
100
101
10100
n
n +1
n (n + 1)
๏
100
๏
n
We also observe that the number of the term
corresponds to the number of rows in the arrays
and that the number of columns in the array is
the number of the term plus one. Thus, the next
three terms are 30, 42, and 56. The 100th term
is 10,100 and the nth term is n (n + 1).
21. A possible pattern is that each successive figure
is constructed by adjoining another pentagon to
the previous figure
(a)
(b) Observe that the perimeter of the first
figure is 5 and that when a new pentagon is
adjoined 4 new sides are added and one
side (where the new pentagon is adjoined)
is lost. Make a table.
Number of
terms
1-unit sides
(perimeter)
1
5
2
5 -1+ 4 = 8
3
8 – 1 + 4 = 11
4
11 – 1 + 4 = 14
(c) and (d) Looking at the terms in the sequence
and noting that the difference of terms is three,
we suspect that the sequence is arithmetic and
conjecture that the nth term is 3n+2. However,
we need to be sure. Looking at the 4th term
(part a) we observe that the pentagons on the
end Contribute 4 sides to the perimeter and the
โmiddleโ pentagons contribute 3 sides. Thus, in
the nth figure there will be 2 โendโ pentagons
that contribute 4 1-unit sides and n – 2
โmiddleโ pentagons that contribute 3(n – 2) 1
unit sides. The total will be 3(n – 2) + 2(4) =
3n + 2 units. Thus the 100th term is
3(100) + 2 = 302.
22. (a) The circled terms will constitute an
arithmetic sequence because the common
difference will be twice the difference in
the original series.
(b) The new sequence will be a geometric
sequence because the ratio will be the
square of the ratio of the original series.
23. When n = 1, then n2 โ n = 12 โ 1 = 0, so the first
term, a1 = 0. When n = 2, then n2 โ n = 22 โ 2 or
2. Thus a1 + a2 = 2; hence a2 = 2. For n = 3, n2 โ
n = 32 โ 3 = 6. Substituting for a1 and a2, we get
a3 = 6 โ 2 =4. For n = 4, 42 โ 4 = 12.
Substituting for a1, a2, and a3, we get 0 + 2 + 4
+ a4 = 12. Hence a4 = 6.
24. (a) Letโs call the missing terms a, and b then
the sequence becomes:
13, a, b, 27
13 ๏ซ a ๏ฝ b ๏ฎ a ๏ฝ b ๏ญ 13
a ๏ซ b ๏ฝ 27 ๏ฎ b ๏ญ 13 ๏ซ b ๏ฝ 27
๏ฎ 2b ๏ฝ 40
๏ฎ b ๏ฝ 20
a ๏ฝ b ๏ญ 13 ๏ฎ a ๏ฝ 7
So 7 and 20 are the missing terms.
(b) Letโs call the missing terms a, and b then
the sequence becomes:
137, a, b,163
137 ๏ซ a ๏ฝ b ๏ฎ a ๏ฝ b ๏ญ 137
a ๏ซ b ๏ฝ 163 ๏ฎ b ๏ญ 137 ๏ซ b ๏ฝ 163
๏ฎ 2b ๏ฝ 300
๏ฎ b ๏ฝ 150
a ๏ฝ b ๏ญ 137 ๏ฎ a ๏ฝ 13
So 13 and 150 are the missing terms.
Copyright ยฉ 2020 Pearson Education, Inc.
Chapter 1 Review 19
(c) Letโs call the missing terms x, and y, then
the sequence becomes:
b, x, y, a
b๏ซ x ๏ฝ y ๏ฎ x ๏ฝ y ๏ญb
x ๏ซ y ๏ฝ a ๏ฎ y ๏ญb๏ซ y ๏ฝ a
๏ฎ 2y ๏ฝ a ๏ซ b
a๏ซb
๏ฎy๏ฝ
2
a๏ซb
a ๏ซ b 2b
๏ญb ๏ฎ x ๏ฝ
๏ญ
x๏ฝ
2
2
2
a ๏ญb
๏ฎx๏ฝ
2
a ๏ญb
a+b
and
.
2
2
25. The first line tells us that since three cylinders
=15, each cylinder must be worth 5. Using that
information, we can use the second line to
figure out that a circle must be worth 4. That
information can be used on the third line to
determine that the cup-like shape must be equal
to 1. So, putting it all together in the fourth line,
we have 4 + 5 + 1 = 10.
26. The pattern involves taking two squares that are
diagonal, summing their numbers, and putting
that sum in the square that is below one of the
diagonal squares and to the left of the other
diagonal square. For example, the top square is
4; the square diagonal to it (to the right) is 2;
their sum is 6, which is the number that appears
in the square below the 4 and to the left of the 2.
You can also see it where 9 + 10 = 19, and 1 +
6 =7. So using this pattern, the question mark
must be equal to 3.
So the missing terms are
27. Since 1, 2, and 3 are already on the triangle, the
numbers 4-9 will be used to fill in the six
question marks. On the left side, 1+2 = 3, so the
two question marks on that side must sum to 14.
Similarly on the bottom, the two question marks
must sum to 13, and on the right side, the two
question marks must sum to 12. So on the right
side with the available numbers, only two
possibilities, 5+7, or 4+8, sum to 12.
If the two numbers on the right side are 5 and 7,
the bottom two numbers (which must sum to
13) are 4 and 9, and the two numbers on the left
side must be 6 and 8. If the two numbers on the
right side are 4 and 8, the two bottom numbers
must be 6 and 7, leaving 5 and 9 to be the
numbers on the right. Below are the two
solutions presented on the triangle:
Copyright ยฉ 2020 Pearson Education, Inc.
CHAPTER 2
INTRODUCTION TO LOGIC AND SETS
Assessment 2-1A: Reasoning and Logic:
An Introduction
1. (a) False statement. A statement is a sentence
that is either true or false, but not both.
(b) False statement. Los Angeles is a city, not
a state.
(c) Not a statement. Questions are not
statements.
(d) True statement.
2. (a) There exists at least one natural number n
such that n + 8 = 11.
(b) There exists at least one natural number n
such that n 2 = 4.
(c) For all natural numbers n, n + 3 = 3 + n.
(d) For all natural numbers n, 5n + 4n = 9n.
3. (a) For all natural numbers n, n + 8 = 11.
(b) For all natural numbers n, n 2 = 4.
(c) There is no natural number x such that
x + 3 = 3 + x.
(d) There is no natural number x such that
5x + 4 x = 9 x.
4. (a) The book does not have 500 pages.
(b) 3 โ
5 ยน 15.
(c) Some dogs do not have four legs.
(d) No rectangles are squares.
(e) All rectangles are squares.
(f) Some dogs have fleas.
5. (a) If n = 4, or n = 5, then n 3, so the statement is true, since it
can be shown to work for some natural
numbers n.
(b) All natural numbers are greater than zero;
so, since the condition is that n > 0 or
n 5 and n > 2, so the
statement is true.
(b) n could equal 5, so the statement is false.
6. (a) p ๏ q is false only if both p and q are
false, so if p is true the statement is true
regardless of the truth value of q.
(b) An implication is false only when p is true
and q is false, so if p is false then the
statement is true regardless of the truth
value of q.
Copyright ยฉ 2020 Pearson Education, Inc.
24 Chapter 2: Introduction to Logic and Sets
7.
p
q
~q
p ๏ ~q ~( p ๏ ~q)
T T
F
T
F
T F
F T
T
F
T
F
F
T
F F
T
T
F
8. (a) q ๏ r.
(b) q ๏ ~ r.
(c) ~ r ๏ ~ q.
(c) The first president of the United States
was not George Washington.
(d) A quadrilateral has three sides of the
same length and the quadrilateral does
not have four sides of the same length.
(e) A rectangle has four sides of the same
length and that rectangle does not have
three sides of the same length.
In both (d) and (e), the negation of the
conditional statement p ๏ฎ q is p ๏ ~ q.
11. (a)
(d) ~(q ๏ r ).
9. (a) True. This statement is a disjunction. The
two parts could be stated as such: p is the
statement โ4 + 6 = 10โ, while q is the
statement โ2 + 3 = 5โ. In this situation p is
true, and q is true. In order for a
disjunction to be true, either p or q (or
both) have to be true; the only way a
disjunction can be false is if both p and q
are false. So therefore, this statement is
true.
(b) Answers may vary. If a team has more than
11 players on the field, it is a penalty and the
play will not count; so, an answer of False
could be given. However, just because it is a
penalty doesnโt mean a team canโt have 12 or
more players on the field; so, you could say
the answer is True. You could also say True
if you consider the time between plays when
players are substituting in for other players;
oftentimes, there are more than 11 players on
the field during substitutions.
(c) True.
(d) False. To see, sketch a drawing where
three sides are the same length, but with the
two angles where the sides intersect being
different measures (in fact, make one a
right angle, the other an obtuse angle). You
should easily make a quadrilateral with a
side length different from the other three.
(e) True. If a rectangle has four sides of the
same length, then by default it must have
three sides the same length. Of course, a
rectangle with four equal sides is a square!
10. (a) By DeMorganโs Laws, the negation of the
disjunction p ๏ q is ~ p ๏ ~ q . So, the
statement would be 4 + 6 ยน 10 and
2+3ยน5
(b) A National Football League team cannot
have more than 11 players on the field
while a game is in progress.
p
q
~ p ~q
p ๏ q ~( p ๏ q) ~ p ๏ ~ q
T T
F
F
T
F
F
T F
F T
F
T
T
F
T
T
F
F
F
F
F
T
T
F
T
T
F
Since the truth values for ~( p ๏ q) are the
same as for ~ p ๏ ~ q, the statements are
logically equivalent.
(b)
p
q
~ p ~q
p ๏ q ~( p ๏ q) ~ p ๏ ~ q
T T
F
F
T
F
F
T F
F T
F
T
T
F
F
F
T
T
T
T
F
T
T
F
T
T
F
Since the truth values for ~ ๏จ p ๏ q ๏ฉ are the
same as for ~ p ๏ ~ q, the statements are
logically equivalent.
12. Megan is a female brunette who owns a car.
She is not single.
13. (a) p ๏ฎ q.
(b) ~ p ๏ฎ q.
(c)
p ๏ฎ ~ q.
(d) q if p, or p ๏ฎ q.
(e) ~ p ๏ฎ ~ q.
(f ) ~q ๏ฎ ~ p.
14. (a) Converse: If x 2 = 9, then x = 3.
Inverse: If x ยน 3, then x 2 ยน 9.
Contrapositive: If x 2 ยน 9, then x ยน 3.
(b) Converse: If classes are canceled, then it
snowed.
Copyright ยฉ 2020 Pearson Education, Inc.
Assessment 2-2A: Describing Sets 25
Inverse: If it does not snow, then classes
are not canceled.
Contrapositive: If classes are not
canceled, then it did not snow.
15. No. This is the inverse; i.e., if it does not rain
then lris can either go to the movies or not
without making her statement false.
16. (a) Valid. Use modus ponens: Hypatia was a
woman ๏ฎ all women are mortal ๏ฎ Hypatia
was mortal.
(b) Valid. Since Dirty Harry was not written
by J.K. Rowling, and she wrote all the
Harry Potter books, then Dirty Harry
cannot be a Harry Potter book.
(c) Not valid.
17. (a) Since all students in Integrated
Mathematics I are in Kappa Mu Epsilon,
and Helen is in Integrated Mathematics I,
then the conclusion is that Helen is in
Kappa Mu Epsilon.
(b) Let p = all engineers need mathematics
and q = Ron needs mathematics.
Then p ๏ฎ q, or if all engineers need
mathematics then Ron needs mathematics.
p is true, but q is false, Ron does not need
mathematics.
So Ron is not an engineer.
(c) Since all bicycles have tires and all tires
use rubber, then the conclusion is all
bicycles use rubber.
18. (a) If a number is a natural number, then it is a
real number.
(b) If a figure is a circle, then it is a closed
figure.
19. DeMorganโs Laws are that:
~( p ๏ q) is the logical equivalent of ~ p ๏ ~ q.
~ ๏จ p ๏ q ๏ฉ is the logical equivalent of ~ p ๏ ~ q.
Thus:
(a) The negation is 3 + 5 = 9 or 3 โ
5 ยน 15.
(b) The negation is I am not going and she is
not going.
Assessment 2-2A: Describing Sets
1. (a) Either a list or set-builder notation may be
used: {a,s,e,m,n, t} or { x | x is a letter in
the word assessment}.
(b) {21, 22, 23, 24,โฆ} or { x | x is a natural
number and x > 20} or { x | x ร N
and x > 20}.
2. (a) P = {p,q,r, s}.
(b) {1, 2} ร {1, 2, 3}. The symbol ร refers to
a proper subset.
(c) {0,1} ร{1, 2, 3}. The symbol ร refers to
a subset.
3. (a) Yes. {1, 2,3, 4,5} ~ {m, n, o, p, q} because
both sets have the same number of
elements and thus exhibit a one-to-one
correspondence.
(b) Yes. {a, b, c, d , e, f ,โฆ, m} ~ {1, 2,3,๏ ,13}
because both sets have the same number of
elements.
(c) No. { x | x is a letter in the word
/ {1, 2,3, 4,๏ ,11}; there are
mathematics} ~
only eight unduplicated letters in the word
mathematics.
4. (a) The first element of the first set can be
paired with any of the six in the second set,
leaving five possible pairings for the
second element, four for the third, three for
the fourth, two for the fifth, and one for the
sixth. Thus there are
6 โ
5 โ
4 โ
3 โ
2 โ
1 = 720 one-to-one
correspondences.
(b) There are
n โ
(n – 1) โ
(n – 2) โ
๏ โ
3 โ
2 โ
1 = n!
possible one-to-one correspondences. The
first element of the first set can be paired
with any of the n elements of the second
set; for each of those n ways to make the
first pairing, there are n – 1 ways the
second element of the first set can be paired
with any element of the second set; which
means there are n – 2 ways the third
element of the first set can be paired with
any element of the third set; and so on. The
Fundamental Counting Principle states that
the choices can be multiplied to find the
total number of correspondences.
5. (a) If x must correspond to 5, then y may
correspond to any of the four remaining
elements of {1, 2,3, 4,5}, z may correspond
to any of the three remaining, etc. Then
1 โ
4 โ
3 โ
2 โ
1 = 24 one-to-one
correspondences.
Copyright ยฉ 2020 Pearson Education, Inc.
26 Chapter 2: Introduction to Logic and Sets
(b) There would be 1 โ
1 โ
3 โ
2 โ
1 = 6 one-toone correspondences.
(c) The set {x, y, z} could correspond to the set
{1,3,5} in 3 โ
2 โ
1 = 6 ways. The set
{u,๏ต} could correspond with the set {2, 4}
in 2 โ
1 = 2 ways. There would then be
6 โ
2 = 12 one-to-one correspondences.
(b) Since A is a subset of A and A is the only
subset of A that is not proper, A has
25 – 1 = 31 proper subsets.
(c) Let B = {b,c,d}. Since B ร A, the
subsets
of B are all subsets of A that do not
contain a and e. There are 23 = 8 of
these subsets. If we join (union) a and e
to each of these subsets there are still 8
subsets.
6. There are three pairs of equal sets:
(i ) A = C . The order of the elements does
not matter.
(ii ) E = H. They are both the null set.
Alternative. Start with {a,e}. For each
element b, c, and d there are two options:
include the element or do not include the
element. So there are 2 โ
2 โ
2 = 8 ways
to create subsets of A that include a and
e.
(iii ) I = L. Both represent the numbers
1,3,5, 7,โฆ.
7. (a) Assume an arithmetic sequence with
a1 = 201, an = 1100, and d = 1. Thus
1100 = 201 + (n – 1) โ
1; solving,
n = 900. The cardinal number of
the set is therefore 900.
(b) Assume an arithmetic sequence with
a1 = 1, an = 101, and d = 2. Thus
101 = 1 + (n – 1) โ
2; solving, n = 51.
The cardinal number of the set is therefore
51.
(c) Assume a geometric sequence with
a1 = 1, an = 1024, and r = 2. Thus
1024 = 1 โ
2n-1 ๏ 210 = 2n-1 ๏ n – 1
= 10 ๏ n = 11. The cardinal number of
the set is therefore 11.
(d) If k = 1, 2,3,๏ ,100, the cardinal number
of the set
3
{x | x = k , k = 1, 2,3,๏ ,100} = 100,
since there are 100 elements in the set.
8.
A represents all elements in U that are not in A,
or the set of all college students with at least
one grade that is not an A.
9. (a) A proper subset must have at least one less
element than the set, so the maximum
n( B ) = 7.
(b) Since B ร C , and n( B) = 8 then C
could have any number of elements in it,
so long as it was greater than eight.
10. (a) If C ร D and D ร C , then the sets arc
equal; so n(D) = 5.
(b) Answers vary. For example, the sets are
equal and/or the sets are equivalent.
11. (a) A has 5 elements, thus 25 = 32 subsets.
12. If there are n elements in a set, 2n subsets can
be formed. This includes the set itself. So if
there are 127 proper subsets, then there are 128
subsets. Since 27 = 128, the set has 7
elements.
13. In roster format,
A = {3, 6,9,12,๏}, B = {6,12,18, 24,๏}, and
C = {12, 24,36,๏}. Thus, B ร A, C ร A, and
C ร B.
Alternatively: 12n = 6(2n) = 3(4n).
Since 2n and 4n are natural number
C ร A, C ร B, and B ร A.
14. (a) ร . There are no elements in the empty set.
(b) ร. 1024 = 210 and 10 ร N .
(c) ร. 3(1001) – 1 = 3002 and 1001 ร N .
(d) ร. For example, x = 3 is not an element
because for 3 = 2n , n ร N .
15. (a) ร . 0 is not a set so cannot be a subset of
the empty set, which has only one subset,
ร.
(b) ร . 1024 is an element, not a subset.
(c)
ร . 3002 is an element, not a subset.
(d) ร . x is an element, not a subset.
16. (a) Yes. Any set is a subset of itself, so if
A = B then A ร B.
(b) No. A could equal B; then A would be a
subset but not a proper subset of B.
Copyright ยฉ 2020 Pearson Education, Inc.

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