Math Module for Grade 10

1. Module
1
Searching for Patterns in
Sequences,
Arithmetic, Geometric and
Others
What this module is all about
This module will teach you how to deal with a lot of number patterns. These
number patterns are called sequences. Go over the lessons and have fun in working
with the exercises.
What you are expected to learn
It is expected that you will be able to demonstrate knowledge and skill related
to sequences and apply these in solving problems. Specifically, you should be able
to:
a) list the next few terms of a sequence given several consecutive terms.
b) derive by pattern searching, a mathematical expression (rule) for
generating the sequences.
c) generate the next few terms of sequences defined recursively.
d) describe an arithmetic sequence by any of the following:
• giving the first few terms
• giving the formula for the nth term
• drawing the graph
How much do you know

2. A. Write the first five terms of the sequence.
1. an =
n
n 3+
3. an = 4n – 3
2. an = 2n - 1
B. Find the indicated term for the sequence.
4. an = -9n + 2; a8 6. an =
52
73
−
+
n
n
; a14
5. an = (n + 1)(2n + 3); a5
C. Find the general term, an, for the given terms of the sequence.
7. 4, 8, 12, 16,… 9. ,...
12
1
,
6
1
,
2
1
8. -10, -20, -30, -40,…
D. Find the first four terms of the sequence defined recursively.
10. a1= -1, an = 3an – 1 11. a1 = 5, an =
1
1
−
−
n
an
E. Find the common difference and the next three terms of the given
arithmetic sequence.
12. 2, 8, 14, 20,… 14. 1, 2, 3, 4,…
13. 7, 8.5, 10, 11.5,… 15. 24, 21, 18, …
What you will do
Lesson 1
Sequences and its Kinds
It is a common experience to be confronted with a set of numbers arranged in
some order. The order and arrangement may be given to you or you have to discover
a rule for it from some data. For example, the milkman comes every other day. He
came on July 17; will he come on Aug 12? Consider that you are given the set of
dates
17, 19, 21,…
2

3. arranged from left to right in the order of increasing time. Continuing the set we have
17, 19, 21, …, 29, 31, 2, 4, ….,28, 30…
so that the answer to our question is yes.
Any such ordered arrangement of a set of numbers is called a SEQUENCE.
Look at this second example. Lorna, a 2nd
year student in a certain public
school, is able to save the money her ninongs and ninangs gave her last Christmas.
She then deposits her savings of P1,000 in an account that earns 10% simple
interest. The total amount of interest she earned in each of the 1st
4 years of her
saving is shown below:
Year 1 2 3 4
Total amount 10 20 30 40
The list of numbers 10, 20, 30, 40 is called a sequence. The list 10,20,30,40
is ordered because the position in this list indicates the year in which that total
amount of interest is earned.
Now, each of the numbers of a sequence is called a term of the sequence.
The first term in the sequence 10, 20, 30, 40 is 10, the second term is 20, while the
third term is 30 and the fourth term is 40. It is also good to point out that the
preceding term of a given term is the term immediately before that given term. For
example, in the given sequence 20 is the term that precedes 30.
Examples of other sequences are shown below. These sequences are
separated into two groups. A finite sequence contains a finite number of terms. An
infinite sequence contains an infinite number of terms.
Finite sequence Infinite sequence
1, 1, 2, 3, 5, 8 1, 3, 5, 7, …
1, 2, 3, 4, 5, …, 8 1, ,
8
1
,
4
1
,
2
1
…
1, -1, 1, -1 1, 1, 2, 3, 5, 8, …
In general:
A sequence is a set of numbers written in a specific order:
a1, a2, a3, a4, a5, a6,………, an
3

4. The number a1 is called the 1st
term, a2 is the 2nd
term, and in general, an is the nth
term. Note that each term of the sequence is paired with a natural number.
Given at least the first 3 terms of a sequence, you can easily find the next term
in that sequence by simply discovering a pattern as to how the 3rd
term is derived
from the 2nd
term, and the 2nd
from the 1st
term. You will find that either a constant
number is added or subtracted or multiplied or divided to get the next term or a
certain series of operations is performed to get the next term. This may seem hard at
first but with practice and patience in getting them, you will find that it’s very exciting.
Examples:
Find the next term in each sequence.
1. 17, 22, 27, 32, …
2.
11
1
,
8
1
,
5
1
,
2
1
…
3. 5, 10, 20, 40,…
4. 3, -3, 3, -3,…
Solutions:
1. Notice that 5 is added to 17 to get 22, the same is added to 22 to get 27, and
the same (5) is added to 27 to get 32. So to get the next term add 5 to the
preceding term, that is, 32 + 5 = 37. The next term is 37.
2. Notice that 1 is the numerator of all the fractions in the sequence while the
denominators- 2, 5, 8, 11 form a sequence. 3 is added to 2 to get 5, 3 is also
added to 5 to get 8. So that 3 is added to 11 to get 14. The next term is
therefore 1/14.
3. For this example, 2 is multiplied to 5 to get 10, 2 is multiplied to 10 to get
20 and 2 is also multiplied to 20 to get 40. So the next term is 80, the result of
multiplying 40 by 2.
4. It is easy to just say that the next term is 3 since the terms in the sequence is
alternately positive and negative 3. Actually the first, second, and third terms
were multiplied by -1 to get the second, third and fourth terms respectively.
Try this out
A. Write F if the sequence is finite or I if the sequence is infinite.
1. 2, 3, 4, 5, ….., 10
2. 7, 10, 13, 16, 19, 22, 25
4

5. 3. 4, 9, 14, 19, …
4. 2, 6, 18, 54
5. 3, 9, 27, 81, …., 729, …
6. -2, 4, -8, 16, …..
7. 100, 97, 94, 91, …, -2
8.
64
1
,
32
1
,
16
1
,
8
1
,
4
1
9. 1, 4, 9, 16, 25, …., 144
10.
25
4
,
16
3
,
9
2
,
4
1
B. Answer the puzzle.
Why are Policemen Strong?
Find the next number in the sequences and exchange it for the letter which
corresponds each sequence with numbers inside the box to decode the answer to
the puzzle.
A 2, 5, 11, 23, __ N 2, 6, 18, 54, __
B 2, 4, 16, __ O 20, 19, 17, __
C 7, 13, 19, __ P 2, 3, 5, 7, 9, 11, 13, 15, __
D 19, 16, 13, __ R 13, 26, 39, __
E 4, 8, 20, 56, __ S 5, 7, 13, 31, __
F 2, 2, 4, 6, 10, 16, __ T 1, 1, 2, 4, 7, 13, 24, __
H 1, 1, 2, 4, 7, 13, __ U 1, 1, 1, 2, 3, 4, 6, 9, 13, __
I 3, 6, 12, 24, __ Y 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, __
L 10, 11, 9, 12, 8, __
Lesson 2
Finding the Terms of a Sequence
Frequently, a sequence has a definite pattern that can be expressed by a rule
or formula. In the simple sequence
24 14 13 10 19 17 44 52 47 26 26 48 25
256 164 25 47 19 85 164 44 24 164 6 25 47 162
5

6. 2, 4, 6, 8, 10, ….
each term is paired with a natural number by the rule an = 2n. Hence the sequence
can be written as
2, 4, 6, 8,… 2n,…
1st
term 2nd
term 3rd
term 4th
term nth term
a1 a2 a3 a4 an
Notice how the formula an = 2n gives all the terms of the sequence. For
instance, substituting 1, 2, 3, and 4 for n gives the 1st
four terms:
a1 = 2(1) = 2 a3 = 2(3) = 6
a2 = 2(2) = 4 a4 = 2(4) = 8
To find the 103rd
term of this sequence, use n=103 to get a103 = 2(103) = 206.
Examples:
1. Find the 1st
four terms of the sequence whose general term is given by
an = 2n – 1.
Solution:
To find the first, second, third and fourth terms of this sequence, simply
substitute 1, 2, 3, 4 for n in the formula an = 2n-1.
If the general term is an = 2n – 1, then the
1st
term is a1 = 2(1) – 1 = 1
2nd
term is a2 = 2(2) – 1 = 3
3rd
term is a3 = 2(3) – 1 = 5
4th
term is a4 = 2(4) – 1 = 7.
The 1st
four terms of this sequence are the odd numbers 1, 3, 5, and 7. The
whole sequence can be written as
1, 3, 5, …, 2n – 1
Since each term in this sequence is larger than the preceding term, we say
that the sequence is an increasing sequence.
A sequence is increasing if an + 1 > an for all n.
6

7. 2. Write the 1st
4 terms of the sequence defined by an =
1
1
+n
.
Solution:
Replacing n with 1, 2, 3, and 4, respectively the 1st
four terms are:
1st
term = a1 =
11
1
+
=
2
1
2nd
term = a2 =
12
1
+
=
3
1
3rd
term = a3 =
13
1
+
=
4
1
4th
term = a4 =
14
1
+
=
5
1
The sequence defined by an =
1
1
+n
can be written as
2
1
,
3
1
,
4
1
,
5
1
, ……,
1
1
+n
Since each term in the sequence is smaller than the term preceding it, the
sequence is said to be a decreasing sequence.
A sequence is decreasing if an + 1 < an for all n.
3. Find the 1st
5 terms of the sequence defined by an = n
n
2
)1(−
.
Solution:
Again by simple substitution,
1st
term = a1 = 1
1
2
)1(−
= -
2
1
2nd
term = a2 = 2
2
2
)1(−
=
4
1
3rd
term = a3 = 3
3
2
)1(−
= -
8
1
4th
term = a4 = 4
4
2
)1(−
=
16
1
7

8. 5th
term = a5 = 5
5
2
)1(−
= -
32
1
The sequence defined by an = n
n
2
)1(−
can be written as
-
2
1
,
4
1
, -
8
1
,
16
1
, -
32
1
,…, n
n
2
)1(−
Notice that the presence of (-1) in the sequence has the effect of making
successive terms alternately negative and positive.
It is often useful to picture a sequence by sketching its graph w/ the n values
as the x- coordinates and the an values as the y- coordinates.
The corresponding graphs of Examples 1 - 3 are given below
Example 1
0
1
2
3
4
5
6
7
8
0 1 2 3 4 5
n
a
n
8

9. You can also find a specific term, given a rule for the sequence, as seen in the
following example.
4. Find the 13th
and 100th
terms of the sequence whose general term is
given by
An = 2
)1(
n
n
−
Solution: For the 13th
term, replace n with 13 and for the 100th
term, replace n w/
100:
13th
term = a13 = 2
13
13
)1(−
=
169
1−
100th
term = a100 = 2
100
100
)1(−
=
10000
1
Example 2
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4 5
n
a
n
Example 3
-0.6
-0.4
-0.2
0
0.2
0.4
0 1 2 3 4 5 6
n
a
n
9

10. Try this out
A. Write the 1st
4 terms of the sequence whose nth term is given by the formula.
1. an = n + 1 2. an = 2 – 2n
3. an = n - 1 4. an = 2 n
5. an = 2n + 1 6. an = n² + 1
7. an = 3n - 1 8. an =
1+n
n
9. an = 1 - 2n 10. an = n –
n
1
11. an = 3n
12. an = (-1) n+1
n
13. an = n² - 1 14. an = (-1) n
2 n
15. an = n² -
n
1
16. an =
n
n 12
−
17. an =
1
)1( 1
+
− +
n
n
18. an =
1
)1(
2
1
+
− +
n
n
19. an = 2
1
3
1
+






n
20. an =
3
1
n³ + 1
21. an = (1)n
(n²+2n+1)
B. Find the indicated term of the sequence whose nth term is given by the formula.
22. an = 3n + 4 a12
23. an= n(n -1) a11
24. an= (-1) n - 1
n² a15
25. an= (
2
1
)n
a8
26. an = 2n - 5 a10
27. an =
1+n
n
a12
28. an = (-1) n - 1
(n - 1) a25
29. an = (
3
2
) n
a5
30. an = (n + 2)(n + 3) a17
10

11. 31. an = (n + 4)(n + 1) a7
32. an = 2
12
)1(
n
n+
−
a6
33. an =
4
)1( 2
+
−
n
n
a16
34. an =
2
3
n² - 2 a8
35. an =
3
1
n + n² a6
Lesson 3
Finding the nth Term of a Sequence
In Lesson 2, some terms of a sequence were found after being given the
general term. In this lesson, the reverse is done. That is, given some terms of the
sequence, try to find the formula for the general term.
Examples:
1. Find a formula for the nth term of the sequence 2, 8, 18, 32,…
Solution:
Solving a problem like this involve some guessing. Looking over the first 4
terms, see that each is twice a perfect square:
2 = 2(1)
8 = 2(4)
18 = 2(9)
32 = 2(16)
By writing each sequence with an exponent of 2, the formula for the nth term
becomes obvious:
a1 = 2 = 2(1)²
a2 = 8 = 2(2)²
a3 = 18 = 2(3)²
a4 = 32 = 2(4)²
.
.
.
an = 2(n)² = 2n²
11

12. The general term of the sequence 2, 8, 18, 32,…. is an = 2n².
2. Find the general term for the sequence 2,
8
3
,
27
4
,
14
5
,….
Solution:
The first term can be written as
1
2
. The denominators are all perfect cubes
while the numerators are all 1 more than the base of the cubes of the
denominators:
a1 = 2/1 = 3
1
11 +
a2 = 3/8 = 3
2
12 +
a3 = 4/27 = 2
3
13 +
a4 = 5/64 = 3
4
14 +
Observing this pattern, recognize the general term to be an = 3
1
n
n +
3. Find the nth term of a sequence whose first several terms are given
2
1
,
4
3
,
6
5
,
8
7
, . . .
Solution:
Notice that the numerators of these fractions are the odd numbers and
the denominators are the even numbers. Even numbers are in the form
usually written in the form 2n, and odd numbers are written in the form 2n – 1
(an odd number differs form an even number by 1). So, a sequence that
has these numbers for its first four terms is given by an =
n
n
2
12 −
.
4. Find the nth term of a sequence whose first several terms are given
-2, 4, -8, 16, -32,…
Solution:
12

13. These numbers are powers of 2 and they alternate in sign, so a
sequence that agrees with these terms is given by an = (-1)n
2n
.
Note: Finding the nth term of a sequence from the 1st
few terms is not always
automatic. That is, it sometimes takes a while to recognize the pattern. Don’t
be afraid to guess the formula for the general term. Many times an incorrect
guess leads to the correct formula.
Some pointers on how to find the general term of a sequence is given below.
Pointers on How to Find the General Term of a Sequence
1. Study each term of the sequence as it compares to its term number. Then
answer the following questions:
a. Is it a multiple of the term number?
b. Is it a multiple of the square or cube of the term number? If each
term is a multiple of the term number, there will be a common
number.
2. Examine the sequence. Does it increase or decrease?
a. If it increases slowly, consider expressions that involve the term
number plus or minus a constant like: n + 2 or n – 3.
b. If it increases moderately, think about multiples of the term number
plus or minus a constant like: 2n or 3n – 1.
c. If the sequence increases very rapidly, try powers of the term
number plus or minus a constant like: n2
or n2
+ 1.
3. If the sequence consists of fractions, examine how the denominator and
numerator change as separate sequences. For example: an = 2
1
n
n +
yields
,...
25
6
,
16
5
,
9
4
,
4
3
,
1
2
Also, though not all sequences can be defined by a formula, like for the
sequence of prime numbers, be assured that the sequences discussed or given here
are all obvious sequences that one can find a formula or rule for them.
13

14. Try this out
A. Write the formula for the nth term of the sequence:
1. The sequence of the natural numbers.
2. The sequence of the negative even integers.
3. The sequence of the odd natural numbers.
4. The sequence of the negative odd numbers.
5. The sequence of the multiples of 7.
6. The sequence of the positive even integers that are divisible by 4.
7. The sequence of the negative integers less than -5.
8. The sequence of the positive odd integers greater than 9.
9. The sequence for which a1 is 8 and each term is 6 more than the
preceding term.
10. The sequence for which a1 is 3 and each term is 10 less than the
preceding term.
B. Determine the general term for each of the following sequences:
1. 2, 3, 4, 5,…
2. 3, 6, 9, 12,…
3. 4, 8, 12, 16, 20,…
4. 3, 4, 5, 6,…
5. 7, 10, 13, 16,…
6. 4, 9, 14, 19,…
7. 3, 12, 27, 48,…
8. 2, 16, 54, 128,…
9. 4, 8, 16, 32,…
10. 1, 8,27, 64,…
11. 1, 4, 9, 16,…
Lesson 4
Recursively Defined Sequences
Some sequences do not have simple defining formulas like those in Lesson 3.
The nth term of a sequence may depend on some or all of the terms preceding it. A
sequence defined in this way is called recursive.
One has to be sure that the notations: an and an – 1 is understood first before
going to the examples below. To illustrate: if the 24th
term is an, then it can be written
as a24. So that an – 1 means a24 – 1 = a23 or the 23rd
term.
14
12. 3, 9, 27, 81,…
13. -2, 4, -8, 16,…
14. -3, 9, -27, 81,…
15.
32
1
,
16
1
,
8
1
,
4
1
,…
16.
81
1
,
27
1
,
9
1
,
3
1
,…
17.
25
4
,
16
3
,
9
2
,
4
1
,…
18.
32
1
,
28
3
,
10
2
,
4
1
,…

15. Examples:
1. Find the first five terms of the sequence defined recursively by a1 = 1
and an = 3(an – 1 + 2).
Solution:
The defining formula asks that the preceding term, an – 1, be identified first
before one can get the nth term, an. Thus, one can find the second term from the
first term, the third term from the second term, the fourth term from the third term,
and so on. Since the first term is given as a1 = 1, proceed as follows:
a2 = 3(a1 + 2) = 3(1 + 2) = 9
a3 = 3(a2 + 2) = 3(9 + 2) = 33
a4 = 3(a3 + 2) = 3(33 + 2) = 105
a5 = 3(a4 + 2) = 3(105 + 2) = 321
Thus, the first five terms of this sequence are
1, 9, 33, 105, 321,…
2. Find the first 11 terms of the sequence defined recursively by F1 = 1,
F2 = 1 and Fn = Fn – 1 + Fn – 2
Solution:
To find Fn, the preceding terms, Fn – 1 and Fn – 2 have to be found first. Since
the first two terms are given, then
F3 = F2 + F1 = 1 + 1 = 2
F4 = F3 + F2 = 2 + 1 = 3
F5 = F4 + F3 = 3 + 2 = 5
By this time it should be clear as to what is happening here. Each term is
simply the sum of the first two terms that precede it, so that its easy to write down as
many terms as one pleases. Here are the first 11 terms:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…
The sequence in Example 2 is called the Fibonacci sequence, named after
the 13th
century Italian mathematician who used it to solve a problem about the
breeding of rabbits. The sequence also occurs in a lot of other applications in nature.
15

16. 8
5
3
2
1
1
The Fibonacci sequence
In the branching of a tree
To better understand the difference between recursively defined sequences
and the general sequences encountered in the last three lessons, study the
examples below.
3. Find the 120th
term of the sequence defined by an = 2n + 3.
Solution:
Replacing n with 120, then
120th
term = a120 = 2(120) + 3 = 243
The 120th
term is 243.
Notice that the 120th
term from the example above is found without finding first
the preceding term, that is, the 119th
term! Whereas with the recursively defined
sequence one has to find all the terms preceding the required term,
4. Find the 4th
term of the recursively defined sequence an =
2
1−na
where a1 = -3.
Solution:
First term = a1 = -3
Second term = a2 =
2
12−a
=
2
1a
=
2
3−
Third term = a3 =
2
13−a
=
2
2a
=
2
2
3−
=
4
3−
Fourth term = a4 =
2
14−a
=
2
3a
=
2
4
3−
=
8
3−
The fourth term is -
8
3
.
16

17. Try this out
Find the first five terms of the given recursively defined sequences:
1. an = 2(an – 1 – 2) and a1 = 3
2. an =
2
1−na
and a1 = -8
3. an = 2an – 1 + 1 and a1 = 1
4. an =
11
1
−+ na
and a1 = 1
5. an = an – 1 + an – 2 and a1 = 1, a2 = 2
6. an = an – 1 + an – 2 + an – 3 and a1 = a2 = a3 = 1
Lesson 5
Arithmetic Sequences
There are two main types of sequences. These are the arithmetic sequences
and the geometric sequences. This lesson will show what arithmetic sequences are.
The more detailed lesson on arithmetic sequences will be discussed in the next
module.
Look at the following sequences.
1. 4, 7, 10, 13,…
2. 33, 38, 43, 48,…
3. -2, -6, -10, -14,…
4. 100, 98, 96, 94,…
5.
2
1
, 1, 1
2
1
, 2, …
Can you give the next two terms of the above sequences? How did you get
the next terms?
If you get 16 and 19 for a, then you are correct. Notice that a constant
number , 3, is added to each term to get the next term. In b, 5 is added to the
preceding term after the first, while in c, -4 is added to get the next term, in d, -2 is
added to the preceding term and in e, ½ is added to get the next term.
Notice that a constant or common number is added to the preceding term to
get the next term in each of the sequences above. All these sequences are called
17

18. arithmetic sequences. The constant number is called the common difference and is
represented as d.
To find the common difference, d, simply subtract the first term from the
second term, a2 – a1, or the second term from the third term, a3 – a2, or the third term
from the fourth term, a4 – a3; or in general,
d = an – an – 1
Examples:
1. Determine if the sequence is arithmetic or not. If it is, find the common
difference and the next three terms. Then graph.
-11, -4, 3, 10,…
Solution:
To find out if the sequence is arithmetic, there must be a common
difference between any two terms in the sequence. So that
d = a2 – a1 = -4 – (-11) = 7
= a3 – a2 = 3 – (-4) = 7
= a4 – a3 = 10 – 3 = 7
The sequence is arithmetic and the common difference is 7. The next
three terms are obtained by adding 7 to the preceding term, so that
a5 = a4 + 7 = 10 + 7 = 17
a6 = a5 + 7 = 17 + 7 = 24
a7 = a6 + 7 = 24 + 7 = 31
Example 1
-20
-10
0
10
20
30
40
0 2 4 6 8
n
an
18

19. 2. Write the first five terms of the arithmetic sequence with first term 5 and
common difference -2.
Solution:
The second term is found by adding -2 to the first term 5, getting 3. For the
next term, add -2 to 3, and so on. The first five terms are
5, 3, 1, -1, -3.
Remark: There is another way of finding the specified term of an arithmetic sequence
but it will be discussed in the next module. The same thing is true for the general
term of any arithmetic sequence.
Try this out
Determine whether the sequence is arithmetic or not. If it is, find the common
difference and the next three terms.
1. 2, 5, 8, 11,…
2. 2, -4, 6, -8, 10,…
3. -6, -10, -14, -18,…
4. 40, 42, 44, 46,…
5. 1.2, 1.8, 2.4,…
6. 1, 5, 9, 13,…
7. ,...
5
1
,
4
1
,
3
1
,
2
1
8. ,...8,7,6,5
9. 98, 95, 92, 89,…
10. 1,
3
5
,
3
4
, 2,…
Let’s Summarize
1. A sequence is a list of numbers in which order is important.
a1, a2, a3, a4, …, an, …
Each number in the list corresponds to each natural number.
2. A sequence may either be finite or infinite. A finite sequence has a specific
number of terms. An infinite sequence has an endless number of terms.
19

20. 3. To find the terms of a sequence given its rule, simply replace n with the
number of the specific term needed to be found.
4. Most sequences have a general term or rule that describes all the terms in
the sequence. There is no specific way of finding the general term of a
given sequence.
5. A sequence is defined recursively when the nth term can be found only
when the preceding term is found.
6. An arithmetic sequence is a sequence where each term is found by adding
a constant number, called the common difference, to the preceding term.
7. The Fibonacci sequence is a special recursively defined sequence.
What have you learned
A. Write the first five terms of the sequence.
1. an =
n
n 7+
3. an = 5n – 2
2. an = 3n - 1
B. Find the indicated term for the sequence.
4. an = -7n + 3; a8 6. an =
53
72
−
+
n
n
; a14
5. an = (n + 2)(2n - 3); a5
C. Find the general term, an, for the given terms of the sequence.
7. 3, 7, 11, 15,… 9. ,...
16
1
,
8
1
,
4
1
,
2
1
8. 0, -4, -8, -12,…
E. Find the first four terms of the sequence defined recursively.
10. a1= -1, an = 5an – 1 11. a1 = 6, an =
13
1
−
−
n
an
F. Find the common difference and the next three terms of the given
arithmetic sequence.
12. 1, 10, 19, 28,… 14. 1, 3, 5, 7,…
13. 5.5, 7, 8.5, 10,… 15. 43, 39, 35, …
20

21. 21

22. Answer Key
How much do you know
1
4,
5
8
,
4
7
,2,
2
5 9.
an =
nn +2
1
2 1, 2, 4, 8, 16 10 -1, -3, -9, -27
3 1, 5, 9, 13, 17 11
5, 5,
6
5
,
2
5
4 -70 12 d = 6; 26, 32, 38
5 78 13 d = 1.5; 13, 14.5, 16
6 49/23 14 d = 1; 5, 6, 7
7 an = 4n 15 d = -3; 15, 12, 9
8 an = -10n
Lesson 1
A.
B.
Because they can hold up traffic.
Lesson 2
A.
1 2, 3, 4, 5 12 1, -2, 3, -4
2 0, -2, -4, -6 13 0, 3, 8, 15
3 0, 1, 2, 3 14 -2, 4, -8, 16
4 2, 4, 8, 16 15
0,
4
63
,
3
26
,
2
7
5 3, 5, 7, 9 16
0,
4
15
,
3
8
,
2
3
6 2, 5, 10, 17 17
5
1
,
4
1
,
3
1
,
2
1
−−
7 2, 5, 8, 11 18
17
1
,
10
1
,
5
1
,
2
1
−−
1 F 3 I 5 I 7 F 9 F
2 F 4 F 6 I 8 F 10 F
22

23. 8
5
4
,
4
3
,
3
2
,
2
1 19
243
2
,
81
2
,
27
2
,
9
2
9 -1, -3, -5, -7 20
3
67
,10,
3
11
,
3
4
10
0,
4
15
,
3
8
,
2
3 21 4, 9, 16, 25
11 3, 9, 27, 81
B.
22 40 29
243
32
23 110 30 380
24 225 31 88
25
256
1 32
36
1−
26 15 33
20
1
27
13
12 34 94
28 24 35 38
Lesson 3
A.
B.
1 an = n + 1 10 an = n3
2 an = 3n 11 an = n2
3 an = 4n 12 an = 3n
4 an = n + 2 13 an = (-2)n
5 an = 3n + 4 14 an = (-3)n
6 an = 5n – 1 15
an = 1
2
1
+n
7 an = 3n2
16
an = n
3
1
8 an = 2n3
17
an = 2
)1( +n
n
1 an = n 6 an = 4n
2 an = -2n 7 an = -(n + 5)
3 an = 2n – 1 8 an = 2n + 9
4 an = -2n + 1 9 an = 6n + 2
5 an = 7n 10 an = -10(n + 1)+ 3
23

24. 9 an = 2n + 1
18
an =
13 +n
n
Lesson 4
1 3, 2, 0, -4, -12 4
1,
8
5
,
5
3
,
3
2
,
2
1
2
-8, -4, -2, -1, -
2
1 5 1, 2, 3, 5, 8
3 1, 3, 7, 15, 31 6 1, 1, 1, 3, 5
Lesson 5
1 Arithmetic d = 3 14, 17, 20
2 No
3 Arithmetic d = -4 -22, -26, -30
4 Arithmetic d = 2 48, 50, 52
5 Arithmetic d = 0.6 3, 3.6, 4.2
6 Arithmetic d = 4 17, 21, 25
7 No
8 No
9 Arithmetic d = 3 86, 83, 80
10 Arithmetic d = 1/3
3,
3
8
,
3
7
What have you learned
1
8,
5
12
,
4
11
,
3
10
,
2
9 9.
an = n
2
1
2 1, 3, 9, 27, 81 10 -1, -5, -25, -125
3 3, 8, 13, 18, 23 11
6,
220
3
,
20
3
,
5
6
4 -53 12 d = 9; 37, 46, 55
5 49 13 d = 1.5; 11.5, 13, 14.5
6 35/37 14 d = 2; 9, 11, 13
7 an = 4n – 1 15 d = -4; 31, 27, 23
8 an = 4(1 – n)
24