Find the next sequence of entries in the 11th row of the Pascal’s
triangle.
An ice cream parlor has 12 flavors of ice cream and with
toppings chocolate, nuts, chips, raisins, M&M and crushed.
How many different sets of
a. Exactly 3 toppings could you have?
b. Exactly 4 toppings
c. Any number of toppings ( up to 6 available)?
Arithmetic
Sequences
and Series
Sequences
List with commas
3, 8, 13, 18
Series
“Indicated sum”
3 + 8 + 13 + 18
An Arithmetic Sequence is
defined as a sequence in
which there is a common
difference between
consecutive terms.
Which of the following sequences are
arithmetic? Identify the common
difference.
3,  1, 1, 3, 5, 7, 9, . . .
YES
15.5, 14, 12.5, 11, 9.5, 8, . . .
d 2
YES d  1.5
84, 80, 74, 66, 56, 44, . . . NO
8, 6,  4, 2, 0, . . .
NO
50,  44,  38,  32,  26, . . . YES d  6
26,  21,  16,  11,  6, . . .
The general form of an ARITHMETIC sequence.
First Term:
a1
Second Term:
a2  a1  d
Third Term:
a3  a1  2d
Fourth Term:
a4  a1  3d
Fifth Term:
a5  a1  4d
nth Term:
an  a1   n 1 d
Formula for the nth term of an ARITHMETIC
sequence.
an  a1   n 1 d
an  The nth term
a1  The 1st term
n  The term number
d  The common difference
Given: 79, 75, 71, 67, 63, . . .
Find: a32
IDENTIFY
a1  79
SOLVE
an  a1   n  1 d
d  4
a32  79   32  1 4 
n  32
a32  45
Given: 79, 75, 71, 67, 63, . . .
Find: What term number is -169?
IDENTIFY
a1  79
d  4
an  169
SOLVE
an  a1   n  1 d
169  79   n  1 4 
n  63
Given:
a10  3.25
Find: a1
a12  4.25
Why?!!?
What’s the real question?
Letting 3.25 and 4.25 in new
position to find the difference
a1  3.25
a3  4.25
n3
The Difference
SOLVE
an  a1   n  1 d
4.25  3.25   3  1 d
d  0.5
The common difference is always the difference between any term and the term
that proceeds that term.
Given:
a10  3.25
Find: a1
a12  4.25
IDENTIFY
Use the original position of the
term to find the first term
SOLVE
an  a1   n  1 d
a10  3.25
d  0.5
3.25  a1  10  1 0.5
n  10
a1  1.25
You can use either the two given 3.25 or 4.25 to find the first term.
-1.25 -0.75 -0.25
0.25
0.75
1.25
1.75
2.25
2.75
3.25
3.75
4.25
When a garbage truck starts collecting rubbish it first stops at a corner
store where it collects 86 kg of rubbish. It then travels down a long
suburban street where it picks up 40 kg of rubbish at each house.
a. How much garbage would be carried by the truck after: i 15 pick-ups? ii
27 pick-ups?
b. The maximum amount of garbage that can be carried by the truck is
1500 kg. After picking up from the corner store, what is the maximum
number of houses it can pick up from before it is fully loaded?
Answers: i.
50
  73  2 p   71 69  67  . . .   25   27
p 1
71  69  67  . . .   25   27 
 27    25  . . .  67  69  71
44  44  44  . . .  44  44  44
50 Terms

50  71   27  
2
 1100
71 + (-27) Each sum
is the same.
a1   a1  d    a1  2d   . . .   a1   n  1 d 
 a   n 1 d   . . .   a
1
1
 2d    a1  d   a1
 a  a   n 1 d     a  a   n 1 d    . . .   a  a   n 1 d  
1
s
1
1
n  a1  an 

2
1
1
1
S  Sum

n  Number of Terms


a1  First Term


an  Last Term

S
Find the sum of the terms of this
arithmetic series. 35
 29  3k 

n  a1  an 
k 1
2
n  35
a1  26
a35  76
35  26  76 
S
2
S  875
Find the sum of the terms of this arithmetic
series.
151  147  143  139  . . .   5
n  a1  an 
S
2
n  40
a1  151
a40  5
What term is -5?
an  a1   n  1 d
5  151   n  1 4 
n  40
40 151  5 
S
2
S  2920
Substitute an  a1   n 1 d
n  a1  an 
S
2
S
S
n  a1  a1   n  1 d 
2
n  2a1   n  1 d 
2
n  # of Terms

 a1  1st Term
 d  Difference

36
Find the sum of this series
  2.25  0.75 j 
j 0
 2.25  3  3.73  4.5  . . .
S
n  2a1   n  1 d 
n  37
2
37  2  2.25   37  1 0.75 
a1  2.25
S
d  0.75
S  582.75
2
35
  45  5i 
n  a1  an 
S
2
n  35 a1  40 an  130
35  40  130 
S
2
S  1575
i 1
S
n  2a1   n  1 d 
2
n  35 a1  40 d  5
S
35  2  40    35  1 3
S  1575
2
An introduction…………
1, 4, 7, 10, 13
35
2, 4, 8, 16, 32
62
9, 1,  7,  15
12
9,  3, 1,  1/ 3
20 / 3
6.2, 6.6, 7, 7.4
27.2
,   3,   6
3  9
1, 1/ 4, 1/16, 1/ 64 85 / 64
9.75
, 2.5, 6.25
Arithmetic Sequences
Geometric Sequences
ADD
To get next term
MULTIPLY
To get next term
Arithmetic Series
Sum of Terms
Geometric Series
Sum of Terms
Find the next four terms of –9, -2, 5, …
Arithmetic Sequence
2  9  5  2  7
7 is referred to as the common difference (d)
Common Difference (d) – what we ADD to get next term
Next four terms……12, 19, 26, 33
Find the next four terms of 0, 7, 14, …
Arithmetic Sequence, d = 7
21, 28, 35, 42
Find the next four terms of x, 2x, 3x, …
Arithmetic Sequence, d = x
4x, 5x, 6x, 7x
Find the next four terms of 5k, -k, -7k, …
Arithmetic Sequence, d = -6k
-13k, -19k, -25k, -32k
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
d  common difference
nth term of arithmetic sequence  an  a1  n  1 d
sum of n terms of arithmetic sequence  Sn 
n
 a1  an 
2
Given an arithmetic sequence with a15  38 and d  3, find a1.
x
a1  First term
38
an  nth term
15
n  number of terms
NA Sn  sum of n terms
-3
d  common difference
an  a1  n  1 d
38  x  15  1 3 
X = 80
Find S63 of  19,  13, 7,...
-19 a1  First term
353
??
an  nth term
n  number of terms
63
x
Sn  sum of n terms
6
d  common difference
an  a1  n  1 d
??  19   63  1 6 
??  353
n
 a1  an 
2
63

 19  353 
2
Sn 
S63
S63  10521
Try this one: Find a16 if a1  1.5 and d  0.5
1.5 a1  First term
x
16
an  nth term
n  number of terms
NA Sn  sum of n terms
0.5
d  common difference
an  a1  n  1 d
a16  1.5  16  1 0.5
a16  9
Find n if an  633, a1  9, and d  24
9
a1  First term
633 an  nth term
x
n  number of terms
NA Sn  sum of n terms
24
d  common difference
an  a1  n  1 d
633  9   x  1 24
633  9  24x  24
X = 27
Find d if a1  6 and a29  20
-6
a1  First term
20 an  nth term
29
n  number of terms
NA Sn  sum of n terms
x
d  common difference
an  a1  n  1 d
20  6   29  1 x
26  28x
13
x
14
Find two arithmetic means between –4 and 5
-4, ____, ____, 5
-4
a1  First term
5
an  nth term
n  number of terms
4
NA
x
Sn  sum of n terms
d  common difference
an  a1  n  1 d
5  4   4  1 x 
x 3
The two arithmetic means are –1 and 2, since –4, -1, 2, 5
forms an arithmetic sequence
Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
1
a1  First term
4
an  nth term
5
NA
x
n  number of terms
Sn  sum of n terms
d  common difference
an  a1  n  1 d
4  1   5  1 x 
3
x
4
The three arithmetic means are 7/4, 10/4, and 13/4
since 1, 7/4, 10/4, 13/4, 4 forms an arithmetic sequence
Find n for the series in which a1  5, d  3, Sn  440
5
a1  First term
y
an  nth term
x
n  number of terms
440 Sn  sum of n terms
3
d  common difference
an  a1  n  1 d
y  5   x  1 3
x
440   5  5   x  1 3 
2
x  7  3x 
440 
2
880  x  7  3x 
0  3x 2  7x  880
Graph on positive window
X = 16
n
Sn   a1  an 
2
x
440   5  y 
2
The sum of the first n terms of an infinite sequence
is called the nth partial sum.
Sn  n (a1  an)
2
Example 6. Find the 150th partial sum of the arithmetic sequence, 5,
16, 27, 38, 49, …
a1  5
d  11
 c  5  11  6
an  11n  6  a150  11150  6  1644
S150
150

 5  1644   75 1649   123,675
2
Example 7. An auditorium has 20 rows of seats. There are 20 seats in
the first row, 21 seats in the second row, 22 seats in the third row, and
so on. How many seats are there in all 20 rows?
d 1
c  20  1  19
an  a1   n 1 d  a20  20  19 1  39
20
S 20   20  39   10  59   590
2
Example 8. A small business sells $10,000 worth of sports memorabilia
during its first year. The owner of the business has set a goal of
increasing annual sales by $7500 each year for 19 years. Assuming that
the goal is met, find the total sales during the first 20 years this business
is in operation.
a1  10,000
d  7500
c  10,000  7500  2500
an  a1   n 1 d  a20  10,000  19  7500  152,500
20
S20  10,000  152,500   10 162,500   1,625,000
2
So the total sales for the first 2o years is $1,625,000
Geometric progressions
We shall now move on to the other type of sequence we want
to explore.
Consider the sequence
2, 6, 18, 54, ... .
Here, each term in the sequence is 3 times the previous term.
And in the sequence 1, −2, 4, −8, ... , each term is ___ times
the previous term.
Sequences such as these are called geometric progressions,
or GPs for short.
Exercises
(a) Write down the first five terms of the geometric progression
which has first term 1 and common ratio 1/2.
(a) Find the 10th and 20th terms of the GP with first term 3 and
common ratio 2.
The sum of a geometric series
Find the sum of the geometric series
2 + 6 + 18 + 54 + ...
where there are 16 terms in the series.
For this series, we have a = 2, r = 3 and n = 6.
Answer
Find the sum of the geometric series
8, −4, 2, −1 + ...
where there are 5 terms in the series.
Solution For this series, we have a = 8, r = and n = 5.
Answer 5 ½
How many terms are there in the geometric progression
2, 4, 8, ..., 128?
Solution
In this sequence a = 2 and r = 2. We also know that the n-th
term is 128.
Answer n = 7
Example
How many terms are there in the geometric progression
2, 4, 8, ..., 128?
Answer n = 7.
Use the rule to determine the 8th and 11th term of the
geometric sequence
100,50,25,12.5,...
The 8th term is ----
A fisherman harvested 350 kg of fishes on Monday. From Monday
to Friday, the amount of fishes, he harvested increased by 10%
per day. What’s the total amount of fishes did the fisherman
harvest in the first five days? Round your answer to the nearest
whole number.
A population of a bacteria doubles its numbers every minute.
If we start off with five bacteria, how many will we have at the
start of the 31st minute (that is, after half an hour)?
Find the sum of the arithmetic series.
1)-4 - 1 + 2 + 5 + 8 + 11 + 14
2)13 + 15 + 17 + 19 + ... + 31
3) 247 + 245 + 243 + 241 + ... + 229
4) -16 - 11 - 6 - 1 + ... + 39
Find the sum of the geometric series.
1. 4/3 + 16/3 + 64 /3 + 256/ 3 + 1024 /3
2. 3 + 9 + 27 + 81 + 243
3. 3/ 2 + 3/ 8 + 3 /32 + 3 /128 + 3 /512
The sum of the numbers in arithmetic progression is 24. If
the first number is decreased by 1 and the second
decreased by 2, the three numbers are in a geometric
progression. Find the three numbers.
Find the sum of the geometric series.
1. 4/3 + 16/3 + 64 /3 + 256/ 3 + 1024 /3
2. 3 + 9 + 27 + 81 + 243
3. 3/ 2 + 3/ 8 + 3 /32 + 3 /128 + 3 /512
4. How many terms are there in the geometric progression
2.3, 10.35, 46.575,+ ... + 158,590,804.2 ?
5. A population of a bacteria doubles its numbers every minute.
If we start off with five bacteria, how many will we have at the
start of the 31st minute (that is, after half an hour)?
7. In a geometric sequence, the 4th term is 24 and the 9th term
is 768. Find three terms of the sequence.