This document defines arithmetic sequences and series. It provides examples of arithmetic sequences where each term is obtained by adding a constant to the previous term. Formulas are given for finding the nth term and the sum of the first n terms of an arithmetic sequence. Examples are worked through of finding missing terms, the nth term, and the sum of sequences. Arithmetic means are also defined and formulas are provided for finding the average of two numbers and the average of a data set.

1. ARITHMETIC SEQUENCE
and SERIES

2. CONSIDER THE FOLLOWING LISTS OF NUMBERS:
2, 4, 6, 8, 10, . . .
Each term is obtained by adding 2 to the previous
term.
15, 12, 9, 6, 3, . . .
Each term is obtained by adding -3 to the previous
term.

3. FIND THE MISSING NUMBER
1. 23, 38, 53, ____, 83, 98
63 68 73 78
2. 45, 37, ____, 21, 13, 5
27 28 29 30

4. It is a sequence of
numbers where each
term after the first
term is obtained by
adding the same
constant(always the
same).
CONSTANT
-it is called common
difference, it denoted
with letter d.

5. EXAMPLES!
5, 10, 15, 20, 25, . . . D = 5
3, 7, 11, 15, . . . D = 4
20, 18, 16, 14, . . . D = 4

6. Formula for the nth term of an Arithmetic
Sequence
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝑾𝒉𝒆𝒓𝒆;
𝒂 𝟏 = 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒊𝒓𝒔𝒕 𝒕𝒆𝒓𝒎
𝒂 𝒏 = 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒏𝒕𝒉 𝒕𝒆𝒓𝒎
𝒏 = 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒆𝒓𝒎 𝒕𝒐 𝒇𝒊𝒏𝒅
𝒅 = 𝒕𝒉𝒆 𝒄𝒐𝒎𝒎𝒐𝒏 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆

7. FIND THE NTH TERM OF THE
ARITHMETIC SEQUENCE
5, 8, 11, 14, . . .
We know that 𝒂 𝟏 = 𝟓 𝒅 = 𝟑
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝒂 𝒏 = 𝟓 + 𝟑𝒏 − 𝟑
𝒂 𝒏 = 𝟓 + 𝒏 − 𝟏 𝟑
𝒂 𝒏 = 𝟐 + 𝟑𝒏

8. FIND THE 18TH TERM OF
THE ARITHMETIC
SEQUENCE
21, 24, 27, 30, 33, . . .
We know that 𝒂 𝟏 = 𝟐𝟏, 𝒅 = 𝟑, 𝒏 = 𝟏𝟖
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝒂 𝒏 = 𝟐𝟏 + (𝟏𝟕) 𝟑
𝒂 𝒏 = 𝟐𝟏 + 𝟏𝟖 − 𝟏 𝟑
𝒂 𝒏 = 𝟕𝟐
𝑻𝒉𝒖𝒔, 𝟕𝟐 𝒊𝒔 𝒕𝒉𝒆 𝟏𝟖𝒕𝒉
𝒕𝒆𝒓𝒎 𝒐𝒇 𝒕𝒉𝒆
𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒆

9. IN THE ARITHMETIC
SEQUENCE BELOW WHICH
TERM IS 124?
We know that 𝒂 𝟏 = 𝟏𝟎, 𝒅 = 𝟔, 𝒂 𝒏 = 𝟏𝟐𝟒
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟐𝟒 = 𝟏𝟎 + 𝟔𝒏 − 𝟔
𝟏𝟐𝟒 = 𝟏𝟎 + 𝒏 − 𝟏 𝟔
𝒏 = 𝟏𝟐𝟎
𝑻𝒉𝒖𝒔, 𝟏𝟐𝟒 𝒊𝒔 𝒕𝒉𝒆 𝟐𝟎𝒕𝒉
𝒕𝒆𝒓𝒎 𝒐𝒇 𝒕𝒉𝒆
𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒆
𝟏𝟐𝟒 − 𝟒 = 𝟔𝒏

10. We know that 𝒂 𝟏 = 𝟕, 𝒂 𝟓 = 𝟏𝟗,
𝒅 = 𝟑
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟗 = 𝟕 + 𝟓 − 𝟏 𝒅
𝟏𝟗 = 𝟕 + 𝟒𝒅
𝟏𝟗 − 𝟕 = 𝟒𝒅
𝟏𝟐 = 𝟒𝒅
𝒂 𝟏𝟓 = 𝟕 + 𝟏𝟓 − 𝟏 𝟑
𝒂 𝟏𝟓 = 𝟒𝟗
𝑻𝒉𝒖𝒔, 𝒕𝒉𝒆 𝟏𝟓𝒕𝒉
𝐭𝐞𝐫𝐦 𝐢𝐬 𝟒𝟗.

11. We know that 𝒂 𝟑𝟑 = 𝟖𝟎, 𝒅 = 𝟐, 𝒏 = 𝟑𝟑
𝒂 𝟏 = 𝟏𝟔
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝟖𝟎 = 𝒂 𝟏 + 𝟑𝟑 − 𝟏 𝟐
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝒕𝒉𝒆𝒓𝒆 𝒂𝒓𝒆
𝟏𝟔 𝒔𝒆𝒂𝒕𝒔 𝒊𝒏 𝒕𝒉𝒆
𝒇𝒊𝒓𝒔𝒕 𝒓𝒐𝒘.
𝟖𝟎 = 𝒂 𝟏 + 𝟑𝟐 𝟐
𝟖𝟎 = 𝒂 𝟏 + 𝟔𝟒
𝟖𝟎 − 𝟔𝟒 = 𝒂 𝟏

12. •1, 2, 3, 4, . . . And 1+2+3+4+…
•2, 4, 6, 8, . . . And 2+4+6+8+…
Each indicated sum of the terms of
an arithmetic sequence is an
ARITHMETIC SERIES.

13. 𝑺 𝒏 =
𝒏
𝟐
(𝒂 𝟏+𝒂 𝒏)
𝑺 𝒏 =
𝒏
𝟐
[𝟐𝒂 𝟏+(݊−1)݀]

14. 𝑺 𝒏 =
𝒏
𝟐
(𝒂 𝟏+𝒂 𝒏)
Find the sum of the first 100 positive integers.
𝒂 𝟏 = 𝟏, 𝒂 𝒏 = 𝟏𝟎𝟎, 𝒏 = 𝟏𝟎𝟎
𝑺 𝟏𝟎𝟎 =
𝟏𝟎𝟎
𝟐
(𝟏 + 𝟏𝟎𝟎)
𝑺 𝟏𝟎𝟎 = 𝟓, 𝟎𝟓𝟎

15. Find the sum of the first
12 terms of the
arithmetic sequence 50,
47, 44, 41, 38, . . .
𝒂 𝟏 = 𝟓𝟎, 𝒅 = −𝟑, 𝒏 = 𝟏𝟐
𝑺 𝟏𝟐 =
𝟏𝟐
𝟐
[2(50) + (12−1) − 𝟑]
𝑺 𝟏𝟐 = 𝟒𝟎𝟐
𝑺 𝟏𝟐 = 𝟔(𝟏𝟎𝟎 − 𝟑𝟑)
𝑺 𝒏 =
𝒏
𝟐
[𝟐𝒂 𝟏+(݊−1)݀]

16. Find the sum of the first
18 terms of the
arithmetic sequence 3,
5, 7, 9, 11, . . .
𝒂 𝟏 = 𝟑, 𝒅 = 𝟐, 𝒏 = 𝟏𝟖
𝑺 𝟏𝟖 =
𝟏𝟖
𝟐
[2(3) + (18−1)𝟐]
𝑺 𝟏𝟖 = 𝟒𝟎
𝑺 𝒏 =
𝒏
𝟐
[𝟐𝒂 𝟏+(݊−1)݀]
𝑺 𝟏𝟖 = 𝟗(𝟔 + 𝟑𝟒)

17. Find the sum of the
first 30 natural
numbers.
𝒂 𝟏 = 𝟏, 𝒂 𝒏 = 𝟑𝟎, 𝒏 = 𝟑𝟎
𝑺 𝟑𝟎 =
𝟑𝟎
𝟐
(𝟏 + 𝟑𝟎)
𝑺 𝟑𝟎 = 𝟒𝟔𝟓
𝑺 𝒏 =
𝒏
𝟐
(𝒂 𝟏+𝒂 𝒏)
𝑺 𝟑𝟎 = 𝟏𝟓(𝟑𝟏)

18. QUIZ!
IN YOUR
NOTEBOOK
ANSWER
PAGE 27
TRY IT 2
TO BE
SUBMITTE
D TODAY!!

19. It is the terms
between any two
terms.
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅

20. Insert 3 arithmetic
means between 18
and 30.
𝒂 𝟏 = 𝟏𝟖, 𝒂 𝒏 = 𝟑𝟎, 𝒏 = 𝟓
𝟑𝟎 = 𝟏𝟖 + 𝟓 − 𝟏 𝒅
𝒅 = 𝟑
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝟑𝟎 = 𝟏𝟖 + 𝟒𝒅
𝟑𝟎 − 𝟏𝟖 = 𝟒𝒅
𝟏𝟐 = 𝟒𝒅 𝟏𝟖, 𝟐𝟏, 𝟐𝟒, 𝟐𝟕, 𝟑𝟎

21. Insert 4 arithmetic
means between 7 and
37.
𝒂 𝟏 = 𝟕, 𝒂 𝒏 = 𝟑𝟕, 𝒏 = 𝟔
𝟑𝟕 = 𝟕 + 𝟔 − 𝟏 𝒅
𝒅 = 𝟔
𝒂 𝒏 = 𝒂 𝟏 + 𝒏 − 𝟏 𝒅
𝟑𝟕 = 𝟕 + 𝟓𝒅
𝟑𝟕 − 𝟕 = 𝟓𝒅
𝟑𝟎 = 𝟓𝒅 𝟕, 𝟏𝟑, 𝟏𝟗, 𝟐𝟓, 𝟑𝟏, 𝟑𝟕

22. The arithmetic mean of a number (or
average) of the number x and y given by 𝐱 =
𝒙+𝒚
𝟐
.
The arithmetic mean (or average) of a set of
numbers 𝒙 𝟏, 𝒙 𝟐, 𝒙 𝟑, . . . 𝒙 𝒏,is given by
𝐱 =
𝒙 𝟏, 𝒙 𝟐, 𝒙 𝟑, ...𝒙 𝒏,
𝒏
=
⅀𝒙
𝒏
.

23. FIND THE ARITHMETIC MEAN BETWEEN 8
AND 20.
𝒙 =
𝒙 + 𝒚
𝟐
𝒙 =
𝟖 + 𝟐𝟎
𝟐
𝒙 =
𝟐𝟖
𝟐
𝒙 = 𝟏𝟒

24. FIND THE ARITHMETIC MEAN BETWEEN -5
AND 7.
𝒙 =
𝒙 + 𝒚
𝟐
𝒙 =
−𝟓 + 𝟕
𝟐
𝒙 =
𝟐
𝟐
𝒙 = 𝟏

25. DOMINIC’S SCORE IN FIVE MATH TEST ARE 94, 90,
96, 93 AND 95. WHAT IS HIS AVERAGE SCORE?
𝒙 =
⅀𝒙
𝒏
=
𝟗𝟒 + 𝟗𝟎 + 𝟗𝟔 + 𝟗𝟑 + 𝟗𝟓
𝟓
=
𝟒𝟔𝟖
𝟓
𝒙 = 𝟗𝟑. 𝟔

26. WHAT IF DOMINIC’S SCORES WERE 88, 86, 87,
AND 84?
𝒙 =
⅀𝒙
𝒏
=
𝟖𝟖 + 𝟖𝟔 + 𝟖𝟕 + 𝟖𝟒
𝟒
=
𝟑𝟒𝟓
𝟒
𝒙 = 𝟖𝟔. 𝟐𝟓

27. ASSIGNMENTS!
IN YOUR NOTEBOOK
ANSWER PAGE 29
PRACTICE and
APPLICATION
I. # 1-5
II.# 11-15
III.# 21-25
TO BE
SUBMITTED
ON SUNDAY
(OCTOBER 11,
2020)!!