Arithmetic Mean and Series Prepared by: Darwin Joseph Santos Mark Joseph Salazar Cluadine Doma Maricon Hollon Randolph Sampang

1. Arithmetic Series
and
Arithmetic Mean
Prepared By:
Salazar, Mark Joseph
Sampang, Randolph Brian
Santos, Darwin Joseph
Doma, Cluadine
Hollon, Maricon

2. Arithmetic Series

3. Arithmetic Series
 A series such as (3 + 7 + 11 + 15 + ··· + 99
or 10 + 20 + 30 + ··· + 1000) which has a
constant difference between terms.
first term is a1
common difference is d
number of terms is n
sum of an arithmetic series is Sn
 An arithmetic series is the sum of an
arithmetic sequence.

4. Arithmetic Series
 Formula:
or
~when an is
given

5. Arithmetic Series
 Example #1:
3 + 7 + 11 + 15 + ··· + 99 has a1 = 3 and d = 4. To
find n, use the explicit formula for an arithmetic
sequence.
We solve:
3 + (n – 1)·4 = 99 to get n = 25.

6. Arithmetic Series
 Example #2:
Find the sum of the first 12 positive even
integers.
positive even integers: 2, 4, 6, 8, ...
n = 12; a1 = 2, d = 2
We are missing a12, for the sum formula so we
will use
= 12/2[2(2) + (12 – 1)2]
= 6[4 + 22]
= 6(26)
= 156

7. Arithmetic Series
 Activity:
Find the sum of each arithmetic series.
1. Find the sum of the sequence
-8, -5, -2, ..., 7
2. Find the sum of the first 10 positive integers
3. Find the sum of the first 20 terms of the
sequence 4, 6, 8, 10, ...
Answers:
1. -3
2. 55
3. 460

8. Arithmetic Mean

9. Arithmetic Mean
 The numbers between
arithmetic extremes are
called arithmetic mean,
found in an arithmetic
sequence wherein each
term is obtained by adding
a fixed value called the
common difference.
Example:
4, 7, 10, 13, 16
The arithmetic means
are 7, 10 and 13
9, 15, 21
The arithmetic mean
is 15

10. Let’s Try!
① Insert 3 arithmetic
means between 1
and 17
1, _ , _ , _ ,
17
a5 = 1 + (5-1)d
17 = 1 + (5-1)d
17 = 1 + 4d
17 - 1 = 4d
16 = 4d
d = 4
an = a1 + (n-1)d a2 = a1 + d
a2 = 1 + 4
a2 = 5
a3 = a1 + 2d
a3 = 1 + (2)4
a3= 9
a4 = a1 + 3d
a4= 1 + (3)4
a4= 13
② Insert arithmetic
means between 95
and 185
95, _ , 185
an = a1 + (n-1)d
185 = 95 + (3-1)d
185 = 95 + 2d
185 - 95 = 2d
90 = 2d
d = 45
a2 = a1 + d
a2 = 95 + 45
a2 = 140

11. Word Problem
③ John recruited 2 persons for the networking
business. After a week, he recruited 5 persons again
and on the 5th week of recruitment, he recruited
another 14 persons for the networking business. If
this continues, how many persons did John already
recruited after the 6th week of recruitment?
an = ?
a1 = 2
d = 3
n = 6
Sn = n/2 [2a1 + (n - 1)d]
S6= 6/2 [2(2) + (6 - 1)3]
S6= 3 [4 + (5)3]
S6 = 3 [4 + (15)]
S6 = 3 [19]
S6 = 57
John already recruited 57 persons after 6 weeks of
recruitment