powerpoint presentation

1. BY: MS. G. MARTIN

2. LET’S LEARN!
The terms between any two non-consecutive terms of an
arithmetic sequence are called arithmetic means.
In the sequence 10, 20, 30, 40, 50, 60, 70, the terms 30, 40 and 50, are
called arithmetic means between 20 and 60.
Remember:
If A1, A2, …, An-1, An is an arithmetic sequence, then the numbers A2, …,
An-1, are the arithmetic means between A1 and An.

3. Example 1. Insert four arithmetic means between 18 and 43.
Steps Solutions
1. Identify the given terms.
18, ___, ___, ___, ___, 43
We have A1, = 18 and A6 = 43.
2. Solve for the common difference using the
formula
An = A1 + (n – 1)d
An = A1 + (n – 1)d
A6 = A1 + (6 – 1)d
43 = 18 + 5d
43 – 18 = 5d
25 = 5d
5 = d
3. Find the four terms between 18 and 43 using the
common difference.
(So, the four arithmetic means between 18
and 43 are 23, 28, 33 and 38)
If A1 = 18 and d = 5, then
A2 = A1 + d = 18 + 5 = 23
A3 = A2 + d = 23 + 5 = 28
A4 = A3 + d = 28 + 5 = 33
A5 = A4 + d = 33 + 5 = 38

4. Example 2. Insert five arithmetic means 8 and eleven.
Steps Solution
1. Identify the given terms.
8, ___, ___, ___, ___, ___, 11
Let A1 = 8 and A7 = 11 since there are five
terms between 8 and 11.
2. Solve for the common difference using
the formula
An = A1 + (n – 1)d.
If A1 = 8 and A7 = 11, we have
An = A1 + (n – 1)d
A7 = A1 + (7 – 1)d
11 = 8 + 6d
11 – 8 = 6d
3 = 6d
1
2
= d

5. 3. Find the five terms between 8 and
11 using the common difference.
If A1 = 8 and d =
1
2
, we have
A2 = A1 + d = 8 +
1
2
=
𝟏𝟕
𝟐
A3 = A2 + d =
17
2
+
1
2
=
18
2
= 9
A4 = A3 + d = 9 +
1
2
=
𝟏𝟗
𝟐
A5 = A4 + d =
19
2
+
1
2
= 10
A6 = A5 + d = 10 +
1
2
=
𝟐𝟏
𝟐
So, the five arithmetic means between 8 and 11 are
𝟏𝟕
𝟐
, 9,
𝟏𝟗
𝟐
, 10 and
𝟐𝟏
𝟐
.

6. Example 3. Insert three arithmetic means between 4 and 12.
Steps Solution
1. Identify the given terms.
4, ___, ___, ___, 12
A1 = 4, A5 = 12 and n = 5
2. Substitute the given in the formula of
arithmetic sequence.
An =A1 + (n – 1)d
12 = 4 + (5 – 1)d
12 = 4 + 4d
12 – 4 = 4d
8 = 4d
2 = d
3. Solve for the arithmetic means using the
common difference.
A1 = 4 and d = 2, then
A2 = A1 + d = 4 + 2 = 6
A3 = A2 + d = 6 + 2 = 8
A4 = A3 + d = 8 + 2 = 10

7. Example 4. How many numbers divisible by 8 are between 4 and 1,000?
Solution:
The common difference is 8.
Given: A1 = 8 and An = 992 (since 992 is the last number divisible by 8 before 1,000)
Substitute the given in the formula of arithmetic sequence.
An = A1 + (n – 1)d
992 = 8 + (n – 1)8
992 = 8 + 8n – 8
992 = 8n
124 = n

8. Thank You!