1) The document discusses geometric sequences and their properties including defining the common ratio, finding the nth term, and identifying whether example sequences are geometric. 2) It provides examples of calculating the common ratio, next terms, and determining if sequences are geometric. 3) An example problem is worked out finding the number infected on the 6th day of a measles outbreak where the number doubles every day.

1. Grade 10 – Mathematics
Quarter I
GEOMETRIC SEQUENCE

2. Objectives:
• define geometric sequence;
• identify the common ratio and the next term of
the sequence;
• find the nth term of the geometric sequence; and
• solve problems involving geometric sequence.

3. Find the ratio of the second number to the first number.
2, 81 𝑟 =
8
2
= 𝟒
-3, 92 𝑟 =
9
−3
= −𝟑
1,
1
2
3
𝑟 =
1
2
1
=
𝟏
𝟐

4. A sequence is geometric if there exist
a number r, called the common ratio.
The common ratio, r can be
determined by dividing any term in
the sequence by the term that
precedes it.

5. Identify the common ratio and the next term in the
following sequences.
1, 2, 4, 8, …1 𝑟 =
2
1
= 2
The next term is 16
since 8(2) = 16.
80, 20, 5, …2 𝑟 =
20
80
=
1
4
The next term is
5
4
since 5(
1
4
) =
5
4
.
2, -8, 32, -1283 𝑟 =
−8
2
= −4 The next term is 512
since -128(-4) = 512.

6. State whether each of the following sequences is
geometric or not.
1
2
3
5, 20, 80, 320, …
7 2, 5 2, 3 2, 2, …
5, -10, 20, -40, …
4 10
3
,
10
6
,
10
9
,
10
15
20
5
= 4
80
20
= 4
320
80
= 4
5 2
7 2
=
5
7
3 2
5 2
=
3
5
−10
5
= −2
20
−10
= −2
ൗ10
6
ൗ10
3
=
1
2
ൗ10
9
ൗ10
6
=
2
3
ൗ10
15
ൗ10
9
=
3
5

7. The nth term of a geometric
sequence is given by
𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏
, 𝑟 ≠ 0.
where 𝑎1 is the first term
and r is the common ratio.

8. What is the 10th term of the geometric
sequence 8, 4, 2, 1, …?
𝑟 =
4
8
=
1
2
𝑎 𝑛 = 8
𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏
𝑎10 = 8
1
2
10−1
= 8
1
2
9
𝑎10 =
8
512
= 8
1
512
=
𝟏
𝟔𝟒

9. Find the missing term in 3, 12, 48, ____, _____
𝑟 =
12
3
= 4 48(4) = 192
192
192(4) = 768
768
Find the missing term in __, __, 32, 64, 128, …
𝑟 =
64
32
= 2
32
2
= 16
16
16
2
= 8
8

10. During the initial phase of an outbreak of measles, the
number of infections can grow geometrically. If there
were 4, 8, 16, … on the first three days of an outbreak of
measles, how many will be infected on the 6th day?
𝑟 =
8
4
= 2
𝑎 𝑛 = 4
𝑛 = 6
𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏
𝑎6 = 4 2 6−1
𝑎6 = 4 2 5
𝑎6 = 4(32)= 128
There will be 128 people infected with measles on the 6th day.