This document discusses geometric sequences and geometric means. It defines a geometric sequence as a sequence where each term after the first is the product of the preceding term and a fixed number called the common ratio. It provides the formula for calculating the nth term and the sum of the first n terms of a geometric sequence. The terms between the first and last term of a geometric sequence are called the geometric means. It includes sample problems demonstrating how to find specific terms, the common ratio, the first term, geometric means, and the sum of terms for various geometric sequences.

1. GEOMETRIC
SEQUENCES
and
GEOMETRIC
MEANS

2. GEOMETRIC SEQUENCE
 Consider the sequence 2, 6, 18, 54, 162,…,
in which each term (after the first) can be
found by multiplying the preceding term
by 3. That is,
the second term = the first term x 3
the third term = the second term x 3
and so forth. A sequence like this is given a
special name.

3.  A geometric sequence is a sequence in
which every term after the first is the product
of the preceding term and the fixed number
called the common ratio of the sequence.
 We will use the following notations for
geometric sequence:
𝑎1 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
𝑎 𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚
𝑟 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜
𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑎1 𝑡𝑜 𝑎 𝑛
𝑆 𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛 𝑡𝑒𝑟𝑚𝑠

4.  For example, the geometric sequence 1, ½, ¼, 1/8,
1/16,…, where 𝑎1 = 1 and 𝑟 = 1/2 (to obtain each
succeeding term multiply the preceding one by ½.
Thus,
𝑎2 = 𝑎1 ∙ 𝑟 = 1 ∙
1
2
=
1
2
𝑎3 = 𝑎2 ∙ 𝑟 =
1
2
∙
1
2
=
1
4
𝑎4 = 𝑎3 ∙ 𝑟 =
1
4
∙
1
2
=
1
8
𝑎5 = 𝑎4 ∙ 𝑟 =
1
8
∙
1
2
=
1
16
and so on. Notice that if we divide any term by the
preceding one, the quotient or ratio is always ½. This is
why we call 𝑟 as the common ratio of the sequence

5. FORMULA FOR THE nth TERM
 As with arithmetic sequences, there is a formula for
calculating the nth term of a geometric
sequence. Let us calculate several terms of an
arbitrary geometric sequence.
1𝑠𝑡 𝑡𝑒𝑟𝑚 = 𝑎1 = 𝑎1 𝑟0
2𝑛𝑑 𝑡𝑒𝑟𝑚 = 𝑎2 = 𝑎1 𝑟1
3𝑟𝑑 𝑡𝑒𝑟𝑚 = 𝑎3 = 𝑎1 𝑟2
4𝑡ℎ 𝑡𝑒𝑟𝑚 = 𝑎4 = 𝑎1 𝑟3
5𝑡ℎ 𝑡𝑒𝑟𝑚 = 𝑎5 = 𝑎1 𝑟4
Thus, we have
𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 = 𝑎 𝑛 = 𝑎1 𝑟 𝑛−1

6. Sample Problem
1. Find the 7th and 10th terms of the following geometric
sequence.
9, 3, 1,
1
3
,
1
9
, … …
Answer:
𝑎1 = 9 and 𝑟 =
1
3
𝑎 𝑛 = 𝑎1 𝑟 𝑛−1
𝑎7 = 9
1
3
7−1
= 9
1
3
6
=
1
81
𝑎10 = 9
1
3
10−1
= 9
1
3
9
=
1
2187
Therefore,
1
81
and
1
2187
are the 7th and 9th terms of the
sequence, respectively.

7. SUM OF THE FIRST n TERMS
 We now derive a formula for calculating the sum of the first
𝑛 terms of a geometric sequence. Let 𝑆 𝑛 denote the sum of
the first 𝑛 terms. The series is
𝑆 𝑛 = 𝑎1 + 𝑎1 𝑟 + 𝑎1 𝑟2 + 𝑎1 𝑟3 + ⋯ + 𝑎1 𝑟 𝑛−2 + 𝑎1 𝑟 𝑛−1 eq. 1
Multiply the equation by 𝑟, we have
𝑟𝑆 𝑛 = 𝑎1 𝑟 + 𝑎1 𝑟2 + 𝑎1 𝑟3 + ⋯ + 𝑎1 𝑟 𝑛−2 + 𝑎1 𝑟 𝑛−1 + 𝑎1 𝑟 𝑛 eq. 2
Subtracting eq. 1 and eq. 2, and solving for 𝑆 𝑛, we have
𝑆 𝑛 =
𝑎1−𝑎1 𝑟 𝑛
1−𝑟
𝑜𝑟 𝑆 𝑛 =
𝑎1 1−𝑟 𝑛
1−𝑟

8. Sample Problem
1. Find the 8th term and the sum of the first 8 terms of the
geometric sequence -2, 1, -1/2, ¼… .
Answer:
𝑎1 = −2 and 𝑟 =
−1
2
𝑎 𝑛 = 𝑎1 𝑟 𝑛−1
𝑎8 = −2(
−1
2
)8−1
= −2
−1
2
7
=
1
64
𝑆 𝑛 =
𝑎1−𝑎1 𝑟 𝑛
1−𝑟
𝑆8 =
−2− −2
−1
2
8
1−
−1
2
= −
85
64
Therefore,
1
64
and −
85
64
are the 8th term and the sum of the first 8
terms, respectiveley.

9. GEOMETRIC MEANS
 The terms between 𝑎1 and 𝑎 𝑛 of a
geometric sequence are called
geometric means of 𝑎1 and 𝑎 𝑛 . If the
geometric sequences are 𝑎1 to 𝑎5, then
the geometric means are 𝑎2, 𝑎3 and 𝑎4.

10. Sample Problem
1. Insert four geometric means between 5 and -160.
Answer: Since we must insert four numbers between 5 and -160, there
are six numbers in the sequence. This means that 𝑎1 = 5 and 𝑎6 = −160
and we must find 𝑎2, 𝑎3, 𝑎4 and 𝑎5.
𝑎 𝑛 = 𝑎1 𝑟 𝑛−1
−160 = 5𝑟6−1
−160 = 5𝑟5
𝑟 = 2
𝑎2 = 𝑎1 𝑟 = 5 −2 = −10
𝑎3 = 𝑎2 𝑟 = −10 −2 = 20
𝑎4 = 𝑎3 𝑟 = 20 −2 = −40
𝑎5 = 𝑎4 𝑟 = −40 −2 = 80
Thus, the four geometric means between 5 and -160 are -10, 20, -40
and 80.

11. 2. Find 𝑎1 and 𝑟 for a geometric sequence that has
𝑎2 = 10 and 𝑎5 = 80.
Answer:
80 = 𝑎5 = 𝑎1 𝑟5−1
10 = 𝑎2 = 𝑎1 𝑟2−1
80
10
=
𝑎5
𝑎2
=
𝑎1 𝑟4
𝑎1 𝑟1
80
10
= 𝑟3
, 𝑟3
= 8, 𝑟 = 2
then, 10 = 𝑎2 = 𝑎1 𝑟
10 = 𝑎1 2 𝑎1 = 5
Therefore, 5 and 2 are 𝑎1 and 𝑟, respectively.

12. Break a leg!
1. Write the first five terms and the sum of
the first five of the geometric sequence
having 𝑎1 = 32 and 𝑟 =
1
4
.
2. If 𝑎2 =
1
3
, 𝑎5 = −9, solve 𝑟 and 𝑎1.

13. THANK YOU VERY MUCH!!!
PROF. DENMAR ESTRADA MARASIGAN