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