1. The sequence is a geometric sequence with first term a1 = 3 and ratio r = 3. The specific formula is an = 3(3)n-1. The sum of the first 20 terms is 1743392200 and the infinite sum is 3072. 2. The sequence is a geometric sequence with first term a1 = 1/90 and ratio r = 3. The specific formula is an = (3/90)(3)n-1. The sum of the first 20 terms is 174339220/9 and the infinite sum is 1/30. 3. The sequence results from monthly deposits of $1000 at 1% interest per month. The specific formula is an = 1000

1. Example A. The sequence 2, 6, 18, 54, … is a geometric
sequence because 6/2 = 18/6 = 54/18 = … = 3 = r.
Since a1 = 2, set them in the general formula of the
geometric sequence an = a1r n – 1 , we get the specific formula
for this sequence an = 2(3n – 1).
Geometric Sequences
If a1, a2 , a3 , …an is a geometric sequence such that the
terms alternate between positive and negative signs,
then r is negative.
Example B. The sequence 2/3, –1, 3/2, –9/4, … is a geometric
sequence because
–1/(2/3) = (3/2) / (–1) = (–9/4) /(3/2) = … = –3/2 = r.
Since a1 = 2/3, the specific formula is
an = ( )n–12
3 2
–3
Use the general formula of geometric sequences
an = a1*rn–1 a to find the specific formula.

2. Example C. Given that a1, a2 , a3 , …is a geometric sequence
with r = –2 and a5 = 12,
a. find a1
By that the general geometric formula
an = a1r n – 1, we get
a5 = a1(–2)(5 – 1) = 12
a1(–2)4 = 12
16a1 = 12
a1 = 12/16 = ¾
3
4
an= (–2)n–1
Geometric Sequences
To use the geometric general formula to find the specific
formula, we need the first term a1 and the ratio r.
b. find the specific equation.
Set a1 = ¾ and r = –2 into the general formula an = a1rn – 1 ,
we get the specific formula of this sequence

3. set n = 9, we get
c. Find a9.
3
4
a9= (–2)9–1
a9 = (–2)8 = (256) = 192
3
4
Geometric Sequences
3
4
Since an= (–2)n–1,
3
4
Example D. Given that a1, a2 , a3 , …is a geometric sequence
with a3 = –2 and a6 = 54,
a. find r and a1
Given that the general geometric formula an = a1rn – 1,
we have
a3 = –2 = a1r3–1 and a6 = 54 = a1r6–1
–2 = a1r2 54 = a1r5
Divide these equations: 54
–2
=
a1r5
a1r2

4. 54
–2
=
a1r5
a1r2
–27
3 = 5–2
–27 = r3
–3 = r
Put r = –3 into the equation –2 = a1r2
Hence –2 = a1(–3)2
–2 = a19
–2/9 = a1
Geometric Sequences
b. Find the specific formula and a2
Use the general geometric formula an = a1rn – 1,
set a1 = –2/9, and r = –3 we have the specific formula
–2
9
an = (–3)n–1
–2
9
(–3) 2–1
To find a2, set n = 2, we get
–2
9
a2 =
3
2
3= (–3) =

5. Geometric Sequences
2
3
– 3
2
an= ( ) n–1
To find n, set an = =
2
3
– 3
2
( ) n – 1–81
16
– 3
2
= ( ) n – 1–243
32
Compare the denominators to see that 32 = 2n – 1.
Since 32 = 25 = 2n – 1
n – 1 = 5
n = 6
= a
1 – rn
1 – r
The Sum of the First n Terms of a Geometric Sequence
a + ar + ar2 + … +arn–1
Example E. Find the geometric sum :
2/3 + (–1) + 3/2 + … + (–81/16)
We have a = 2/3 and r = –3/2, and an = –81/16. We need the
number of terms. Put a and r in the general formula to get the
specific formula

6. Therefore there are 6 terms in the sum,
2/3 + (–1) + 3/2 + … + (–81/16)
S =
2
3
1 – (–3/2)6
1 – (–3/2)
=
2
3
1 – (729/64)
1 + (3/2)
=
2
3
–665/64
5/2
–133
48
Geometric Sequences
Set a = 2/3, r = –3/2 and n = 6 in the formula
1 – rn
1 – r
S = a
we get the sum S
=

7. Infinite Sums of Geometric Sequences
The Sum of Infinitely–Many Terms of a Geometric Sequence
Given a geometric sequence a, ar, ar2 … with| r | < 1
 a rn = a + ar + ar2 + … = a
1 – rn=0
∞
then
google source
15 cm2
Assuming the ratio of 1.15
is the cross–sectional areas of
the successive chambers so
the areas of the chambers form
a geometric sequence,
starting with the first area of
15 cm2 with r = 1/1.15.
Hence the approximate total
area is the infinite sum:
15
1 – (1/1.15)n=0
∞
15 + 15(1/1.15) + 15(1/1.15)2 + 15(1/1.15)23 + ...
=  15(1/1.15)n =

8. Geometric Sequences
2. –2, 4, –8, 16,..1. 1, 3, 9, 27,..
4. 3/64, 9/32, 27/64, 81/128,..3. 1/90, 1/30, 1/10, 3/10,..
6. 2.3, 0.23, 0.023, 0.0023,..5. 4/3, – 2/3, 1/3, –1/6,..
8. a3 = –17,.., r = 1/2,7. a2 = 3/16,.., r = –2,
10. a5 = 4, r = –1/39. a4 = –2, r = 2/3
12. a3 = 125, a6 = –111. a4 = 0.02, a7 = 20
15. a2 = 0.3, a4 = 0.003
Exercise A. For each geometric sequence below
a. identify the first term a1 and the ratio r
b. find a specific formula for an and find a10
c. find the sum  an
d. if –1 < r < 1, find the sum  an. Use a calculator if needed.
n=1
20
n=1
∞
16. a4 = –0.21, a8 = – 0.000021
13. a4 = –5/2, a8 = –40 14. a3 = 3/4, a6 = –2/9

9. Geometric Sequences
2. –2 + 6 –18 + .. + 486
3. 6 – 3 + 3/2 – .. + 3/512
1. 3 + 6 + 12 + .. + 3072
4. 4/3 + 8/9 + 16/27 + 32/81
5. We deposit $1,000 at the beginning of each month at
1% monthly interest rate for 10 months. How much is there
in total right after the last or the 10th deposit?
6. Find a formula for the total right after the kth deposit.
B. For each sum below, find the specific formula of the
terms, write the sum in the  notation, then find the sum.

10. 1. 𝑎1 = 1, 𝑟 = 3, 𝑎 𝑛 = 3 𝑛−1, 𝑎10 = 19683, 𝑛=1
20
𝑎 𝑛 = 1743392200
3. 𝑎1 =
1
90
, 𝑟 = 3, 𝑎 𝑛 =
3 𝑛−1
90
, 𝑎10 =
2187
10
, 𝑛=1
20
𝑎 𝑛 = 174339220/9
5. 𝑎1 =
4
3
, 𝑟 =
1
2
, 𝑎 𝑛 =
23−𝑛 −1 𝑛−1
3
, 𝑎10 = −
1
384
,
𝑛=1
20
𝑎 𝑛 = 349525/393216 , 𝑛=1
∞
𝑎 𝑛 = 8/9
7. 𝑎1 =
−3
32
, 𝑎 𝑛 =
−3(−2) 𝑛−1
32
, 𝑎10 = 48, 𝑛=1
20
𝑎 𝑛 = 1048575/32
(Answers to the odd problems) Exercise A.
9. 𝑎1 =
−27
4
, 𝑎 𝑛 =
−27 ∙2 𝑛−3
3 𝑛−1 , 𝑎10 = −1152,
𝑛=1
20
𝑎 𝑛 = −3485735825/172186884 , 𝑛=1
∞
𝑎 𝑛 = −81/4
11. 𝑎1 =
1
50000
, 𝑟 = 10, 𝑎 𝑛 =
10 𝑛−1
50000
, 𝑎10 = 20000,
𝑛=1
20
𝑎 𝑛 = 11111111111111111111/50000
Geometric Sequences

11. 15. 𝑎1 = 3, 𝑟 =
1
10
, 𝑎 𝑛 =
3
10 𝑛−1 , 𝑎10 = 3 × 10−9
. 𝑛=1
20
𝑎 𝑛 = 3.333333. . , 𝑛=1
∞
𝑎 𝑛 = 10/3
13. 𝑎1 =
−5
16
, 𝑟 = 2, 𝑎 𝑛 = −5 ∙ 2 𝑛−5
, 𝑎10 = −160,
𝑛=1
20
𝑎 𝑛 = −5242875/16
3. 𝑛=1
∞
6
−1
2
𝑛−1
= 41. 𝑛=1
∞
3𝑛 = ∞
Exercise B.
5. 𝑛=1
10
1000
1
10
𝑛−1
= 1111.111111
Geometric Sequences