# How to Find a Certain Term in a Geometric Sequence?

The formula for the nth term involves four variables: an, a1, r, and n. If we know the value of any three of them, we can find the value of the fourth.

Explicit geometric sequences also have a formula for finding any term in a sequence.

an=a1 r(n-1)

an= the term in the sequence you are trying to find (n represents the desired term number)
a1= the first term in the sequence
r= the common ratio

Example 1. What is the 15th term of the geometric sequence -9, 27, -81, … ?
Solution:
Calculate the common ratio. The common ratio is -3.
an=a1 r(n-1)
a15=-9⋅(-3)(15-1)
=-9⋅(-3)14
=-43,046,721

The 15th term of the sequence is -43,046,721.

Ex2. What is the 10th term of the geometric sequence 6, -24, 96, … ?
Solution:
Calculate the common ratio. The common ratio is -4.
an=a1 r(n-1)
a10=6⋅(-4)(10-1)
=6⋅(-4)9
=-1,572,864

The 10th term of the sequence is -1,572,864.

Ex3. Find the eleventh term of the sequence 3, -6, 12, -24, …
(A). 1024 (B). 3072 (C). 33 (D). -6144
Solution:
Calculate the common ratio. The common ratio is -2.
an=a1 r(n-1)
a11=3(-2)(11-1)
=3(-2)10
a11=3072

The eleventh term of the sequence is 3072. Choice B is the correct answer.

Ex4 — Ex10: Find the specified term for each geometric sequence or sequence with the given characteristics
Ex4. a9 for 60, 30, 15, … .
Solution:
First, find the common ratio.

30÷60=½
15÷30=½

Use the formula for the nth term of a geometric sequence to find a9. Ex5. a4 for 7, 14, 28, … .
Solution:
First, find the common ratio.

14÷7=2
28÷14=2

Use the formula for the nth term of a geometric sequence to find a4.
an=a1 r(n-1)
a4=7⋅2(4-1)
a4=7⋅8
a4=56

Ex6. a5 for 3, 1, ⅓, … .
Solution:
First, find the common ratio.

1÷3=⅓
⅓÷1=⅓

Use the formula for the nth term of a geometric sequence to find a5. Ex7. a6 for 540, 90, 15, … .
Solution:
First, find the common ratio.

90÷540=⅙
15÷90=⅙

Use the formula for the nth term of a geometric sequence to find a6. Ex8. a7 if a3=24 and r=0.5.
Solution:

a7=a3 r(7-3)
a7=24⋅½4
a7=1.5

Ex9. a6 if a3=32 and r=-0.5.
Solution:

a6=a3 r(6-3)
a6=32⋅(-½)3
a6=32⋅(-⅛)
a6=-4

Ex10. a8 if a1=4096 and r=¼.
Solution:
Use the formula for the nth term of a geometric sequence to find a8. Ex11 — Ex14: Find the indicated term of the geometric sequence

Ex11. The 10th term; 6, 18, 54, 162, …
Solution:
First develop a formula for an; first term is 6 and common ratio is 3.

an=x(3n)
a1=x(31)
6=3x
2=x

Now, find the 10th term: a10=2(310)=118098

Ex12. The 9th term; 5, 20, 80, 160, …
Solution:
First develop a formula for an; first term is 5 and common ratio is 4 formula for an=5/4 (4n).
Now, find the 9th term: a9=5/4 (49)=327680

Ex13. The 14th term; 200, 100, 50, 25, …
Solution:
First develop a formula for an; first term 200, common ratio ½.

an=xn)
a1=x1)
200=½x
2⋅200=2⋅½x
x=400

formula for an=400(½n).
Now, find the 14th term: a14=400(½14)=25/1024

Ex14. The 20th term; 39366, 13122, 4374, …
Solution:
First develop a formula for an; first term 39366 common ratio ⅓.

an=x(⅓n)
a1=x(⅓1)
39366=⅓x
3⋅39366=3⋅⅓x
x=118098
an=118098(⅓n)

now I can find a20=118098(⅓20)=2/59049

Ex15. Find 4th term in the geometric sequence 5, 10, 20, …
Solution:

a=5, r=10/5=2, an=?
an=ar(n-1)
a4=5⋅2(4-1)=5×8=40

Ex16. Determine the 10th term of the sequence: 3, 6, 12, …
Solution:

a=3, r=6/3=12/6=2
an=ar(n-1)
a10=3⋅2(10-1)=3⋅29=3⋅512=1536

Ex17. Find the 7th term in the following sequence: 6, 18, 54, 162, …
Solution:
Finding the common ratio can be harder than finding the common difference. One way to find it is the divide each term by the term before it.
18÷6=3, 54÷18=3, 162÷54=3. So the common ratio is 3.

a7=6⋅3((7-1)))
a7=6⋅36=6×729
a7=4,374

So the 7th term of the sequence is 4,374.

Ex18. Find the 8th term in the following sequence: 96, 48, 24, 12, 6, …
Solution:
To find the common ratio, divide each term by the one before it.
48÷96=½, 24÷48=½, 12÷24=½. The common ratio is ½. The 8th term of the sequence is 0.75.

Ex19. What is the 10th term of the sequence 4, 12, 36, 108, 324, … ?
Solution:
The sequence 4, 12, 36, 108, 324, … is a geometric sequence with a1=4 and r=3.Therefore,

an=a1 r(n-1)
a10=4⋅39=78732

Use a calculator for the computation.

Ex20: Find the Next Term

Test-Taking Tip
Since the terms of this sequence are increasing, the missing term must be greater than 125. You can immediately eliminate 75 as a possible answer.

Multiple-Choice Test Item
Find the missing term in the geometric sequence: 8, 20, 50, 125, _
(A). 75 (B). 200 (C). 250 (D). 312.5
Solution:
Since 20/8=2.5, 50/20=2.5, and 125/50=2.5, the sequence has a common ratio of 2.5.
Solve the Test Item
To find the missing term, multiply the last given term by 2.5:

125(2.5)=312.5.

Ex21. The first term of a geometric series is 1 and the common ratio is 9. What is the 8th term of the sequence?
Solution:

an=a1 r(n-1)
a8=1⋅9(8-1)
=1⋅97
=4,782,969

The 8th term of the sequence is 4,782,969.

Ex22. The first term of a geometric series is 2 and the common ratio is 4. What is the 14th term of the sequence?
Solution:

an=a1 r(n-1)
a14=2⋅4(14-1)
=2⋅413
=134, 217, 728

The 14th term of the sequence is 134, 217, 728.

Ex23: Finding a missing term
Find the first term of a geometric sequence whose fourth term is 8 and whose common ratio is ½.
Solution:
Let a4=8, r=½, andn n=4 in the formula an=a1 r(n-1).

8=a1⋅½(4-1)
8=a1⋅⅛
64=a1

So the first term is 64.

Ex24. If r=4 and a8=100, what is the first term of the geometric sequence?
Solution:
Substitute a8=100, r=4, and n=8 into the formula for the nth term of a geometric sequence to find the a1. Ex25: Find a Term Given the Fourth Term and the Ratio
Find the tenth term of a geometric sequence for which a4=108 and r=3.
Solution:
First, find the value of a1. Formula for nth term, an=a1r(n-1).

n=4, r=3
a4=a1⋅3(4-1)
a4=108
108=27a1

Divide each side by 27. Formula for nth term, an=a1r(n-1).
n=10, a1=4, r=3
an=a1r(n-1)
a10=4⋅3(10-1)
a10=78, 732

Ex26. Determine the 12th term of a geometric sequence whose 8th term is 192 and the common ratio is 2.
Solution: We have Ex27. Find the eighth term of a geometric sequence for which a3=81 and r=3.
Solution:
Because a3=81, the third term in the sequence is 81. To find the eighth term of the sequence, you need to find the 1st term of the sequence. Use the nth term of a Geometric Sequence formula.

an=a1 r(n-1)
a3=a1⋅3(3-1)
81=a1⋅9
a1=9

Then the first term a1 is 9.
Use a1 to find the eighth term of the sequence.
an=a1 r(n-1)
a8=a1⋅3(8-1)
=9⋅37
=19,683

The eighth term of the geometric sequence is 19,683.

Ex28. Find the 12th term of a geometric sequence whose 8th term is 192 and the common ratio is 2.
Solution:
Common ratio, r=2
Let a be the first term of the geometric sequence. Ex29: Find a Particular Term
Find the eighth term of a geometric sequence for which a1=-3 and r=-2.
Solution:
Formula for nth term, an=a1r(n-1).

n=8, a1=-3, r=-2
a8=(-3)⋅(-2)(8-1)
a8=(-3)⋅(-128)
a8=384

The eighth term is 384.

Ex30. Find the sixth term of a geometric sequence with a first term of 9 and a common ratio of 2.
Solution:
Use the formula for the nth term of a geometric sequence to find the a6.

an=a1 r(n-1)
a6=9⋅2(6-1)
a6=288

Ex31: Using the Formula for the General Term of a Geometric sequence
Find the eighth term of the geometric sequence whose first term is -4 and whose common ratio is -2.
Solution:
To find the eighth term, a8, we replace n in the formula with 8, a1 with -4, and r with

an=a1 r(n-1)
a8=-4(-2)(8-1)=-4(-2)7=-4(-128)=512

The eighth term is 512. We can check this result by writing the first eight terms of the Growth sequence:
-4, 8, -16, 32, -64, 128, -256, 512.

Ex32. (Biology) A certain bacteria grows at a rate of 3 cells every 2 minutes. If there were 260 cells initially, how many are there after 21 minutes?
Solution:
Given, a1=260, n=21 and r=3/2 Ex33. (Biology) A virus goes through a computer, infecting the files. If one file was infected initially and the total number of files infected doubles every minute, how many files will be infected in 20 minutes?
Solution:
Given, a1=1, r=200% or 2, n=20.

an=a1 r(n-1)
a20=1⋅2(20-1)
=524,288

Ex34. (Short Response) Elisa has a savings account. She withdraws half of the contents every year. After 4 years, she has \$2000 left. How much did she have in the savings account originally?
Solution:
Given n=4, a4=2000, r=0.5 She have invested (a0) \$32, 000.

Ex35. (Sense-Making)
A certain drug has a half-life of 8 hours after it is administered to a patient. What percent of the drug is still in the patient’s system after 24 hours?
Solution:
Given a1=100% or 1 and r=½ and n=4.

an=a1 r(n-1)
a4=1⋅½3
=⅛
=0.125 or 12.5%

After 24 hours, 12.5% of the drug is still in the patient’s system.