Finding the nth Term Given the
Common Ratio and any Term
or Two Terms
Lori Jordan
Kate Dirga
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Printed: May 4, 2015
AUTHORS
Lori Jordan
Kate Dirga
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Chapter 1. Finding the nth Term Given the Common Ratio and any Term or Two Terms
1
C HAPTER
Finding the nth Term Given
the Common Ratio and any Term or
Two Terms
Here you’ll write an nth term rule for a geometric sequence given the common ratio and any term or any two terms
in the sequence.
A bacteria sample doubles every hour. After four hours there are 64 bacteria in the sample. What is the nth term rule
for the geometric sequence represented by this situation?
Guidance
We will be using the general rule for the nth term in a geometric sequence and the given term(s) to determine the
first term and write a general rule to find any other term.
Example A
Consider the geometric sequence in which the common ratio is − 45 and a5 = 1280. Find the first term in the sequence
and write the general rule for the sequence.
Solution: We will start by using the term we know, the common ratio and the general rule, an = a1 rn−1 . By plugging
in the values we know, we can then solve for the first term, a1 .
4
4
a5 = a1 −
5
4
4
1280 = a1 −
5
1280
4 = a1
− 54
3125 = a1
Now, the nth term rule is an = 3125 − 45
n−1
.
Example B
Find the nth term rule for a sequence in which a1 = 16 and a7 =
Solution: Since a7 =
ratio:
1
4
1
4
and we know the first term, we can write the equation
1
4
= 16r6 and solve for the common
1
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1
4
1
r 64
1
6
64
1
2
= 16r6
= r6
=
√
6
r6
=r
1 n−1
2
The nth term rule is an = 16
Example C
Find the nth term rule for the geometric sequence in which a5 = 8 and a10 = 41 .
Solution: Using the same method at the previous example, we can solve for r and a1 . Then, write the general rule.
Equation 1: a5 = 8, so 8 = a1 r4 , solving for a1 we get a1 =
Equation 2: a10 = 41 , so
1
4
8
.
r4
= a1 r9 , solving for a1 we get a1 =
1
4
r9
.
1
8
4
=
r4 r9
1
8r9 = r4
4
1 4
9
8r
4r
=
8r4
8r4
1
r5 =
32
r
√
5
1
5
r5 =
32
1
r=
2
Thus, a1 =
8
4
( 21 )
=
8
1
16
= 81 · 16
1 = 128.
The nth term rule is an =
3
8
(2)n−1 .
∗
Note: In solving the equation above for r we divided both sides by r4 . In general it is not advisable to divide both
sides of an equation by the variable because we may lose a possible solution, r = 0. However, in this case, r 6= 0
since it is the common ratio in a geometric sequence.
Intro Problem Revisit We are given that a4 = 64 and because the sample doubles every hour we know that the
common ratio is 2. We can therefore plug the known values into the equation an = a1 rn−1 to get a1 .
a4 = a1 rn−1 .
64 = a1 (2)3
64 = 8a1
8 = a1
2
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Chapter 1. Finding the nth Term Given the Common Ratio and any Term or Two Terms
Therefore, there are 8 bacteria in the sample to begin with and the nth term rule is an = 8 · 2n−1 .
Guided Practice
1. Find the first term and the nth term rule for the geometric sequence given that r = − 12 and a6 = 3.
16
2. Find the common ratio and the nth term rule for the geometric sequence given that a1 = − 625
and a6 = − 52 .
3. Find the nth term rule for the geometric sequence in which a5 = 6 and a13 = 1536.
Answers
1. Use the known quantities in the general form for the nth term rule to find a1 .
5
1
3 = a1 −
2
32
1
32
−
· 3 = a1 −
· −
1
32
1
a1 = −96
Thus, an = −96 − 12
n−1
2. Again, substitute in the known quantities to solve for r.
5
16
− = −
r5
2
625
5
625
−
−
= r5
2
16
3125
= r5
32
r
√
5
5 3125
= r5
32
5
r=
2
16
So, an = − 625
5 n−1
2
3. This time we have two unknowns, the first term and the common ratio. We will need to solve a system of equations
using both given terms.
Equation 1: a5 = 6, so 6 = a1 r4 , solving for a1 we get a1 =
6
.
r4
Equation 2: a13 = 1536, so 1536 = a1 r12 , solving for a1 we get a1 =
1536
.
r12
Now that both equations are solved for a1 we can set them equal to each other and solve for r.
3
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6
1536
= 12
4
r
r
6r12 = 1536r4
6r12 1536r4
=
6r4
6r4
8
r = 256
√
√
8
8
r8 = 256
r=2
Now use r to find a1 : a1 =
The nth term rule is an =
6
(24 )
3
8
=
6
16
= 38 .
(2)n−1 .
Explore More
Use the given information to find the nth term rule for each geometric sequence.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
r = 32 and a8 = 256
81
r = − 34 and a5 = 405
8
r = 65 and a4 = 3
r = − 21 and a7 = 5
r = 76 and a0 = 1
a1 = 11
8 and a7 = 88
a1 = 24 and a4 = 81
a1 = 48 and a4 = 43
6
a1 = 343
216 and a5 = 7
a6 = 486 and a10 = 39366
a5 = 648 and a10 = 19683
4
a3 = 23 and a5 = 23
a5 = 43 and a10 = − 128
3
Use a geometric sequence to solve the following word problems.
14. Ricardo’s parents want to have $100,000 saved up to pay for college by the time Ricardo graduates from high
school (16 years from now). If the investment vehicle they choose to invest in claims to yield 7% growth per
year, how much should they invest today? Give your answer to the nearest one thousand dollars.
15. If a piece of machinery depreciates (loses value) at a rate of 6% per year, what was its initial value if it is 10
years old and worth $50,000? Give your answer to the nearest one thousand dollars.
4