THE GENERAL TERM FORMULA
Suppose the first term of a geometric sequence is u₁ and the common ratio is r.
Then u₂=u₁ r, u₃=u₁⋅r², u₄=u₁⋅r³, and so on. Hence, term number n, the power of r is one less than the term number (n-1).

For a geometric sequence with first term u₁ and common ratio r, the general term or nth term is u_n=u₁⋅r_(n-1).

Example 1. Consider the sequence 8, 4, 2, 1, ½, …
a) Show that the sequence is geometric.
b) Find the general term u_n.
c) Hence, find the 12th term as a fraction.
Solution:
a) 4/8=½; 2/4=½; 1/2=½; ½/1=½
Consecutive terms have a common ratio of ½. ∴ the sequence is geometric with u₁=8 and r=½.
b) u_n=u₁⋅r_(n-1)

Worked Example 1.
State whether the sequence t_n: {2, 6, 18, … } is geometric by finding the ratio of successive terms. If it is geometric find the next term in the sequence, t₄, and the nth term for the sequence, t_n.
Steps:

(3) Compare the ratios and make your conclusion.
since

the sequence is geometric with the common ratio r=3.
(4) Because the sequence is geometric, find the fourth term by multiplying the preceding (third) term by the common ratio.
t₄=t₃×r
t₄=18×3=54
(5) Write the general formula for the nth term of the geometric sequence.
t_n=ar_(n-1)
(6) Identify the values of a and r. a=2; r=3
(7) Substitute the values of a and r into the general formula.
t_n=2×3_(n-1)

Worked Example 2.
Find the nth term and the 10th term in the geometric sequence where the first term is 3 and the third term is 12.
Steps:
(1) Write the general formula for the nth term in the geometric sequence.

t_n=ar_(n-1)
(2) State the value of a (the first term in the sequence) and the value of the third term.
a=3; t₃=12
(3) Substitute all known values into the general formula.
12=3×r^(3-1)
(4) Solve for r (note that there are two possible solutions)
12÷3=r²
r=√4=±2
(5) Substitute the values of a and r into the general equation. Because there are two possible values for r, you must show both expressions for the nth term of the sequence.
So t_n=3×2_(n-1) or t_n=3×(-2)_(n-1)
(6) Find the 10th term by substituting n=10 into each of the two expressions for the nth term.
When n=10, t₁₀=3×2^(10-1)⋅(using r=2)
=3×2⁹=1536
or t₁₀=3×(-2)^(10-1) (using r=-2)
=3×(-2)⁹=-1536

Worked Example 3.
The fifth term in a geometric sequence is 14 and the seventh term is 0.56. Find the common ratio, r, the first term, a, and the nth term for the sequence.
Steps:
(1) Write the general rule for the nth term of a geometric sequence.

t_n=ar_(n-1)
(2) Use the information about the fifth term to form an equation. Label it [1].
When n=5, t_n=14
14=ar^(5-1)
14=ar⁴ … [1]
(3) Similarly, use information about the seventh term to form an equation. Label it [2].
When n=7, t_n=0.56
0.56=ar^(7-1)
0.56=ar⁶ … [2]
(4) Solve the equations simultaneously: divide equation [2] by equation [1] to eliminate a.

(5) Solve for r.
r²=0.04=(±0.2)²
r=±0.2
(6) Because there are two solutions, we have to perform two sets of computations. Consider the positive value of r first. Substitute the value of r into either of the two equations, say equation [1], and solve for a.
If r=0.2
Substitute r into [1]:
a×(0, 2)⁴=14
0.0016a=14
a=14÷0.0016=8750
(7) Substitute the values of r and a into the general equation to find the expression for the nth term.
The nth term is:
t_n=8750×(0.2)_(n-1)
(8) Now consider the negative value of r.
If r=-0.2
(9) Substitute the value of r into either of the two equations, say equation [1], and solve for a. (Note that the value of a is the same for both values of r.)
Substitute r into [1]
a=(-0.2)⁴=14
0.0016a=14
a=14÷0.0016=8750
(10) Substitute the values of r and a into the general formula to find the second expression for the nth term of the sequence.
The nth term is:
t_n=8750×-(0.2)_(n-1)

The first term and the common ratio determine all of the terms of a geometric sequence.

Example 2: Finding the nth term
Write a formula for the nth term of the geometric sequence

6, 2, ⅔, 2/9, … .
Solution:
We can obtain the common ratio by dividing any term after the first by the term preceding it. So
r=2÷6=⅓
Because each term after the first is ⅓ of the term preceding it, the nth term is given by
a_n=6⋅⅓_(n-1).

Example 3: Finding the nth term
Find a formula for the nth term of the geometric sequence

2, -1, ½, -¼, … .
Solution:
We obtain the ratio by dividing a term by the term preceding it:
r=-1÷2=-½
Each term after the first is obtained by multiplying the preceding term by -½. The formula for the nth term is
a_n=2(-½)_(n-1).

Example 4: Write an Equation for the nth Term
Write an equation for the nth term of the geometric sequence 3, 12, 48, 192, … .
Solution:
In this sequence, a₁=3 and r=4.. Use the nth term formula to write an equation.
Formula for nth term, a_n=a₁⋅r_(n-1).

a₁=3, r=4
a_n=3⋅4_(n-1)
An equation is a_n=3⋅4_(n-1).

Example 5. Find the 10th and nth terms of the G.P. 5, 25, 125, … .
Solution:
Here a=5 and r=5. Thus,

a₁₀=5⋅5^(10-1)=5¹⋅5⁹=5¹⁰ and a_n=ar_(n-1)
∴a_n=5⋅5_(n-1)=5ⁿ.

Example 6:
Find the formula for the nth term of the geometric sequence
1) 2, 6, 18, …
2) 486, 162, 54, … .
Solution:
1) Here a=2 and r=3, so a_n=2×3_(n-1).
2) Here a=486 and r=⅓, so a_n=486×⅓_(n-1).

Example 7: Write an equation for the nth term of each geometric sequence, and find the indicated term.
a) Find the fifth term of -6, -24, -96, … .
Solution:
Calculate the common ratio.

Use the formula a_n=ar_(n-1) to write an equation for the nth term of the geometric series. The common ratio is 4,
So r=4. The first term is -6, so a₁=-6. Then, a_n=-6⋅4_(n-1).
a₅=-6⋅4^(5-1)=-1536
The 5th term of the sequence is -1536.

b) Find the seventh term of -1, 5, -25, … .
Solution:
Calculate the common ratio.

Use the formula a_n=ar_(n-1) to write an equation for the nth term of the geometric series. The common ratio is -5,
so r=-5. The first term is -1, so a₁=-1. Then, a_n=-1⋅(-5)_(n-1).
a₇=-1⋅(-5)^(7-1)
=-1⋅(-5)⁶
=-1⋅(-1)⁶⋅5⁶
=-15, 625
The 7th term of the sequence is -15, 625.

c) Find the tenth term of 72, 48, 32, … .
Solution:
Calculate the common ratio.

Use the formula a_n=ar_(n-1) to write an equation for the nth term of the geometric series. The common ratio is ⅔,
so r=⅔.The first term is 72, so al a₁=72. Then, a_n=72⋅⅔_(n-1)

The 10th term of the sequence is 4096/2187.

d) Find the ninth term of 112, 84, 63, … .
Solution:
Calculate the common ratio.

Use the formula a_n=ar_(n-1) to write an equation for the nth term of the geometric series. The common ratio is ¾,
so r=¾. The first term is 112, so a₁=112.Then, a_n=112⋅¾_(n-1)

The 9th term of the sequence is 45, 927/4096.

Example 8: Write an equation for the nth term of each geometric sequence.
a) -3, 6, -12, …
Solution:

r=6÷(-3) or-2
a₁=-3
a_n=a₁⋅r_(n-1)
a_n=(-3)⋅(-2)_(n-1)
Answer:
a_n=(-3)⋅(-2)_(n-1)

b) 2, 4, 8, 16, 32, …
Solution:
I have to multiply by 2 to get from one number to the next. Since the first number is 2 the formula, there isn’t any algebra needed to find a number to put in front of the 2.
Answer: a_n=2ⁿ

c) 6, 12, 24, 48, …
Solution:
Each number is a multiple of 2 apart. I need a 2ⁿ in the formula. I will need a number in front of the 2ⁿ to get the correct answer. I find the number by solving

x(2¹)=5
x=3
Answer: a_n=3(2ⁿ)

d) l, 2, 4, 8, 16 …
Solution:
Each number is a multiple of 2 apart. I need a 2ⁿ in the formula. I will need a number in front of the 2ⁿ to get the correct answer. I find the number by solving

x(2¹)=1
x=½
Answer: a_n=½(2ⁿ) it would also be correct to write 2_(n-1)

Example 9: Find a formula for the nth term of the geometric sequence with first term a₁ and common ratio (r)
a) a₁=2; r=3

a_n=x(3ⁿ)
a₁=x(3¹)
2=3x
x=⅔
Answer: a_n=⅔⋅3ⁿ

b) a₁=6; r=2

a_n=x(2ⁿ)
a₁=x(2¹)
6=2x
x=3
Answer: a_n=3⋅2ⁿ

c) a₁=1; r=1

a_n=x(½ⁿ)
a₁=x(½¹)
1=½x
2⋅1=2⋅½x
x=2
Answer: a_n=2(½ⁿ) or a_n=2^(1-n)

d) a₁=4; r=½

a_n=x(½ⁿ)
a₁=x(½¹)
4=½x
x=8
Answer: a_n=8(½ⁿ) or a_n=2^(3-n)

Example 10: Write an explicit formula and a recursive formula for finding the nth term of each geometric sequence.
a) 36, 12, 4, … .
Solution:
First, find the common ratio.

12÷36=⅓
4÷12=⅓
For an explicit formula, substitute a₁=36 and r=⅓. In the nth term formula.
a_n=a₁⋅r_(n-1)
=36⋅⅓_(n-1)
For a recursive formula, state the first term a₁.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=36, a_n=⅓⋅a_(n-1)

b) 64, 16, 4, … .
Solution:
First, find the common ratio.

16÷64=¼
4÷16=¼
For an explicit formula, substitute a₁=64 and r=¼ in the nth term formula.
a_n=a₁⋅r_(n-1)
=64⋅¼_(n-1)
For a recursive formula, state the first term a₁.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=64, a_n=¼a_(n-1)

c) -2, 10, -50, … .
Solution:
First, find the common ratio.

10÷-2=-5
-50÷10=-5
For an explicit formula, substitute a₁=-2 and r=-5 in the nth term formula.
a_n=a₁⋅r_(n-1)
=-2⋅(-5)_(n-1)
For a recursive formula, state the first term a₁.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=-2, a_n=-5a_(n-1)

d) 4, -12, 36, … .
Solution:
First, find the common ratio.

-12÷4=-3
36÷(-12)=-3
For an explicit formula, substitute a₁=4 and r=-3 in the nth term formula.
a_n=a₁⋅r_(n-1)
=4⋅(-3)_(n-1)
For a recursive formula, state the first term a₁.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=42, a_n=-3a_(n-1)

e) 4, 8, 16, … .
Solution:
First, find the common ratio.

8÷4=2
16÷8=2
For an explicit formula, substitute a₁=4 and r=2 in the nth term formula.’
a_n=a₁⋅r_(n-1)
=4⋅2_(n-1)
For a recursive formula, state the first term a₁.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=4, a_n=2a_(n-1)

f) 20, 30, 45, … .
Solution:
First, find the common ratio.

30+20=1.5
45+30=1.5
For an explicit formula, substitute a₁=20 and r=1.5 in the nth term formula.
a_n=a₁⋅r_(n-1)
=20⋅(1.5)_(n-1)
For a recursive formula, state the first term all. Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=20, a_n=1.5a_(n-1)

g) 15, 5, 5/3, …
Solution:
First, find the common ratio.

For an explicit formula, substitute a₁=15 and r=⅓ in the nth term formula.
a_n=a₁⋅r_(n-1)
=15⋅⅓_(n-1)
For a recursive formula, state the first term all.
Then indicate that the next term is the product of the first term a_(n-1) and r.
a₁=15, a_n=⅓a_(n-1)

h) 1/32, 1/16, ⅛, … .
Solution:
First, find the common ratio.

For an explicit formula, substitute a₁=1/32 and r=2 in the nth term formula.

For a recursive formula, state the first term a₁.

Then indicate that the next term is the product of the first term a_(n-1) and r.

a₁=1/32, a_n=2a_(n-1)

Question 1. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Solution:
It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a geometric sequence.
Here, a=30 and r=2.

a₃=ar²=30⋅2²=120
Therefore, the number of bacteria at the end of 2nd hour will be 120.
a₅=ar⁴=30⋅2⁴=480
The number of bacteria at the end of 4th hour will be 480.
a_(n+1)=arⁿ=30⋅2ⁿ
Thus, number of bacteria at the end of nth hour will be 30⋅2ⁿ.

Example 11. A geometric sequence has u₂=-6 and u₅=162. Find its general term.