**THE GENERAL TERM FORMULA**

Suppose the first term of a geometric sequence is

*u*

_{1}and the common ratio is

*r*.

Then

*u*

_{2}=

*u*

_{1}

*r*,

*u*

_{3}=

*u*

_{1}⋅

*r*

^{2},

*u*

_{4}=

*u*

_{1}⋅

*r*

^{3}, and so on. Hence, term number

*n*, the power of

*r*is one less than the term number (

*n*-1).

For a

geometric sequencewithfirst termandu_{1}common ratio, thergeneral termorisnth termu=_{n}u_{1}⋅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

*u*

_{1}=8 and

*r*=½.

b)

*u*=

_{n}*u*

_{1}⋅

*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,

*t*

_{4}, and the

*n*th 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.

*t*

_{4}=

*t*

_{3}×

*r*

*t*

_{4}=18×3=54

(5) Write the general formula for the

*n*th 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*=2×3

_{n}_{(n-1)}

Worked Example 2.

Find the *n*th 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 *n*th 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;

*t*

_{3}=12

(3) Substitute all known values into the general formula.

*r*

^{(3-1)}

(4) Solve for

*r*(note that there are two possible solutions)

*r*

^{2}

*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

*n*th term of the sequence.

*t*=3×2

_{n}_{(n-1)}or

*t*=3×(-2)

_{n}_{(n-1)}

(6) Find the 10th term by substituting

*n*=10 into each of the two expressions for the

*n*th term.

*n*=10,

*t*

_{10}=3×2

^{(10-1)}⋅(using

*r*=2)

=3×2

^{9}=1536

or

*t*

_{10}=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 *n*th term for the sequence.

Steps:

(1) Write the general rule for the *n*th 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].

*n*=5,

*t*=14

_{n}14=

*ar*

^{(5-1)}

14=

*ar*

^{4}… [1]

(3) Similarly, use information about the seventh term to form an equation. Label it [2].

*n*=7,

*t*=0.56

_{n}0.56=

*ar*

^{(7-1)}

0.56=

*ar*

^{6}… [2]

(4) Solve the equations simultaneously: divide equation [2] by equation [1] to eliminate

*a*.

(5) Solve for

*r*.

*r*

^{2}=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*.

*r*=0.2

Substitute

*r*into [1]:

a×(0, 2)

^{4}=14

0.0016

*a*=14

*a*=14÷0.0016=8750

(7) Substitute the values of

*r*and

*a*into the general equation to find the expression for the

*n*th term.

The

*n*th term is:

*t*=8750×(0.2)

_{n}_{(n-1)}

(8) Now consider the negative value of

*r*.

*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

0.0016

*a*=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

*n*th term of the sequence.

The

*n*th term is:

*t*=8750×-(0.2)

_{n}_{(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

*n*th term of the geometric sequence

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

*n*th term is given by

*a*=6⋅⅓

_{n}_{(n-1)}.

Example 3: **Finding the nth term**

Find a formula for the

*n*th term of the geometric sequence

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

*n*th term is

*a*=2(-½)

_{n}_{(n-1)}.

Example 4: **Write an Equation for the nth Term**

Write an equation for the

*n*th term of the geometric sequence 3, 12, 48, 192, … .

Solution:

In this sequence,

*a*

_{1}=3 and

*r*=4.. Use the

*n*th term formula to write an equation.

Formula for

*n*th term,

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}.

*a*

_{1}=3,

*r*=4

*a*=3⋅4

_{n}_{(n-1)}

An equation is

*a*=3⋅4

_{n}_{(n-1)}.

Example 5. Find the 10th and *n*th terms of the G.P. 5, 25, 125, … .

Solution:

Here *a*=5 and *r*=5. Thus,

*a*

_{10}=5⋅5

^{(10-1)}=5

^{1}⋅5

^{9}=5

^{10}and

*a*=

_{n}*ar*

_{(n-1)}

∴

*a*=5⋅5

_{n}_{(n-1)}=5

^{n}.

Example 6:

Find the formula for the *n*th 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*=486×⅓

_{n}_{(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

*n*th term of the geometric series. The common ratio is 4,

So

*r*=4. The first term is -6, so

*a*

_{1}=-6. Then,

*a*=-6⋅4

_{n}_{(n-1)}.

*a*

_{5}=-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

*n*th term of the geometric series. The common ratio is -5,

so

*r*=-5. The first term is -1, so

*a*

_{1}=-1. Then,

*a*=-1⋅(-5)

_{n}_{(n-1)}.

*a*

_{7}=-1⋅(-5)

^{(7-1)}

=-1⋅(-5)

^{6}

=-1⋅(-1)

^{6}⋅5

^{6}

=-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

*n*th term of the geometric series. The common ratio is ⅔,

so

*r*=⅔.The first term is 72, so al

*a*

_{1}=72. Then,

*a*=72⋅⅔

_{n}_{(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

*n*th term of the geometric series. The common ratio is ¾,

so

*r*=¾. The first term is 112, so

*a*

_{1}=112.Then,

*a*=112⋅¾

_{n}_{(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

*a*

_{1}=-3

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

*a*=(-3)⋅(-2)

_{n}_{(n-1)}

Answer:

*a*=(-3)⋅(-2)

_{n}_{(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}*=2

^{n}

c) 6, 12, 24, 48, …

Solution:

Each number is a multiple of 2 apart. I need a 2^{n} in the formula. I will need a number in front of the 2^{n} to get the correct answer. I find the number by solving

^{1})=5

*x*=3

Answer:

*a*=3(2

_{n}^{n})

d) l, 2, 4, 8, 16 …

Solution:

Each number is a multiple of 2 apart. I need a 2^{n} in the formula. I will need a number in front of the 2^{n} to get the correct answer. I find the number by solving

^{1})=1

*x*=½

Answer:

*a*=½(2

_{n}^{n}) 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 a_{1} and common ratio (r)**

a)

*a*

_{1}=2;

*r*=3

*a*=

_{n}*x*(3

^{n})

*a*

_{1}=

*x*(3

^{1})

2=3

*x*

*x*=⅔

Answer:

*a*=⅔⋅3

_{n}^{n}

b) *a*_{1}=6; *r*=2

*a*=

_{n}*x*(2

^{n})

*a*

_{1}=

*x*(2

^{1})

6=2

*x*

*x*=3

Answer:

*a*=3⋅2

_{n}^{n}

c) *a*_{1}=1; *r*=1

*a*=

_{n}*x*(½

^{n})

*a*

_{1}=

*x*(½

^{1})

1=½

*x*

2⋅1=2⋅½

*x*

*x*=2

Answer:

*a*=2(½

_{n}^{n}) or

*a*=2

_{n}^{(1-n)}

d) *a*_{1}=4; *r*=½

*a*=

_{n}*x*(½

^{n})

*a*

_{1}=

*x*(½

^{1})

4=½

*x*

*x*=8

Answer:

*a*=8(½

_{n}^{n}) or

*a*=2

_{n}^{(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.

4÷12=⅓

For an explicit formula, substitute

*a*

_{1}=36 and

*r*=⅓. In the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

=36⋅⅓

_{(n-1)}

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term

*a*

_{(n-1)}and

*r*.

*a*

_{1}=36,

*a*=⅓⋅

_{n}*a*

_{(n-1)}

b) 64, 16, 4, … .

Solution:

First, find the common ratio.

4÷16=¼

For an explicit formula, substitute

*a*

_{1}=64 and

*r*=¼ in the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

=64⋅¼

_{(n-1)}

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term

*a*

_{(n-1)}and

*r*.

*a*

_{1}=64,

*a*=¼

_{n}*a*

_{(n-1)}

c) -2, 10, -50, … .

Solution:

First, find the common ratio.

-50÷10=-5

For an explicit formula, substitute

*a*

_{1}=-2 and

*r*=-5 in the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

=-2⋅(-5)

_{(n-1)}

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term

*a*

_{(n-1)}and

*r*.

*a*

_{1}=-2,

*a*=-5

_{n}*a*

_{(n-1)}

d) 4, -12, 36, … .

Solution:

First, find the common ratio.

36÷(-12)=-3

For an explicit formula, substitute

*a*

_{1}=4 and

*r*=-3 in the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

=4⋅(-3)

_{(n-1)}

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term

*a*

_{(n-1)}and

*r*.

*a*

_{1}=42,

*a*=-3

_{n}*a*

_{(n-1)}

e) 4, 8, 16, … .

Solution:

First, find the common ratio.

16÷8=2

For an explicit formula, substitute

*a*

_{1}=4 and

*r*=2 in the

*n*th term formula.’

*a*=

_{n}*a*

_{1}⋅

*r*

_{(n-1)}

=4⋅2

_{(n-1)}

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term

*a*

_{(n-1)}and

*r*.

*a*

_{1}=4,

*a*=2

_{n}*a*

_{(n-1)}

f) 20, 30, 45, … .

Solution:

First, find the common ratio.

45+30=1.5

For an explicit formula, substitute

*a*

_{1}=20 and

*r*=1.5 in the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*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*.

*a*

_{1}=20,

*a*=1.5

_{n}*a*

_{(n-1)}

g) 15, 5, 5/3, …

Solution:

First, find the common ratio.

For an explicit formula, substitute

*a*

_{1}=15 and

*r*=⅓ in the

*n*th term formula.

*a*=

_{n}*a*

_{1}⋅

*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*.

*a*

_{1}=15,

*a*=⅓

_{n}*a*

_{(n-1)}

h) 1/32, 1/16, ⅛, … .

Solution:

First, find the common ratio.

For an explicit formula, substitute

*a*

_{1}=1/32 and

*r*=2 in the

*n*th term formula.

For a recursive formula, state the first term

*a*

_{1}.

Then indicate that the next term is the product of the first term *a*_{(n-1)} and *r*.

*a*

_{1}=1/32,

*a*=2

_{n}*a*

_{(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 *n*th 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.

*a*

_{3}=

*ar*

^{2}=30⋅2

^{2}=120

Therefore, the number of bacteria at the end of 2nd hour will be 120.

*a*

_{5}=

*ar*

^{4}=30⋅2

^{4}=480

The number of bacteria at the end of 4th hour will be 480.

*a*

_{(n+1)}=

*ar*

^{n}=30⋅2

^{n}

Thus, number of bacteria at the end of

*n*th hour will be 30⋅2

^{n}.

Example 11. A geometric sequence has *u*_{2}=-6 and *u*_{5}=162. Find its general term.