**Geometric sequences**

We shall now move on to the other type of sequence we want to explore.

Consider the sequence

Here, each term in the sequence is 3 times the previous term. And in the sequence

each term is -2 times the previous term. Sequences such as these are called

**geometric sequences**.

Let us write down a general geometric sequence, using algebra. We shall take *a* to be the first term, as we did with arithmetic sequences. But here, there is no common difference. Instead there is a common ratio, as the ratio of successive terms is always constant. So we shall let *r* be this common ratio. With this notation, the general geometric sequence can be expressed as

*a*,

*ar*,

*ar*

^{2},

*ar*

^{3}, … .

So the

*n*-th can be calculated quite easily. It is

*ar*

^{(n-1)}, where the power (

*n*-1) is always one less than the position

*n*of the term in the sequence. In our first example, we had

*a*=2 and

*r*=3, so we could write the first sequence as

^{2}, 2×3

^{3}, … .

In our second example,

*a*=1 and

*r*=-2, so that we could write it as

^{2}, 1×(-2)

^{3}, … .

Key Point

A geometric sequence is a sequence where each new term after the first is obtained by multiplying the preceding term by a constantr, called the Common ratio. If the first term of the sequence isathen the geometric sequence is

a,ar,ar^{2},ar^{3}, … .

where then-th term isar^{(n-1)}.

Example 1. In a geometric sequence, the fifth term is 80 and the common ratio is -2. Determine the first three terms of the sequence.

*a*

_{5}=80 and

*r*=-2

*a*

_{5}=

*ar*

^{4}=

*a*⋅(-2)

^{4}=80

16

*a*=80

*a*=5

∴

*a*

_{1}=5,

*a*

_{2}=5⋅(-2)

^{1}=-10,

*a*

_{3}=5⋅(-2)

^{2}=20

Example 2: **Creating a Geometric Sequence**

Write the sequence in which

*a*=5,

*r*=3,

*n*=5

answer:

*a*

_{1}=

*a*=5

*a*

_{2}=

*a*

*r*=5(3)=15

*a*

_{3}=

*a*

_{2}

*r*=15(3)=45

*a*

_{4}=

*a*

_{3}

*r*=45(3)=135

*a*

_{5}=

*a*

_{4}

*r*=135(3)=405

Therefore, the required sequence is 5, 15, 45, 135, 405.

**Write terms of a geometric sequence.**

Example 3: **Writing the Terms of a Geometric Sequence**

Write the first six terms of the geometric sequence with first term 6 and common ratio ⅓.

Solution:

The first term is 6. The second term is 6⋅⅓, or 2. The third term is 2⋅⅓, or ⅔. The fourth term is ⅔⋅⅓, or 2/9, and so on. The first six terms are

**Use the formula for the general term of a geometric sequence.**

The General Term of a Geometric Sequence

The General Term of a Geometric Sequence

Consider a geometric sequence whose first term is

*a*

_{1}and whose common ratio is

*r*. We are looking for a formula for the general term,

*a*. Let’s begin by writing the first six terms. The first term is

_{n}*a*

_{1}. The second term is

*a*

_{1}⋅

*r*. The third term is

*a*

_{1}

*r⋅*, or

*r**a*

_{1}⋅

*r*

^{2}. The fourth term is

*a*

_{1}⋅

*r*

^{2}⋅

*r*, or

*a*

_{1}⋅

*r*

^{3}, and so on. Starting with

*a*

_{1}and multiplying each successive term by

*r*, the first six terms are

Can you see that the exponent on *r* is 1 less than the subscript of *a* denoting the term number?

Thus, the formula for the *n*th term is

General Term of a Geometric Sequence

Thenth term (the general term) of a geometric sequence with first terma_{1}and common ratiorisa=_{n}a_{1}⋅r^{(n-1)}.

Study Tip

Be careful with the order of operations when evaluating *a*_{1}⋅*r*^{(n-1)}.

First find *r*^{(n-1)}. Then multiply the result by *a*_{1}.

You have seen that each term of a geometric sequence can be expressed in terms of *r* and its previous term. It is also possible to develop a formula that expresses each term of a geometric sequence in terms of *r* and the first term *a*_{1}. Study the patterns shown in the table on the next page for the sequence 2, 6, 18, 54, … .

The three entries in the last column of the table all describe the

*n*th term of a geometric sequence. This leads us to the following formula for finding the

*n*th term of a geometric sequence.

Key Concept:

nth Term of a Geometric Sequence

Thenth termaof a geometric sequence with first term_{n}a_{1}and common ratioris given by

a=_{n}a_{1}⋅r^{(n-1)}.

wherenis any positive integer.

Example 4: **List the first 4 terms of each geometric sequence and find the common ratio ( r).**

a)

*a*=2

_{n}^{n}

Solution:

*a*

_{1}=2

^{1}=2;

*a*

_{2}=2

^{2}=4;

*a*

_{3}=2

^{3}=8;

*a*

_{4}=2

^{4}=16

Each term is a multiple of 2 apart so the common ratio is 2.

Answer: First 4 terms 2, 4, 8, 16 common ratio 2

b)

*a*=3⋅2

_{n}^{n}

Solution:

*a*

_{1}=3⋅2

^{1}=6;

*a*

_{2}=3⋅2

^{2}=12;

*a*

_{3}=3⋅2

^{3}=24;

*a*

_{4}=3⋅2

^{4}=48

Each term is a multiple of 2 apart so the common ratio is 2.

Answer: First 4 terms 6, 12, 24, 48 common ratio 2

c)

*a*=½

_{n}^{n}

Solution:

*a*

_{1}=½

^{1}=½;

*a*

_{2}=½

^{2}=¼;

*a*

_{3}=½

^{3}=⅛;

*a*

_{4}=½

^{4}=1/16

Each term is a multiple of ½ apart so the common ratio is ½.

Answer: first 4 terms ½, ¼, ⅛, 1/16 and the common ratio is ½.

d)

*a*=2

_{n}^{3n}

Solution:

*a*

_{1}=2

^{(3⋅1)}=8;

*a*

_{2}=2

^{(3⋅2)}=64;

*a*

_{3}=2

^{(3⋅3)}=512;

*a*

_{4}=2

^{(3⋅4)}=4096

Each term is a multiple of 8 apart so the common ratio is 8.

Answer: first 4 terms 8, 64, 512, 4096 common ratio 8

e)

*a*=2⋅3

_{n}^{2n}

Solution:

*a*

_{1}=2⋅3

^{(2⋅1)}=18;

*a*

_{2}=2⋅3

^{(2⋅2)}=162;

*a*

_{3}=2⋅3

^{(2⋅3)}=1458;

*a*

_{4}=2⋅3

^{(2⋅4)}=13122

Each term is a multiple of 9 apart so the common ratio is 9. Answer: first 4 terms 18, 162, 1458, 13122 common ratio 9.

Example 5: **Writing the terms**

Write the first five terms of the geometric sequence whose *n*th term is

*a*=3(-2)

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

Solution:

Let

*n*take the values 1 through 5 in the formula for the

*n*th term:

*a*

_{1}=3(-2)

^{(1-1)}=3

*a*

_{2}=3(-2)

^{(2-1)}=-6

*a*

_{3}=3(-2)

^{(3-1)}=12

*a*

_{4}=3(-2)

^{(4-1)}=-24

*a*

_{5}=3(-2)

^{(5-1)}=48

Notice that

*a*=3(-2)

_{n}^{(n-1)}gives the general term for a geometric sequence with first term 3 and common ratio -2. Because every term after the first can be obtained by multiplying the previous term by -2, the terms 3, -6, 12, -24, and 48 are correct.

Example 6. Write down the first four terms of the sequence *u _{n}*=8⋅¾

^{n}and show that the sequence is geometric.

Solution:

Thus,

*u*is a geometric sequence.

_{n}**Geometric sequences**

A

**geometric sequence**has the form

*a*,

*ar*,

*ar*

^{2}

*ar*

^{3}, …

in which each term is obtained from the preceding one by multiplying by a constant, called the

**common ratio**and often represented by the symbol

*r*. Note that

*r*can be pos- itive, negative or zero. The terms in a geometric sequence with negative

*r*will oscillate between positive and negative.

The doubling sequence

is an example of a geometric sequence with first term 1 and common ratio

*r*=2, while

is an example of a geometric sequence with first term 3 and common ratio

*r*=-2.

It is easy to see that the formula for the

*n*th term of a geometric sequence is

*a*=

_{n}*ar*

^{(n-1)}.

**Definition**:

A sequence *a*_{1}, *a*_{2}, …, *a _{n}*, … is called a geometric sequence. If there exists a constant

*r*, such that

*a*

_{(k+1)}=

*r*⋅

*a*;

_{k}*k*=1, 2, 3, … .

Thus, a geometric sequence looks as follows:

*a*,

*ar*,

*ar*

^{2},

*ar*

^{3}, …

where

*a*is called the first term and

*r*is called as the

**common ratio**of the geometric sequence.

The

*n*th term of the geometric sequence, is given by

*a*=

_{n}*ar*

^{(n-1)}

Some other examples of geometric sequences are

**:**

*n*th term or General term(or, last term) of a Geometric sequenceIf *a* is the first term and *r* is the common ratio then the general form of geometric sequence is

*a*,

*ar*,

*ar*

^{2},

*ar*

^{3}, … .

If *a*_{1}= 1st term =*a*

*a*

_{2}= 2nd term =

*ar*

*a*

_{3}= 3rd term =

*ar*

^{2}

…

*a*=

_{n}*n*th term =

*ar*

^{(n-1)}

Which is the

*n*th term of geometric sequence in which:

*a*= first term

*r*= common ratio

*n*= number of terms

*a*=

_{n}*n*th term=last term

Example 7. Determine the first three terms of a geometric sequence whose 8th term is 9 and whose 10th term is 25.

The first three terms of a geometric sequence is:

^{7}, 9⋅⅗

^{6}, 9⋅⅗

^{5}

Example 8. Find a geometric sequence for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution

Let *a* be the first term and *r* be the common ratio of the geometric sequence. According to the given conditions,

*a*

_{5}=4

*a*

_{3}

*ar*

^{4}=4

*ar*

^{2}

*r*

^{2}=4 ∴

*r*=±2

*a*

_{1}+

*a*

_{2}=-4

*a*+

*a*

*r*=-4

*a*(1+

*r*)=-4

For

*r*=2 then

*a*=(-4)÷(1+2)∴

*a*=-4/3. Also

For

*r*=-2 then

*a*=(-4)÷(1-2)∴

*a*=4.

Thus, the required geometric sequence is -4/3, -8/3, -16/3, … or 4, -8, 16, -32, … .

Example 9: Write down the first four terms with the following situation, will the terms be the first four terms of a geometric sequence?

The amount of air present in the cylinder when a vacuum pump removes each time ¼ of their remaining in the cylinder.

Solution:

Given

Let the initial volume of air in a cylinder be *V* liters each time ¾ of air in a remaining i.e., (1-¼).

First time, the air in cylinder is *V*.

Second time, the air in cylinder is ¾*V*.

Third time, the air in cylinder is ¾^{2} *V*.

Therefore, the sequence is *V*, ¾*V*, ¾^{2} *V*, ¾^{3} *V*, … .

Clearly, this series is a geometric sequence. With first term (*a*)=*V*, common ratio (*r*)=¾.

Example 10. *k*-1, 2*k*, and 21-*k* are consecutive terms of a geometric sequence. Find *k*.

Solution:

Equating the common ratio *r*.

Check: If

*k*=7/5 the terms are: ⅖, 7/5, 98/5. ✓ {

*r*=7}

If *k*=3 the terms are: 2, 6, 18. ✓ {*r*=3}