# Geometric Sequence or Geometric Progression

__Introduction__

Geometric sequence is a second type of sequence. One important application of geometric progression is computing interest on savings accounts. Other applications can be found in biology and physics. Geometric sequence also has an application for finding the national income or population for any particular future year provided the values for the current year and the growth rate are known.

__Geometric Sequence__

A **geometric** (**exponential**) **sequence** or **progression** (abbreviated as **G.P**) is a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by a fixed number. The **fixed number** is called the **common ratio **and is usually denoted by $$r$$.

To determine whether or not a sequence of numbers is a G.P., we divide each member by the one which precedes it.

For example, the sequence

- $$2,6,18,54, \ldots $$ is a G.P with $$r = 3$$
- $$8, – 4,2, – 1, \ldots $$ is a G.P with $$r = – \frac{1}{2}$$
- $$0.1,0.01,0.001,0.0001, \ldots $$ is a G.P with $$r = \frac{1}{{10}}$$

In other words, quantities are said to be in geometric progression when they increase or decrease by a **constant factor**.

__The nth term of a geometric sequence__

Let $${a_1}$$ be the first term and $$r$$ be the common ratio. Then, the second term is $${a_1}r$$ and the third term is $${a_1}{r^2}$$. In each of these terms the exponent of $$r$$ is $$1$$ less than the number of the term. Similarly, the $$nth$$ term is the $$\left( {n – 1} \right)th$$ after the first and is found by multiplying $$a$$ by $$\left( {n – 1} \right)$$ factors $$r$$, or by $${r^{n – 1}}$$. Hence, if $${a_n}$$ represents the $$nth$$ term, then

\[{a_n} = {a_1}{r^{n – 1}}\]

__Example__:

Given that $$16,8,4$$ are the first three terms of a geometric sequence, find the fifth term of the sequence.

__Solution__:

Here $${a_1} = 16$$, $$r = \frac{8}{{16}} = \frac{1}{2}$$

By using the formula,

$${a_n} = {a_1}{r^{n – 1}}{\text{ }}$$

$$\therefore $$ The fifth term, $${a_5} = 16{\left( {\frac{1}{2}} \right)^{5 – 1}}$$

$$ = 16{\left( {\frac{1}{2}} \right)^4} = 16\left( {\frac{1}{{16}}} \right) = 1$$

__Example__:

What is the first term of a six-term geometric sequence in which the ratio is $$\sqrt 3 $$ and the sixth term is $$27$$?

__Solution__:

Here $${a_n} = 27$$, $$r = \sqrt 3 $$, $$n = 6$$

To find $${a_1}$$ by the formula

$${a_n} = {a_1}{r^{n – 1}}$$

$${a_6} = {a_1}{\left( {\sqrt 3 } \right)^{6 – 1}}$$

$$27 = {a_1}{\left( {\sqrt 3 } \right)^5}{\text{ }} \Rightarrow \frac{{27}}{{9\sqrt 3 }} = {a_1}{\text{ }} \Rightarrow {a_1} = \sqrt 3 $$