Geometric Sequence or Geometric Progression


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

  1. 2,6,18,54, \ldots is a G.P with r = 3
  2. 8, - 4,2, - 1, \ldots is a G.P with r = - \frac{1}{2}
  3. 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}}



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

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



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

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