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Geometric Progression

A sequence of terms is said to be a geometric progression (G.P.) if the ratio of any term to its preceding term is a constant.


This constant is called the common ratio of the G.P. It is found by dividing any term of the sequence by the preceding term.


Example 7.16: 2, 4, 8, 16... is in geometric progression.


Let ‘a’ be the 1st term of a G.P. and ‘r ’ be the common ratio, then the G.P. can be written as follows:


aarar 2ar 3 … ar (n – 1).


Form a G.P. , when the 1st term is 5 and the common ratio is 2.
Using the 1st term ‘a’ and the common ratio ‘r ’, the G.P. can be formed in the following way:
aarar 2ar 3… ar (n – 1)
Here, a = 5 and r = 2.
Substituting the above values, the required G.P. will be
5, 5(2), 5(2)2, 5(2)3…… 5(2)(n – 1) = 5, 10, 20, 40….. 5(2)(n – 1)

  1. nth term of a geometric progression:
    In terms of the G.P., we see that the exponent of ‘r’ is 1 in the 2nd term, 2 in the third term and 3 in the 4th term. It is one less than the number of the term. So for the nth term, exponent of ‘r’ will be (n – 1).
    Hence, we can write the nth term of a G.P. as
    Tn ar n – 1

Which term of the geometric progression
  1. Sum of n terms of a geometric progression:
    The sum of n terms of a G.P is given by

The sum of 10 terms of a G.P., 3 + 6 + 12 + ... is
Given, a = 3, r = 2 and n = 10
Since r > 1, we can write
The sum of ifinite terms in a decreasing G.P. is  
a = first term of the progression r = common ratio


The sum of infinite series  is


Note: If the product of terms in a G.P. is known and you are asked to find the terms, then the terms should be selected in the following ways while solving problems:

  • When the product of three terms is given, the terms must be selected as
  • When the product of four terms is given, the terms must be selected as
  • When the product of five terms is given, the terms must be selected as


Find three numbers in a geometric progression, whose sum is  and whose product is 729.
We know the product of 3 terms.
Hence the terms shall be taken in the form
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  • The reciprocal of the terms of a G.P. is also a G.P.
  • For given two positive numbers A.M. ≤ G.M

Important formulae:

  • The sum of first n natural numbers is given by
  • The sum of first n odd numbers is given by Σ(2n −1) = n2.
  • The sum of first n even numbers is given by Σ2n =n(n +1)
  • The sum of squares of first n natural numbers is given by
  • The sum of cubes of first n natural numbers is given by

Geometric Mean

Geometric Mean of any two numbers ‘a’ and ‘b’ is given by  


Example: The geometric mean of two numbers 9 and 16 is


When the geometric mean is placed in between the numbers, the resulting numbers will be in G.P.


If a, b, c are in G.P. then, b is called the G.M. between a and c and  


Example: Consider three numbers in a G.P. 2, 4 and 8.


We can see that  


Between two given numbers, we can insert any number of terms such that the series thus formed shall be in G.P. The terms inserted are called the geometric means.


To insert ‘n’ geometric means between two numbers a and b, we take the common ratio as



G1 = ar
G2 = ar 2
G3 = ar 3
and so on up to..
Gn = ar n


The product of n geometric means between the terms a and b is




Insert three geometric means between 2 and 32.
Given a = 2, b = 32 and n = 3.
The common ratio for inserting the geometric means can be calculated as
Using this, the three geometric means will be
G1 = 2 × 2 = 4
G2 = 2 × 22 = 8
G3 = 2 × 23 = 16
To check if the answers are correct, write down the geometric means that you calculated in between the two given numbers. If the resulting numbers are in G.P., then the geometric means you calculated are correct.
In the above example, aG1G2G3b = 2, 4, 8, 16, 32 respectively, which is a geometric progression with a common ratio 2. Hence, we are sure that our calculation is correct.


Let us make use of the terms from the above example and find the product of three G.M.s between 2 and 32.
a = 2, b = 32 and n = 3.
The product of n geometric means is

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