Sum of infinite geometric progression

Geometric Progression

Consider a series with the terms 2, 4, 8, 16, 32, and so on. Now consider another series that contains the terms 6, 18, 54, 162, and so on. You might note that the ratio of the second and the first number or the third and the second number in the above series are the same. In series one, the common ratio is 2 while in series two, the common ratio is 3. The series in which the ratio between any two adjacent numbers is the same is known as geometric progression.

So, a geometric progression can be expressed in the form of a, ar, ar², ar³, ar⁴……ar^n-1

Where a = first number in the series
                r = common ratio

A geometric progression that has an infinite number of terms in the series is known as an infinite geometric series.

Sum of geometric progression:

To understand the sum of an infinite geometric progression, let us first understand the formula and the derivation of the sum of a finite geometric progression.

As mentioned above, a Geometric Progression having ‘n’ number of terms can be expressed as a, ar, ar², ar³, ar⁴……ar^n-1

And the sum of this geometric progression is S =  a+ar+ar²+ ar³+ ar⁴……+ar^n-1     – Eqn (1)

Multiply Equation (1) by the common ratio r, we get

Sr = ar+ar²+ ar³+ ar⁴……+arⁿ                                                                                                – Eqn (2)

Subtracting Eqn (2) from Eqn (1), we get

(1-r) S = a (1- rⁿ)

S = (a (1- rⁿ ))/(1-r) ( where r <1)

When r >1, S = (a ( rⁿ-1 ))/(r-1)

The formula to calculate the sum of an infinite geometric progression is given as S_∞ = a/(1-r)where r≠0 and | r | <1

I.e. The sum of an infinite geometric progression is possible to calculate only when the common ratio is between -1 and 1 and is not 0. If the common ratio is beyond the given range, then the sum tends towards infinity.

We know that when r <1, the sum of the geometric progression is S =(a (1- rⁿ  ))/(1-r).

In a geometric progression where n →∞,

S = (a (1- rⁿ))/(1-r)

I.e S = (a )/(1-r)- (a  rⁿ)/(1-r)

Where n → ∞, the value of (a  rⁿ)/(1-r)→ 0 when r≠0 and | r | <1.

Thus, S = (a )/(1-r)

Solved Examples

1. Calculate the sum of the first 7 numbers of the geometric series 27, 81, 243…..

Solution:

In the given geometric progression, the first term ‘a’ = 27 and the common ratio ‘r’ is 81/27=  

We know that when r>1, S = (a ( rⁿ-1 ))/(r-1)

S = (27 x (3⁷-1))/(3-1)

S = 29,511

2. Calculate the sum of an infinite geometric progression having the numbers 32,16,8,4….

                Solution:

            The first term ‘a’ in the infinite series is 32 and the common ratio ‘r’ is 16/32= 1/2

          We know the sum of an infinite geometric progression when the common ratio r <1 is

                S=  (a )/(1-r)

                S =   32/(1-0.5) 

          s= (32 )/0.5

S = 64