Sum of a Geometric Progression

Sum of a Geometric Progression – Definition, formula, and derivation:

The series in which the ratio of any two consecutive numbers is the same is known as geometric progression. For example, consider a series containing the terms 4, 16, 64, 256, and so on. The ratio of any two consecutive numbers i.e 16 4=64 16=256 64=4. Similarly, consider a series having 81,27,9,3, and so on. The ratio of any two consecutive numbers is 1 3

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

Where a = the first term
r = common ratio
n = number of terms in the GP

Now let us learn how to find the sum of a geometric progression series here.

Sum of a geometric progression formula:

If the geometric progression can be expressed as a, ar, ar², ar³,…ar^n-1, then the sum of the geometric progression Sn = a+ar+ar²+ar³+…..+ar^n-1.

Now the common ratio can either be r = 1 or r>1 or r<1. We know that r≠0 in a geometric progression. 

If r = 1, then Sn = a+a(1)+a(1)²+a(1)³+….+a(1)^n-1 = na.

If r >1, then Sn = (a ( rn-1 ))/r-1 

And when r < 1, Sn = S = (a (1- rn ))/1-r

Where,

a = first term

r = common ratio and 

n = number of terms in the GP

A geometric progression that has an infinite number of terms is called an infinite geometric progression. The sum of an infinite geometric series can be calculated using the formula
S = a/( 1-r) where r0 and | r | <1.

Let us now understand how these formulas are derived. 

Derivation of the formulas:

The sum of a GP can be expressed in the form as Sn = a+ar+ar²+ar³+…..+ar^n-1 (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

S-Sr = (a+ar+ar²+ar³+…..+ar^n-1) – (ar+ar²+ ar³+ ar⁴……+arⁿ )

(1-r) S = a (1- rⁿ)

S = (a (1- rn ))/1-r ( where r <1)

When r >1,  S = (a ( rn-1 ))/r-1 

Now, what happens in the case of an infinite geometric progression? In that case, since the value of n is not fixed and it tends to infinity, how do we calculate the sum of the geometric progression?

Let us understand what a convergent series and what a divergent series are before we dive into the derivation of the sum of an infinite geometric progression.

An infinite series is said to be a convergent series if the sum approaches a finite number.

A divergent series is an infinite series that is not convergent. An infinite series where the numbers do not approach zero is diverging.

When the common ratio in a geometric progression is greater than 1 or less than -1, the geometric progression is a divergent series. The sum of an infinite divergent series can not be determined and it tends to infinity. 

The sum of an infinite geometric progression can be calculated only if it is a convergent series.

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))

Let us consider a GP with a common ratio of 0.5 and a  rn1-r part of the above equation. Now, if we are to calculate the sum of the first 10 terms, 0.510 = 0.00097 and for the first 20 terms, 0.520 = 0.00000095. So, we can say that when the terms increase, the value of rn will approach zero and when n → ∞, the value of a  rⁿ/(1-r)→ 0 when r0 and | r | <1.

Thus, S = a/(1-r)