Sequences: Geometric Progression

A sequence of numbers is said to be a geometric progression if the ratio of any two successive terms in the sequence is always a constant. This constant difference is called “common ratio” of the geometric progression. Usually, the first term and the common ratio of the geometric sequence are denoted by a and r respectively.

i.e. if a₁, a₂, …, a_n is the geometric sequence then

a=a₁ and

r= a₂/a₁ = a₃/a₂ = ….= Any term /previous term.

So, we can re-write the above geometric progression as,

a, ar, ar², ar³,…

the n^th term can be written as T_n = ar^n-1

Example:

2,6,18,… is an geometric progression because 6/2=3,18/6=3….

Here, a=2, r=3.

Here nth term =2 * 3^n-1

Sum of n elements of geometric progression:

It is easy to find the sum of all elements in an geometric sequence using normal addition if the number of elements is less. But some times this is very difficult to find the sum if there are more number of elements in the sequence. For this, let us derive a formula for finding the sum of n elements in an geometric sequence.

Let us consider the sequence a, ar, ar², ar³,…

Let S_n be the sum of n elements of this sequence.

Then,

S_n = a+ ar+ ar²+ ar³+…+ ar^n-2+ ar^n-1 … (1)

Consider r*S_n = ar+ ar²+ ar³+…+ ar^n-2+ ar^n-1+ arⁿ … (2)

(1)-(2) gives,

S_n-rS_n =a- arⁿ

S_n (1-r) = a(1-rⁿ)

S_n = a(1-rⁿ)/(1-r), r ≠1.

Example:

Find the sum of first 5 elements of the sequence 2,6,18,…

Solution:

Here, a=2, r=3, n=5.

S_n = a(1-rⁿ)/(1-r) = 2(1-3⁵)/(1-3) = 2(-242)/(-2) = 242.

 

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