Derivation of Sum of Finite and Infinite Geometric Progression

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Geometric Progression, GP
Geometric progression (also known as geometric sequence) is a sequence of numbers where the ratio of any two adjacent terms is constant. The constant ratio is called the common ratio, r of geometric progression. Each term therefore in geometric progression is found by multiplying the previous one by r.
 

Eaxamples of GP:

  • 3, 6, 12, 24, … is a geometric progression with r = 2
  • 10, -5, 2.5, -1.25, … is a geometric progression with r = -1/2

 

The nth term of geometric progression
Given each term of GP as a1, a2, a3, a4, …, am, …, an, expressing all these terms according to the first term a1 will give us...
$ a_1 = a_1 $

$ a_2 = a_1 r $

$ a_3 = a_2 r = (a_1 r) \, r = a_1 r^2 $

$ a_4 = a_3 r = (a_1 r^2) \, r = a_1 r^3 $

$ \dots $

$ a_m = a_1 r^{\, m - 1} $

$ \dots $

$ a_n = a_1 r^{\, n - 1} $

 

Where
a1 = the first term, a2 = the second term, and so on
an = the last term (or the nth term) and
am = any term before the last term
 

Sum of Finite Geometric Progression
The sum in geometric progression (also called geometric series) is given by
$ S = a_1 + a_2 + a_3 + a_4 + \, \dots \, + a_n $

$ S = a_1 + a_1 r + a_1 r^2 + a_1 r^3 + \, \dots \, + a_1 r^{\, n - 1} $   →   Equation (1)
 

Multiply both sides of Equation (1) by r will have
$ Sr = a_1 r + a_1 r^2 + a_1 r^3 + a_1 r^4 + \, \dots \, + a_1 r^{\, n} $   →   Equation (2)
 

Subtract Equation (2) from Equation (1)
$ S - Sr = a_1 - a_1 r^{\, n} $

$ (1 - r)S = a_1 (1 - r^{\, n}) $

$ S = \dfrac{a_1 (1 - r^{\, n})}{1 - r} $

The above formula is appropriate for GP with r < 1.0
 

Subtracting Equation (1) from Equation (2) will give
$ Sr - S = a_1 r^{\, n} - a_1 $

$ (r - 1)S = a_1 (r^{\, n} - 1) $

$ S = \dfrac{a_1 (r^{\, n} - 1)}{r - 1} $

This formula is appropriate for GP with r > 1.0.
 

Sum of Infinite Geometric Progression, IGP
The number of terms in infinite geometric progression will approach to infinity (n = ∞). Sum of infinite geometric progression can only be defined at the range of -1.0 < (r ≠ 0) < +1.0 exclusive.
 

From
$ S = \dfrac{a_1 (1 - r^{\, n})}{1 - r} $

$ S = \dfrac{a_1 - a_1 r^{\, n}}{1 - r} $

$ S = \dfrac{a_1}{1 - r} - \dfrac{a_1 r^{\, n}}{1 - r} $
 

For n → ∞, the quantity (a1 rn) / (1 - r) → 0 for -1.0 < (r ≠ 0) < +1.0, thus,

$ S = \dfrac{a_1}{1 - r} $

 

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