# Derivation of Sum of Finite and Infinite Geometric Progression

**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 n ^{th} term of geometric progression**

Given each term of GP as a

_{1}, a

_{2}, a

_{3}, a

_{4}, …, a

_{m}, …, a

_{n}, expressing all these terms according to the first term a

_{1}will give us...

$a_1 = a_1$

$a_2 = a_1 r$

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

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

$\dots$

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

$\dots$

Where

a_{1} = the first term, a_{2} = the second term, and so on

a_{n} = the last term (or the n^{th} term) and

a_{m} = 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)

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

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

The above formula is appropriate for GP with r < 1.0

Subtracting Equation (1) from Equation (2) will give

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

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

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 (a_{1} r^{n}) / (1 - r) → 0 for -1.0 < (r ≠ 0) < +1.0, thus,

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