# 2. Geometric Progressions

by M. Bourne

A Geometric Progression (GP) is formed by **multiplying** a
starting number (*a*_{1}) by a number *r*,
called the **common ratio.**

### Example 1

The progression `5, 10, 20, 40, 80, 160`, has first term `a_1= 5`, and common ratio `r = 2`.

In this example, we started with `5` and multiplied by `2` each time to get the next number in the progression.

## Formula for the `n`-th term of a GP

The *n*-th term of a geometric progression is given by:

a_{n}=a_{1}r^{n−1}

### Example 2

Find the 50^{th} term of the
geometric progression 5, 10, 20, 40, 80, ...

Continues below ⇩

## The Sum of a Geometric Progression

The sum to *n* terms of a GP means:

a_{1}+a_{1}r+a_{1}r^{2}+a_{1}r^{3}+ ... +a_{1}r^{n-1}

We can show (using Proof by Induction) that this sum is equivalent to:

`S_n=(a_1(1-r^n))/(1-r)\ (r!=1)`

### Example 3

(We first saw this story in the Chapter Introduction).

[Image source.]

A king once promised a prince anything he wanted because he saved the princess's life. The prince requested one grain of rice on the first square of a chess board, `2` on the second, `4` on the third, `8` on the fourth square, etc.

How much rice is there if one grain of rice weighs `20\ "mg"`?

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