Chapter Contents

• Series and the Binomial Theorem
• 1. Arithmetic Progressions
• 2. Geometric Progressions
• 3. Infinite Geometric Series
• 4. The Binomial Theorem
• Series and the Binomial Theorem Problem Solver

• Comments, Questions?

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2. Geometric Progressions

by M. Bourne

A Geometric Progression (GP) is formed by multiplying a starting number (a₁) 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 = a₁rⁿ⁻¹

Explanation

Example 2

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

Answer

The Sum of a Geometric Progression

The sum to n terms of a GP means:

a₁ + a₁r + a₁r² + a₁r³ + ... + a₁r^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).

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"`?

Answer