# Geometric Sequences

How to solve the geometric sequence problems: formula, proof, examples, and their solutions.

## Formula

A geometric sequence is a sequence

whose ratios of the adjacent terms

are the same (= *r*).

*a*_{n} = *a*_{1}⋅*r*^{n - 1}*a*_{n}: *n*th term*a*_{1}: first term*r*: common ratio

## Proof

If the common ratio is *r*, then*a*_{1} = *a*_{1}*a*_{2} = *a*_{1}⋅*r*^{1}*a*_{3} = *a*_{1}⋅*r*^{2}

...*a*_{n} = *a*_{1}⋅*r*^{n - 1}.

## Example 1

Find *a*_{1} and *r*.*a*_{1} = 2, *r* = 3

*a*_{1} = 2, *r* = 3*a*_{n} = 2⋅3^{n - 1}

## Example 2

Find *a*_{1} and *r*.*a*_{1} = 320, *r* = 1/2

*a*_{1} = 320, *r* = 1/2*a*_{n} = 320⋅(1/2)^{n - 1}

320 = 2_{6}⋅5

Prime factorization

(1/2)^{n - 1} = 2^{-n + 1}

Negative exponent

*a*_{k} = 5⋅2^{-k + 7}

And it says *a*_{k} = 5/8.

So *a*_{k} = 5⋅2^{-k + 7} = 5/8.

## Example 3

Write *a*_{2} and *a*_{5}

by using *a*_{1} and *r*.*a*_{2} = *a*_{1}⋅*r* = -6*a*_{5} = *a*_{1}⋅*r*^{4} = 48

The goal is to find *a*_{1} and *r*.

Find *r*

from *a*_{1}⋅*r*^{4} / *a*_{1}⋅*r* = 48 / -6.*r* = -2.

Put *r* = -2 into *a*_{1}⋅*r* = -6.

Then *a*_{1} = 3.

*a*_{1} = 3, *r* = -2

So *a*_{n} = 3⋅(-2)^{n - 1}.