# Geometric Sequences and Sums

## Sequence

A Sequence is a set of things (usually numbers) that are in order.

## Geometric Sequences

In a **Geometric Sequence** each term is found by **multiplying** the previous term by a **constant**.

### Example: 1, 2, 4, 8, 16, 32, 64, 128, 256, ...

This sequence has a factor of 2 between each number.

Each term (except the first term) is found by **multiplying** the previous term by **2**.

**In General** we write a Geometric Sequence like this:

{a, ar, ar^{2}, ar^{3}, ... }

where:

**a**is the first term, and**r**is the factor between the terms (called the**"common ratio"**)

### Example: {1,2,4,8,...}

The sequence starts at 1 and doubles each time, so

**a=1**(the first term)**r=2**(the "common ratio" between terms is a doubling)

And we get:

{a, ar, ar^{2}, ar^{3}, ... }

= {1, 1×2, 1×2^{2}, 1×2^{3}, ... }

= {1, 2, 4, 8, ... }

But be careful, **r** should not be 0:

- When
**r=0**, we get the sequence {a,0,0,...} which is not geometric

## The Rule

We can also calculate **any term** using the Rule:

x_{n} = ar^{(n-1)}

(We use "n-1" because ar^{0} is for the 1st term)

### Example: 10, 30, 90, 270, 810, 2430, ...

This sequence has a factor of 3 between each number.

The values of **a** and **r** are:

**a = 10**(the first term)**r = 3**(the "common ratio")

The Rule for any term is:

x_{n} = 10 × 3^{(n-1)}

So, the **4th** term is:

x_{4} = 10×3^{(4-1)} = 10×3^{3} = 10×27 = 270

And the **10th** term is:

x_{10 }= 10×3^{(10-1)} = 10×3^{9} = 10×19683 = 196830

A Geometric Sequence can also have **smaller and smaller** values:

### Example: 4, 2, 1, 0.5, 0.25, ...

This sequence has a factor of 0.5 (a half) between each number.

Its Rule is **x _{n} = 4 × (0.5)^{n-1}**

## Why "Geometric" Sequence?

Because it is like increasing the dimensions in **geometry**:

^{2}

^{3}

Geometric Sequences are sometimes called Geometric Progressions (G.P.’s)

## Summing a Geometric Series

**To sum these:**

a + ar + ar^{2} + ... + ar^{(n-1)}

(Each term is ar^{k}, where k starts at 0 and goes up to n-1)

**We can use this handy formula:**

**a** is the first term

**r** is the **"common ratio"** between terms

**n** is the number of terms

*What is that funny Σ symbol?* It is called Sigma Notation

And below and above it are shown the starting and ending values:

It says "Sum up * n* where

*goes from 1 to 4. Answer=*

**n****10**

The formula is easy to use ... just "plug in" the values of **a**, **r** and **n**

### Example: Sum the first 4 terms of 10, 30, 90, 270, 810, 2430, ...

This sequence has a factor of 3 between each number.

The values of **a**, **r** and **n** are:

**a = 10**(the first term)**r = 3**(the "common ratio")**n = 4**(we want to sum the first 4 terms)

So:

Becomes:

You can check it yourself:

10 + 30 + 90 + 270 = 400

And, yes, it is easier to just add them *in this example*, as there are only 4 terms.

But imagine adding 50 terms ... then the formula is much easier.

## Using the Formula

Let's see the formula in action:

### Example: Grains of Rice on a Chess Board

On the page Binary Digits we give an example of grains of rice on a chess board. The question is asked:

When we place rice on a chess board:

- 1 grain on the first square,
- 2 grains on the second square,
- 4 grains on the third and so on,
- ...

... **doubling** the grains of rice on each square ...

**... how many grains of rice in total?**

So we have:

**a = 1**(the first term)**r = 2**(doubles each time)**n = 64**(64 squares on a chess board)

So:

Becomes:

= \frac{1−2^{64}}{−1} = 2^{64} − 1

= 18,446,744,073,709,551,615

Which was exactly the result we got on the Binary Digits page (thank goodness!)

And another example, this time with **r** less than 1:

### Example: Add up the first 10 terms of the Geometric Sequence that halves each time:

### { 1/2, 1/4, 1/8, 1/16, ... }

The values of **a**, **r** and **n** are:

**a = ½**(the first term)**r = ½**(halves each time)**n = 10**(10 terms to add)

So:

Becomes:

Very close to 1.

*(Question: if we continue to increase n, what happens?)*

## Why Does the Formula Work?

Let's see **why** the formula works, because we get to use an interesting "trick" which is worth knowing.

**First**, call the whole sum

**"S"**: S = a + ar + ar

^{2}+ ... + ar

^{(n−2)}+ ar

^{(n−1)}

**Next**, multiply

**S**by

**r**:S·r = ar + ar

^{2}+ ar

^{3}+ ... + ar

^{(n−1)}+ ar

^{n}

Notice that S and S·r are similar?

Now **subtract** them!

**Wow! All the terms in the middle neatly cancel out.**

(Which is a neat trick)

By subtracting S·r from S we get a simple result:

S − S·r = a − ar^{n}

Let's rearrange it to find S:

**S**and

**a**:S(1−r) = a(1−r

^{n})

**(1−r)**:S = a \frac{(1−r^{n})}{(1−r)}

Which is our formula (ta-da!):

## Infinite Geometric Series

So what happens when n goes to **infinity**?

We can use this formula:

But **be careful**:

**r** must be between (but not including) **−1 and 1**

and **r should not be 0** because the sequence {a,0,0,...} is not geometric

So our infnite geometric series has a **finite sum** when the ratio is less than 1 (and greater than −1)

Let's bring back our previous example, and see what happens:

### Example: Add up ALL the terms of the Geometric Sequence that halves each time:

### { \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, ... }

We have:

**a = ½**(the first term)**r = ½**(halves each time)

And so:

= \frac{½×1}{½} = 1

Yes, adding **\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...** etc equals **exactly 1**.

Don't believe me? Just look at this square: By adding up we end up with the whole thing! |

## Recurring Decimal

On another page we asked "Does 0.999... equal 1?", well, let us see if we can calculate it:

### Example: Calculate 0.999...

We can write a recurring decimal as a sum like this:

And now we can use the formula:

Yes! 0.999... * does* equal 1.

So there we have it ... Geometric Sequences (and their sums) can do all sorts of amazing and powerful things.