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# Sequences and Series

# Geometric Sequences

It is our experience that people tend to wig out by the time they to geometric sequences. Shmoop to the rescue. What else is new? All you need to remember is that while arithmetic sequences add to get the next number in the sequence, geometric sequences multiply.

That means that this is a geometric sequence:

4, 8, 16, 32, 64, …

In that sequence we are multiplying by 2 each time. This means that 2 is called the **common ratio**.

The common ratio in that geometric sequence is . Your turn.

### Sample Problem

Find the common ratio for the sequence given by

Solution: Remember, the common ratio is just the number we multiply by to get to the next number in a geometric sequence. Here, our numbers are getting smaller. Instead of multiplying by a whole number, we're going to have to multiply by a fraction. This is a completely normal thing for geometric sequences by the way. Here, it is fairly easy to see that we're multiplying by ½.

The other thing it might be helpful to know at this point is the general form of a geometric sequence. It looks like this:

{*a _{n}*} =

*a*

_{1}(

*r*)

^{n-1}

This means that the *n*^{th} term, *a _{n}*, is just the first term multiplying by the common ratio,

*r*, to the

*n*

^{th }power. You're probably going to want to commit that general form equation thingy right there to memory. It will be worth it.

### Sample Problem

List the first four terms and the 20^{th} term of a geometric sequence with a first term of 1 and a common ratio of 2.

Solution: Let's first find our rule for this thing and then use it to get our terms. Since *a*_{1} = 1 and *r *= 2 we can substitute to get…

*a _{n}* =

*a*

_{1}(

*r*)

^{n-1}

*a*= 1(2)

_{n}^{n-1}

Now we can find our first four terms and even our 20^{th} term pretty easily.

*a*_{1}= 1(2)^{1-1 }= 1(2)^{0} = 1(1) = 1

a_{2}= 1(2)^{2-1 }= 1(2)^{1} = 1(2) = 2

a_{3}= 1(2)^{3-1 }= 1(2)^{2} = 1(4) = 4*a*_{4}= 1(2)^{4-1 }= 1(2)^{3} = 1(8) = 8…*a*_{20}= 1(2)^{20-1 }= 1(2)^{19} = 1(1) = 524,288

So our sequence looks like 1, 2, 4, 8,…,524,288,…S, Double U, Double E, T.

Oh, and just one thing worth noting in that example. Notice how the first term ended up with a common ratio to the power of zero. This is exactly what allows our general form to work. Since any number to the zero is just one, it means our first term will always be *a*_{1} times one. This is just *a*_{1}—every single time.

### Sample Problem

Use the geometric sequence given by to find the 5^{th} and 10^{th} terms.

Solution: Oh no. Not the 10^{th} term…You can just go ahead and use the formula on this one.

Wow, that's really small. But that also should make sense since the bigger *n* gets, the smaller the corresponding term will be. We say that the term approaches zero as *n* approaches infinity. Whoa—calculus.

### Sample Problem

Find the rule for the geometric sequence with *a*_{1} = 6 and .

Solution: Finally, a problem we sink our Shmathematical teeth into. Let's start with what we know. Since *a*_{1} = 6 we know that the general form needs to look like…

*a _{n}* = 6(

*r*)

^{n-1}

We also know that *a*_{4} = . By substituting 4 for *n* and for *a*_{4} we could solve for *r*.