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# Sequences and Series

# Arithmetic Sequences

Okay Shmoopers, if you've made it this far, you're ready to get a little more specific with all this sequence and series business. Luckily, we are here to meet your every need and desire. Totally. Or is it just totes?

The first specific type of sequence/series you'll need to know is titled arithmetic. Pronounced more like air-ith-MET-ick than the grade school arithmetic. We wouldn't want you to sound unintelligent talking about such things on your next hot date.

**Arithmetic sequences** are just sequences that go from one term to the next by adding or subtracting the same value. 7, 3, -1, -5, -9, -13,…is an arithmetic sequence. To get to the next term, you just subtract 4.

### Sample Problem

List the first 7 terms of the arithmetic sequence given by {*a _{n}*} = (

*a*

_{n}_{-1}) + 5 whose first term,

*a*

_{1}is -16.

Solution: All we have to do is find seven terms by adding five each time, starting with -16. So easy a caveman can do it. Just mind the whole adding 5 to a negative value. Here on planet earth -16 + 5 ≠ -21.

-16, -11, -6, -1, 4, 9, 14

In this problem, 5 is known as the **common difference**. That term will come back over and over again through this section.

### Sample Problem

Find the recursive rule for the arithmetic sequence given by 13, 10, 7, 4, 1, -2, …

Solution: The first term is 13 followed by 10 then 7. Hopefully you can see the common difference is -3.

Next we must write our recursive rule, which looks a little something like…

{*a _{n}*} =

*a*

_{n}_{-1}– 3

This is actually one of the easiest things you can do with arithmetic sequences because the recursive rule will always take on the form {*a _{n}*} =

*a*

_{n}_{-1}+

*d*.

*d*is the common difference.

### Sample Problem

Find the explicit rule for the arithmetic sequence given by 13, 10, 7, 4, 1, -2, …

Solution: Same sequence, different type of rule. Remember that an explicit rule is a rule that allows you to plug in the term number, *n*, and get out the corresponding number in the sequence. For arithmetic sequences, we will need *a*_{1} in order to do this. In this case *a*_{1} = 13.

The nice thing once again about arithmetic sequences is their explicit rules follow the same general form too. Here that form is {*a _{n}*} =

*a*

_{1}+

*d*(

*n –*1). Plugging in what we know…

{*a _{n}*} = 13 – 3(

*n –*1)

We can check this, too, which is always a plus. #mathpun

If *n *= 5 we should the fifth term, 1. Substituting into our rule we have the following:

*a*_{5} = 13 – 3(5 – 1)*a*_{5} = 13 – 3(4)*a*_{5} = 13 – 12*a*_{5} = 1

That's what we're talking about.

### Sample Problem

Given the arithmetic sequence {*a _{n}*} = (

*a*

_{n}_{-1}) – 3 with

*a*

_{1}is 8, find the 37

^{th}term.

Solution: There are really two ways to attack this problem. One way takes forever and makes you feel like you have the intelligence level of a second grader. That would be the solution you get by subtracting 3 from 8, 36 times. Wait, not 37? No, 36. 8 is the first term. So 5 would be the second after one subtraction and so on. Ultimately, you get an answer here but lose a little bit of pride in process.

The other method makes you feel like you know what's up. It involves changing the recursive rule you are given to an explicit rule. It sounds worse than it is. Just remember that explicit rules look like {*a _{n}*} =

*a*

_{1}+

*d*(

*n –*1).

Since *a*_{1} = 8 and *d* = -3, we can just plug and chug.

The explicit rule would be {*a _{n}*} = 8 – 3(

*n –*1). Plugging in 37 for

*n*gives us

*a*_{37} = 8 – 3(37 – 1) *a*_{37} = 8 – 3(36)*a*_{37} = 8 – 108*a*_{37} = -100