Study Guide

Sequences and Series - Arithmetic Sequences

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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've got your back. 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. For example, 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 seven terms of the arithmetic sequence given by {an} = (an – 1) + 5 whose first term, a1, is -16.

All we need to do is find seven terms by adding 5 each time, starting with -16. So easy a caveman can do it. Just mind the whole adding 5 to a negative value.

-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 this arithmetic sequence:

13, 10, 7, 4, 1, -2, …

The first term is 13, followed by 10, then 7. Hopefully you can see the common difference is -3. We're subtracting 3 each time, or adding -3.

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

{an} = (an – 1) – 3

This is actually one of the easiest things you can do with arithmetic sequences. Here's why: the recursive rule of an arithmetic sequence will always take on the general form {an} = (an – 1) + d, where d is the common difference.

Sample Problem

Find the explicit rule for this arithmetic sequence:

13, 10, 7, 4, 1, -2, …

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'll need a1 in order to do this. In this case, a1 = 13.

The nice thing once again about arithmetic sequences is their explicit rules follow a general form, too: {an} = a1 + d(n – 1), where d is the common difference and a1 is the first term of the sequence. Jot that one down on your knuckles, 'cause it'll come in handy.

For this problem, our first term is 13. And since we're subtracting 3 from each term, our common difference is d = -3. By the way, here's a quick 'n' easy way to find the common difference: just take any term in the sequence and subtract the term right before it. For example, our sequence's second term is 10 and the first is 13, so the common difference is 10 – 13 = -3. It works with any pair of consecutive terms:

7 – 10 = -3
4 – 7 = -3
1 – 4 = -3

And so on. Just remember to start with the later term:

d = an – (an – 1)

Anyway, back to the formula. Plugging in what we know…

{an} = a1 + d(n – 1)

{an} = 13 – 3(n – 1)

It's always nice to simplify things as much as we can, so let's distribute that -3.

{an} = 13 – 3n + 3

{an} = 16 – 3n

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

If n = 5, we should be at the fifth term, 1. Substituting into our shiny new rule, we get the following:

a5 = 16 – 3(5)
a5 = 16 – 15
a5 = 1

That's what we're talking about.

Sample Problem

Given the arithmetic sequence {an} = (an – 1) – 3 with a1 = 8, find the 37th term.

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. The first term is 8, so 5 would be the second after one subtracting, and so on. Ultimately, you get an answer here but lose a little bit of pride in the process.

The other method makes you feel like you know what's up. It involves changing the recursive rule you've been given to an explicit rule. It sounds worse than it is. Just remember that explicit rules look like this:

{an} = a1 + d(n – 1)

Since a1 = 8 and d = -3, we can just plug and chug.

The explicit rule would be {an} = 8 – 3(n – 1), which simplifies to {an} = 11 – 3n. Plugging in 37 for n gives us:

a37 = 11 – 3(37)
a37 = 11 – 111
a37 = -100

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