# Video: Formulas for Arithmetic Sequences

The third term in an arithmetic sequence is 2 and the sixth term is 11. If the first term is 𝑎₁, what is an equation for the 𝑛th term of this sequence.

05:30

### Video Transcript

The third term in an arithmetic sequence is two and the sixth term is 11. If the first term is 𝑎 one, what is an equation for the 𝑛th term of this sequence?

So in order to actually solve this problem, there’s a couple of methods we can use. So I’m actually gonna give an example of each. So the first thing to bear in mind is that actually we’re looking at an arithmetic sequence. And what we have when we’re dealing with an arithmetic sequence is a general rule for any term. And that general rule is that 𝑎 𝑛, so the 𝑛th term, is equal to the first term plus 𝑛 minus one — and 𝑛 is the term position — multiplied by 𝑑, which is our common difference.

Okay, so now we actually have this general rule, let’s actually use this to help us find out what the actual 𝑛th term of our sequence is going to be. So if we take a look at the question, we have two bits of information that are very important. We have the third term, which is equal to two, and the sixth term, which is equal to 11.

So what we can actually do is use these two bits of information to actually form two simultaneous equations, which can actually help us find our 𝑎 and 𝑑, so our first term and our common difference. So first of all, if we use the fact that the third term is equal to two, we can say that two is equal to 𝑎 one, so our first term, plus three minus one 𝑑, and that’s three minus one because actually that was our term number, which is three, and then minus one.

So we can actually just rewrite this as two is equal to 𝑎 one, so our first term, plus two 𝑑, and that’s cause three minus one is two. And then we can label this equation one because this is our first simultaneous equation. And then using the second bit of information, which is that the sixth term is equal to 11, we can say that 11 is equal to 𝑎 one plus six minus one 𝑑. So therefore, we can write this as 11 is equal to 𝑎 one plus five 𝑑, and that’s because we have six minus one which is five. Great! And we’ve also labeled this equation two.

So now to enable us to find our 𝑛th term, what we’ll need to know is our 𝑎 one, so our first term, and our 𝑑, which is our common difference. So what we’re gonna do is actually subtract equation one from equation two to eliminate our 𝑎 one. And when we do this, we get nine is equal to three 𝑑, and that’s because you had 11 minus two, which is nine, and then you had 𝑎 one minus 𝑎 one, which is zero, and then five 𝑑 minus two 𝑑, which is three 𝑑.

And then if we actually divide both sides of the equation by three, we’re gonna get three is equal to 𝑑, so we’ve found the value of 𝑑. So the common difference is going to be three. So now what we can do is actually substitute 𝑑 equals three into equation one. You could actually substitute it into either equation, but we’re gonna choose equation one for this case to find 𝑎 one.

So if we do that, we get two is equal to 𝑎 one plus two multiplied by three. So we’re gonna get two is equal to 𝑎 one plus six. So then if we subtract six from each side of the equation, we get negative four is equal to 𝑎 one. So therefore, we’ve actually found our first term.

So now what we can do is actually write our 𝑛th term. So we can say 𝑎 𝑛, so our 𝑛th term, is equal to negative four, because this is our 𝑎 one value, then plus three multiplied by 𝑛 minus one, and that’s because our 𝑑 value, our common difference, was three.

So now we can expand the parentheses. And when we do that, we’re gonna get our 𝑛th term is equal to negative four plus three 𝑛 because we had three multiplied by 𝑛, which gives us positive three 𝑛 and then minus three, because we had positive three multiplied by negative one. So therefore, we get 𝑎 𝑛, so our 𝑛th term, is equal to three 𝑛 minus seven, and that’s because we had negative four minus three, which gave us negative seven.

Okay, great! So we’ve solved the problem. But what I said is that I’ll actually show you another method you could have used to actually find this. So we’re gonna do that now just to check our answer. Okay, so to use this method, what I’ve actually done is that I’ve written out our term numbers: one, two, three, four, five, six. And I’ve actually put in the information we’ve got.

So we know that the third term is equal to two and the sixth term is equal to 11. And what we can actually see is that there are actually three jumps to get from three to six. So we get from three to four, four to five, and five to six. And we know the difference between two and 11 is actually nine, because you have to add nine onto two to get to 11.

So therefore, what this will give us is our common difference of positive three. So and if we’re trying to find the 𝑛th term, we know that it’s gonna be 𝑎 𝑛, so our 𝑛th term, is equal to three 𝑛. But then we need to work out, well, it’s gonna be three 𝑛, what three 𝑛 plus anything three 𝑛 minus anything.

So to work out the next part of our 𝑛th term, we can actually look at the term we had, which is a third term, which equals two. Well, we can see that three multiplied by three is gonna be equal to nine, so what’d that be is because it’s three 𝑛 it’s three multiplied by our term number, which is three, so three by three, which is nine. But we know that the actual term value is gonna be two because we know the third term is equal to two.

So what do we need to do to get from nine to two? Well, we’re needing to actually subtract seven. So now what we use is we use the negative seven to actually form the second part of our 𝑛th term. So we can say that the 𝑛th term is equal to three 𝑛 minus seven. So then if we check back to previous method, it’s exactly the same.

So great! What we’ve actually done is shown you how to use two methods. And we’ve also checked our answer. So we can say that if the third term of an arithmetic sequence is two and the sixth term is 11 and the first term is 𝑎 one, the equation for the 𝑛th term of that sequence is going to be equal to three 𝑛 minus seven.
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