# Video: Arithmetic Sequences

Defining, identifying, and exploring arithmetic sequences through a series of examples and typical questions. Learn how to calculate the common difference between terms and use it to produce a general formula for the nth term of the sequence.

17:20

### Video Transcript

We’re going to take a look at some arithmetic sequences. We’ll take a look at the definition. And we’ll see some examples and go through some typical arithmetic sequence questions. We’ll be analysing the differences between consecutive terms in the sequence and looking for the value of the zeroth term in order to find the formula for the 𝑛th term of the sequence.

An arithmetic sequence then is a sequence of numbers in which you add the same amount to each term to get the next term, for example, five, eight, 11, 14, and 17. The first term of this sequence is five, the second term is eight, the third term is 11, the fourth term is 14, the fifth term is 17, and so on. So my little orange numbers in circles are just telling me the position of the number within that sequence. And the big blue numbers are the values of those terms. And you can see that as we move through that sequence, every time we move to the next sequence, it’s three bigger than in the previous term in the sequence.

So in this case, we’d say that consecutive terms in the sequence have a common difference of positive three. So it’s that having a common difference of something which makes it in an arithmetic sequence. Then we have a first term. And in that case, our first term has a value of five. Knowing the value of the first term and the common difference means that you can work out the value of any term in that sequence. So when you spot that pattern, that consecutive terms have a common difference, then you can say it’s an arithmetic sequence. So when you know what an arithmetic sequence is, you can answer questions like this.

Write the next three terms in the arithmetic sequence 12, 19, 26, 33.

So we’ve been told the first, second, third, and fourth terms in the sequence. And we wanna work out what the fifth, sixth, and seventh terms are. Well, to get from 12 to 19, we needed to add seven. To get from 19 to 26, we also needed to add seven. And to get from 26 to 33, yet again we needed to add seven. So the common difference is seven. So to get from the fourth to the fifth term, we just need to add seven. And 33 plus seven is 40. Add seven again to get the sixth term. 40 plus seven is 47. And then add seven again to get the seventh term, 54. So our answer is that the next three terms are 40, 47, and 54.

Let’s try another one then.

Write the next three terms in the sequence 3.6, 4.3, 5.0, 5.7, and so on.

This means that our first term is 3.6, our second term is 4.3, the third term is 5.0, and the fourth term is 5.7. We want the next three terms. So that’s the fifth, sixth, and seventh terms. Now when I look at consecutive terms to get from 3.6 to 4.3, I just add 0.7. To get from 4.3 to 5.0, I add 0.7. And to get from 5.0 to 5.7, I just add 0.7. So they’ve got a common difference of 0.7.

So this means we’ve got our arithmetic sequence. So I can just add 0.7 again repeatedly in order to get the fifth, sixth, and seventh terms. And that will give me 6.4, 7.1, and 7.8. So that’s our answer.

And we can see that arithmetic sequences can involve integer numbers. They can involve decimal numbers, fractions. So long as the difference between consecutive terms is always the same, it’s an arithmetic sequence. Okay, let’s try this question then.

Write the 10th term in the arithmetic sequence 23, 19, 15, 11, seven, and so on.

Well, now we’ve been given five terms in the sequence. And we’re not worried about the sixth, seventh, eighth, and ninth, but we want to know what the 10th term is. Now first, let’s look at the common difference. To go from 23 to 19, we have to take away four. From 19 to 15, take away four. 15 to 11, take away four. 11 to seven, take away four. So our common difference is negative four. So it’s even okay for our common differences to be negative.

Now if we think about it, there are two different ways to work out what the 10th term is. We could either work out the sixth, seventh, eighth, and ninth and then work out the 10th. Or we could say that, from seven, we’ve got to take away five lots of negative four for the five steps that take us up to the 10th term.

So doing that the first way, seven take away four is three. Three take away four is negative one. Negative one take away another four is negative five. And negative five take away another four is negative nine. And negative nine take away another four is negative 13. Or as I say, we could’ve started at seven and taking away five lots of four so that’s seven take away 20, which is negative 13. So either way, we’d have got the same answer for our 10th term.

But this is beginning to become a bit cumbersome. What if the question asked us to write the 250th term of the sequence? We wouldn’t wanna work out all the different terms up to the 250th term. So let’s look at a way of coming up with a general formula for the 𝑛th term of the sequence and then using that to predict what the 𝑛th term is going to be.

So let’s type the sequence three, seven, 11, 15, and so on. Let’s label our terms one, two, three, four. And our common difference is positive four. We have to add four to each term to get the next. Now I’m trying to come up with a formula for the general term in this sequence. So I’m gonna put some letters in here. Let’s let the position in our sequence be 𝑛. So if 𝑛 equals one, we’re talking about the first number in the sequence. If 𝑛 is two, we’re talking about the second number, and so on. And I’m gonna use this notation, 𝑡, and then in parentheses 𝑛 to represent the 𝑛th term. So when 𝑛 is one, 𝑡 one is three. When 𝑛 is two, 𝑡 two is equal to seven; 𝑡 three is equal to 11; 𝑡 four, the fourth term, is equal to 15; and so on.

Now if we look at this carefully, every time my position in the sequence goes up by one, the value of the term goes up by four. So the value of the term is going up four times faster than the position. And since 𝑛 is the value of our position, our formula must have positive four times 𝑛 in it somewhere. So let’s work out that 𝑛 is one for the first position, 𝑛 is two for the second, 𝑛 is three for the third, 𝑛 is four for the fourth, and so on. Let’s work out. If we said our formula was just positive four times 𝑛, what values would we generate in our sequence?

Well, positive four times — well, 𝑛 is one in this case for our first term — positive four times one will just be four. When 𝑛 is two, we’re talking about the second term in the sequence. So positive four times two would be eight. For the third term in the sequence, we’d be doing positive four times three, which will be 12. And for the fourth term, we’d be doing positive four times four, which will be 16. So if our formula was positive four times 𝑛, the sequence we would generate would be four, eight, 12, 16. But that’s not the values we’re looking for.

We got four. We wanted three. We got eight. We wanted seven. We got 12. We wanted 11. We got 16. We wanted 15. Well, what have I got to do to each of these numbers in order to turn the number- turn the numbers that I want?

Well, four take away one gives me three. Eight take away one gives me seven. 12 take away one would give me 11. And 16 take away one would give me 15. So if I take these four 𝑛-values and take away one, this process generates exactly the numbers I’m looking for. So the formula for the 𝑛th term of the sequence is that 𝑛th term 𝑡𝑛 is equal to positive four 𝑛. Well, we don’t normally bother writing a positive if it’s positive, so four 𝑛 minus one.

And the 250th term of this sequence, 𝑡 250, will just be four times 250 minus one. Well, four times 250 is 1000, take away one is 999. So the 250th term will be 999. We didn’t have to work out the fifth, the sixth, the seventh, and so on in order to work out the 250th term.

Even better, we can go on to work out the 1527486th term in the sequence just by plugging in that value for 𝑛 in our formula. And when we do that, we get an answer of 6109943. Now that’s saved us a lot of time working and not having to work out millions of terms in order to work out the term that we’re interested in.

Now it’s also worth pointing out there’re different ways of expressing this. Some people say 𝑡𝑛, so 𝑡 with a little 𝑛 underneath it, or 𝑎𝑛 for the 𝑛th term of the sequence. And some people use 𝑡 and then in the parentheses 𝑛. Okay, let’s try out this method then.

Find the 952nd term in the sequence 107, 99, 91, 83, and so on.

So we’ll write out our 𝑛-values, our positions in the sequence, one, two, three, four, and so on, and our 𝑛th terms. So the first term is 107; the second term, 𝑡 two, is 99; the third term, 𝑡 three, is 91; and so on. And in moving from term to term, we’re taking away eight each time. So let’s try negative eight 𝑛 for our formula.

And to do that, we’re just gonna substitute in the values of 𝑛 one, two, three, and four into that formula. So it’s gonna be negative eight times one, negative eight times two, negative eight times three, negative eight times four, and so on. So that gives us negative eight, negative 16, negative 24, and negative 32. Now they’re clearly not the numbers that we’re looking for. We got negative eight, but we wanted 107. We got negative 16, but we wanted 99. We got negative 24, but we wanted 91. And we got negative 32, but we wanted 83. So we clearly got quite the wrong numbers.

But if I add 115 to each of the numbers we got, we do get the numbers that we wanted. So our formula is negative eight 𝑛 plus 115. So the formula for our 𝑛th term, 𝑡𝑛, is negative eight 𝑛 plus 115. And when we substitute in the number 952 to find the 952nd term, the formula gives us negative eight times 952 plus 115, which is negative 7501.

Now all of this is great, but it does mean that we gotta do a lot of working out and write a lot of things down. So we’re gonna look for a slightly shorter way of doing this. So if we plotted our 𝑛-values as 𝑥-coordinates and the corresponding terms themselves as the 𝑦-coordinates, then we get a table of values that looks something like this.

Remember, the 𝑥s are really the 𝑛-values we’ve just been looking at, and the 𝑦s are the 𝑡𝑛s, the 𝑛th terms. And if we plotted those points, they’d look something like that. Now 𝑛 can only take integer values one, two, three, four, and so on because that sequence has only got terms one, two, three, and four, and so on. It doesn’t really make much sense to talk about the 3.75th term for the sequence, for example. But if we did join up those points and then just sort of extend them back to zero, they would cross here at 115. And that’s because every time I increase my 𝑥-coordinate by one, my 𝑦-coordinate decreases by eight. So if I go from this point here and decrease my 𝑥-coordinate by one, my 𝑦-coordinate is gonna go up by eight.

So now we have all the information we need to work out the equation of that line. The slope is negative eight or the gradient, and the 𝑦-intercept is 115. So the equation is 𝑦 equals negative eight 𝑥 plus 115. Now we can use this as a shortcut for working out the formula of our 𝑛th term. So the common difference told us the slope of that line. And the 𝑦-intercept tells us the number that we have to add on to the negative eight 𝑥 to get the 𝑛th term in the formula. Okay, let’s do another question and see that in action and see how much quicker it makes that question to do.

Find the formula for the 𝑛th term of the arithmetic sequence 6.8, 7.9, 9.0, 10.1, and so on.

So we’ve still got our 𝑛s: one, two, three, four, the positions in the sequence. And we’ve still got our terms, our 𝑡𝑛s, 6.8, 7.9, 9.0, 10.1, and so on. Now looking at the common difference from term to term, to get to each consecutive term, we need to add 1.1. So that tells us the multiple of 𝑛. So what we need to do is find the value of our zeroth term, the 𝑦-intercept values. So we’re looking for 𝑛 equals zero.

Now when we were moving along consecutively our terms, we were adding 1.1 to each term. So if we’re going back from the first term to the zeroth term, we’ve got to subtract 1.1. And 6.8 minus 1.1 is 5.7. So the formula for our 𝑛th term is the common difference positive 1.1 times 𝑛. Since it’s positive, we won’t bother writing the positive sign. And we then have to add on 5.7. That’s a positive 5.7. So there’s our formula. The 𝑛th term is 1.1𝑛 plus 5.7.

And if the question asked us now to find the 88th term in the sequence, the calculation we’d have to do to find out 𝑡 88 is 1.1 times 88. So 88 is the value of 𝑛, the 88th term, plus 5.7, which is 102.5. So the 88th term in our sequence would be 102.5.

So to summarise, the formula for the 𝑛th term of an arithmetic sequence is always of this format. The 𝑛th term is equal to something times 𝑛 plus something. Now the 𝑎, the something we’re timesing 𝑛 by, is the common difference of each term in the arithmetic sequence. And the 𝑏-value would be the value of the zeroth term in the sequence. And an arithmetic sequence, 𝑛th term formula, will always be in that format. If it’s got squareds or roots or powers of 𝑛 in it, it can’t be any of those formulae. They’re not linear, so they can’t represent an arithmetic sequence.

Now one more sort of question that you might see in relation to arithmetic sequences is like this. Does 117 belong to the arithmetic sequence five, 18, 31, 44, and so on? And the way you approach these is much the same in the first instances, the ones we’ve just been looking at.

First, we’re gonna analyse the common difference. And in this case, that’s plus 13 each time. So that tells us the multiple of 𝑛 in our formula. And then we’re gonna find the zeroth term of that sequence. So moving back from the first to the zeroth term, we’re moving backwards. So we’re gonna do the opposite of adding 13. That’s taking away 13. Our first term was five, so five take away 13 is negative eight. So our formula for the 𝑛th term then is the common difference times 𝑛. So that’s gonna be 13𝑛. And the zeroth term is the thing that we’re gonna be adding or taking away in this case at the end. So 𝑡𝑛 is 13𝑛 minus eight.

Now remember, we were saying before that the 𝑛-values must all be whole numbers. We only talk about the first term in the sequence, the second, the third, and so on. We don’t talk about the three and a half-th term. So if 117 is in the term, then the- there’s gonna be an 𝑛th term that is 117. And the 𝑛-value corresponding to that would be an integer, a whole number. So there’s gonna be 117 is equal to 13 times some number minus eight. Well, if I add eight to both sides of that equation, I get 125 is equal to 13𝑛. And now if I divide both sides by 13, I get 𝑛 is equal to 125 over 13. So that’s nine and eight-thirteenths. That’s not a whole number. So our answer is 117 is not an exact term in the sequence. It would be the nine and eight-thirteenth term. That’s not an integer, so it’s not a term.

Okay, one last quick question then. Does 5.2 belong to the arithmetic sequence with the general term formula 𝑡𝑛 is equal to 106 minus 1.8𝑛?

And this again boils down to the fact that if the value of 𝑛 that generates a term of 5.2 is an integer, then yes it does belong to the sequence. But if the value of 𝑛 that generates that term isn’t an integer, then it doesn’t. So we’re putting 𝑡𝑛 equal to 5.2. Now in this case, it’s negative 1.8𝑛 there. So I’m gonna add 1.8𝑛 to both sides to give me a positive number of 𝑛s. Then I can subtract 5.2 from each side. And finally divide by 1.8 to leave me with one 𝑛 on the left-hand side. And in this case, that gives me a whole number answer, 56. So our answer is, yes, 5.2 is the 56th term in the sequence.