# Video: Arithmetic Sequences

Tim Burnham

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, eleven, fourteen, and seventeen. The first term of this sequence is five, the second term is eight, the third term is eleven, the fourth term is fourteen, the fifth term is seventeen, 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 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, 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, the 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 twelve, nineteen, twenty-six, thirty-three.

So we’ve been told the first, second, third, and fourth terms in the sequence, and we want to work out what the fifth, sixth, and seventh terms are. Well to get from twelve to nineteen, we needed to add seven. To get from nineteen to twenty-six, we also needed to add seven. And to get from twenty-six to thirty-three, 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 thirty-three plus seven is forty. Add seven again to get the sixth term, forty plus seven is forty-seven. And then add seven again to get the seventh term, fifty-four. So our answer is that the next three terms are forty, forty-seven, and fifty-four.

Let’s try another one then. Write the next three terms in the sequence three point six, four point three, five point zero, five point seven, and so on. This means that our first time is three point six, our second term is four point three, the third term is five point zero, and the fourth term is five point seven. 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 three point six to four point three, I just add an o point seven. To get from four point three to five point zero, I add an o point seven. And to get from five point zero to five point seven, I just add an o point seven. So they’ve got a common difference of o point seven. So this means we’ve got our arithmetic sequence, so I can just add an o point seven again repeatedly in order to get the fifth, sixth, and seventh terms. And that will give me six point four, seven point one, and seven point eight, 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 tenth term in the arithmetic sequence twenty-three, nineteen, fifteen, eleven, 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 tenth term is. Now first, let’s look at the common difference. To go from twenty-three to nineteen, we have to take away four. From nineteen to fifteen, take away four. Fifteen to eleven, take away four. Eleven to seven, take away four. So our common difference is negative four. So it’s even okay our common differences to be negative.

Now if we think about it, there are two different ways to work out what the tenth term is. We can either work out the sixth, seventh, eighth, and ninth and then work out the tenth 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 tenth 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 thirteen. Or as I say, we could have started at seven and taking away five lots of four so that’s seven take away twenty, which is negative thirteen. So either way, we’d have got the same answer for our tenth term. But this is beginning to become a bit cumbersome. What if the question asked us to write the two hundred and fiftieth term of the sequence? We wouldn’t want to work out all the different terms up to the two hundred and fiftieth 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, eleven, fifteen, 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 going to 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 eleven, 𝑡 four the fourth term is equal to fifteen, 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 𝑛 is one for the first position, 𝑛 is two for the second, 𝑛 is three for the third, and 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 would 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 would be twelve. And for the fourth term, we’d be doing positive four times four which would be sixteen. So if our formula was positive four times 𝑛, the sequence we would generate would be four, eight, twelve, sixteen, but that’s not the values we looking.

We got four, we wanted three. We got eight, we wanted seven. We got twelve, we wanted eleven. We got sixteen, we wanted fifteen. Well, what have I got to do to each of these numbers in order to turn them- turn [get] the numbers that I want? Well four take away one gives me three, eight take away one gives me seven, twelve take away one would give me eleven, and sixteen take away one would give me fifteen. 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 the 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 two hundred and fiftieth term of this sequence, 𝑡 two hundred and fifty would just be four times two hundred and fifty minus one.

Well four times two hundred and fifty is a thousand take away one is nine hundred and ninety-nine. So the two hundred and fiftieth term would be nine hundred and ninety-nine. We didn’t have to work out the fifth, the sixth, the seventh, and so on in order to work out the two hundred and fiftieth term. Even better, we can go on to work out the one millionth five hundred and twenty-seven thousand four hundred and eighty-sixth 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 six million one hundred and nine thousand nine hundred and forty-three. 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 we’re interested in. Now it’s also worth noting 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 nine hundred and fifty-second term in the sequence one hundred and seven, ninety-nine, ninety-one, eighty-three, and so on. So we will 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 a hundred and seven, the second term 𝑡 two is ninety-nine, the third term 𝑡 three is ninety-one, and so on. And 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 sixteen, negative twenty-four, and negative thirty-two. Now that clearly not the numbers that we’re looking for. We got negative eight but we wanted a hundred and seven. We got negative sixteen, but we wanted ninety-nine. We got negative twenty-four, but we wanted ninety-one. And we got negative thirty-two, but we wanted eighty-three. So we clearly got quite the wrong numbers. But if I add a hundred and fifteen to each of the numbers we’ve got, we do get the numbers that we wanted. So our formula is negative eight 𝑛 plus a hundred and fifteen. So the formula for our 𝑛th term, 𝑡𝑛, is negative eight 𝑛 plus a hundred and fifteen. And when we substitute in the number nine hundred and fifty-two to find the nine hundred and fifty-second term, the formula gives us negative eight times nine hundred and fifty-two plus one hundred and fifteen, which is negative seven thousand five hundred and one.

Now all of this is great, but it does mean that we’ve got to 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 already the 𝑛-values we’ve just been looking at, and the 𝑦s are the 𝑡𝑛s, the 𝑛th term.

And if we plotted those points, they 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 three point seven fifth 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 a hundred and fifteen. But 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 of the gradient and the 𝑦 intercept is a hundred and fifteen. So the equation is 𝑦 equals negative eight 𝑥 plus a hundred and fifteen. 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 six point eight, seven point nine, nine point zero, ten point one, and so on. So we’ve still got our 𝑛s: one, two, three, and four, the positions in the sequence. And we’ve still got our terms, our 𝑡𝑛s, six point eight, seven point nine, nine point zero, ten point one, and so on. Now looking at the common difference from term to term, to get to each consecutive term, we need to add one point one. 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 one point one to each term. So if we’re going back from the first term to the zeroth term, we’ve got to subtract one point one. And six point eight minus one point one is five point seven. So the formula for our 𝑛th term is the common difference positive one point one times 𝑛. Since it’s positive, we won’t bother writing the positive sign. And we then have to add on five point seven. That’s a positive five point seven. So there’s our formula, the 𝑛th term is one point one 𝑛 plus five point seven. And if the question asked us now to find the eighty-eighth term in the sequence, the calculation we’d have to do to find out 𝑡 eighty-eight is one point one times eighty-eight. So eighty-eight is the value of 𝑛, the eighty-eighth term, plus five point seven, which is a hundred two point five. So the eighty-eighth term in our sequence would be a hundred two point five.

So to summarize 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 times-ing 𝑛 by, is the common difference of each term in our 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 squares 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 our arithmetic sequences is like this. Does one hundred and seventeen belong to the arithmetic sequence five, eighteen, thirty-one, forty-four, 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 thirteen each time, so that tells us the multiple of 𝑛 in our formula. And then we’re going to find the zeroth term of that sequence. So moving back from the first to the zeroth term, we’re moving backwards. So we’re going to do the opposite of adding thirteen. That’s taking away thirteen. Our first term was five, so five take away thirteen is negative eight. So our formula for the 𝑛th term then is the common difference times 𝑛. So that’s going to be thirteen 𝑛. And the zeroth term is the thing that we’re going to be adding or taking away in this case at the end, so 𝑡𝑛 is thirteen 𝑛 minus eight.

Now remember we’re 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 term. So if a hundred and seventeen is in the term, then the-there’s going to be an 𝑛th term that is a hundred and seventeen, and the 𝑛-value corresponding to that would be an integer, a whole number. So there’s going to be a hundred and seventeen is equal to thirteen times some number minus eight. What if I add eight to both sides of that equation? I get a hundred and twenty-five is equal to thirteen 𝑛. And now if I divide both sides by thirteen, I get 𝑛 is equal to a hundred and twenty-five over thirteen, so that’s nine and eight-thirteenths. That’s not a whole number. So our answer is a hundred and seventeen 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 five point two belong to the arithmetic sequence with the general term formula 𝑡𝑛 is equal to a hundred and six minus one point eight 𝑛? And this again boils down to the fact that if the value of 𝑛 that generates a term of five point two 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 five point two. Now in this case, it is negative one point eight 𝑛 there. So I’m gonna add one point eight 𝑛 to both sides to give me a positive number of 𝑛s. Then I can subtract five point two from each side. And finally divide by one point eight to leave me with one 𝑛 on the left-hand side. And in this case, that gives me a whole number answer, fifty-six. So our answer is, yes, five point two is the fifty-sixth term in the sequence.

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