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Hi, my name is Mr. Chan, and in this lesson, we're going to learn how to find terms of a linear sequence.

Let's look at how we find terms of a linear sequence.

So in this example, we've got the nth term of the sequence is 5n add seven.

And we're asked to work out the eighth term of the sequence.

Now, if we're being asked to find the eighth term of the sequence, we're really being asked what number is in the eighth position of the sequence.

And if we think about when we were talking about position of a sequence, that would be the N number.

So in this question, one is asking for the eighth term, we're being we're really being told that n=8.

Okay.

So, all we have to do now is if we know that n is eight, we substitute that back into the nth term rule.

So that calculation will become five multiply by eight, add seven.

Five multiply by eight is 40, add seven, we get 47.

So what we can now say, is that the eighth term of the sequence 5n add seven is 47.

Here's another example.

We're told that the nth term of a sequence is 12n subtract 2n, and we're asked to work out the first five terms of the sequence.

So when we talk about the first five terms of the sequence, we're really looking for the position numbers one, two, three, four, and five.

And the position number remember in a linear sequence, is represented by the N number.

So we're really being asked what is the sequence when n equals one, then two, then three, then four, then five.

So what we can do now is substitute those values of n in one at a time into the sequence.

So we know that the first term we can work out 12 subtract two times one.

All we've done is replaced or substituted the N value into this into the nth rule.

So 12 subtract two times one, that becomes 12 subtract two by equals 10.

So we now know the first term equals 10.

The second term, similarly, we're going to substitute n equals two into the rule.

So that becomes a calculation 12 subtract two times two, 12 subtract four, that equals eight.

The third term, similarly substituting n equals three into the rule, the fourth term, substituting n equals four, and the fifth term substituting n equals five.

So what we have there, are the first five terms of the sequence 10, eight, six, four and two.

Here are some questions for you to try.

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Resume the video once you're finished.

Here are the answers.

So in question one, we're told that the nth term of the sequence is 8n subtract three.

And the function machine does help us because when we're substituting the values of n to find the sequence.

We can think of that like a function machine, where firstly multiplying the value of n by eight, and then subtracting three and that's what the function machine is helping you to do.

In question two, remember, we're substituting the values of n equals one, then two, then three, then four and five, to generate the sequence and find the first five terms. So the answers there.

Hopefully got all those correct.

Here's another question for you to try.

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Here are the answers.

So for this part of the question, finding the 50th term, we can substitute n equals 50.

So when we do that we will find the 50th term in the sequence.

Similarly, when we're trying to find the 100th term in the sequence, you would have to substitute n equals 100 into the nth term rule.

Now, Jack says I will substitute the 50th term and then double it to get the 100th term.

Now, that's quite a common mistake people make because Jack's wrong there, simply because when you're doubling that you will also double the constant term and that leads to an incorrect answer.

So make sure you don't make that mistake.

Here's another example.

We're told the nth term of a sequence is 4n subtract one and we're asked is 122 a term in the sequence? So, if 122 was a term in the sequence, we would expect that to be where in the position number, for example, 37, maybe 38, 39, whatever it would be.

The important part with that is, we would expect the N number for wherever 122 would be to be a whole number or an integer, as we call it.

So, let's look at asking the question is 4n subtract one ever equal to 122? Is 122 ever a term in the sequence? So when we asked that question mathematically, this is what we get.

So we're asking, does the sequence ever equal 122? And when we ask that, we end up with an equation like this 4n subtract one equals 122.

And we can solve this, we can add one to both sides to get 4n equals 123.

And then divide by four on both sides to get n equals 30.

75.

Now, what this means is that 122 would be in the sequence, but it would be in position 30.

75.

But what we know about the n numbers is they're always integers, they would go up in ascending order or descending order, by the same amount each time and n would be a whole number.

So what we can say if we get a decimal value of n, when we solve this equation, is that actually 122 is not in that sequence because it's actually in between two terms. It's in position 30.

75.

We would expect end to be an integer, if it was, so 122 is not in that sequence.

Here's some questions for you to try.

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Here are the answers.

Now question four was asking, how I know that 217 is not in that sequence.

Well, when looking at that sequence four, nine, 14, 19, 24.

I do notice that they are always one less than a number in five times table.

So four is one less than five.

Nine is one less than 10.

14 is one less than 15.

So is always one less than the five times table.

And that's how I know the 217 is not a term in that sequence, because a reasonable answer for that kind of question would be 2017 is three less than a number in the five times table.

It would be three less than 220.

Obviously, that's because I know that the five times table always ends in a five or a zero in the ones.

Here's another question for you to try.

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Here are the answers.

So in this question, we're asked to complete the Venn diagram and complete the sets for sequences 3n subtract two, and 2n add one.

So I would begin this question by writing down the terms in those two sequences with the numbers of 20.

So once I've done that, we have to find the intersection of the two sets, that's the parts that both sets would overlap and share.

So the numbers they have in common, those would be the numbers in the middle of the Venn diagram seven, 13 and 19.

And we're asked what sequence those numbers are the first three terms of, unless you can see they are the terms of 6n add one, or negative 6n add 25.

That's all for this lesson.

Thanks for watching.