Lesson video

Hello.

Mr. Robson here.

Welcome to maths.

Good of you to join me today.

Today we're learning about arithmetic sequences.

More specifically, we're identifying values in arithmetic sequences.

I like this bit of maths, and I think you will too.

So let's get started.

Our learning outcome is we'll be able to identify through the nth term whether a value appears in a given sequence.

Some keywords and terms you're gonna hear today.

An arithmetic or linear sequence is a sequence where the difference between successive terms is a constant.

For example, 20, 30, 40, 50 is an arithmetic sequence, whereas 20, 40, 80, 160 is not.

The nth term of a sequence is the position of a term in a sequence where n stands for the term number.

For example, if n equals 10, this means the 10th term in the sequence.

Two parts to our learning today.

We're gonna start by generating terms in the sequence.

If we know a position-to-term rule for an arithmetic sequence we can find the first five terms using a table of values.

A position-to-term rule might look like this.

T equals five n plus three.

T is the term value.

n is the term number, or the position in the sequence.

When n equals one, the term is five lots of one plus three, it's eight.

We can pop that in our table the values.

When n equals two, the term is five lots of two plus three.

It's 13.

When n equals three, five lots of three plus three, that's 18, and we can keep going to find the fourth term and the fifth term.

The sequence starts 8, 13, 18, 23, 28.

Once we see this sequence, we're reminded of something which could help us generate the first five terms more efficiently.

The coefficient of n is the common difference in the sequence.

A five n sequence has a common additive difference of positive five between successive terms. Three in our position-to-term rule is the constant.

It's also the translation.

What do we mean by translation? Well, in this sequence, five n plus three is a translation of the sequence five n.

Five n is a lovely sequence.

It's your five times table.

When we compare respective terms between the first terms of each sequence, there's a difference of positive three, between the second terms, a difference of positive three, the third terms, a difference of positive three.

In fact, everywhere we look in that sequence, we see that if we take the sequence five n and translate that value by positive three, we reach the sequence five n plus three.

We can now generate the first five terms of this sequence very efficiently by finding the first term and counting on.

For the sequence seven n minus five, the first term when n equals one is gonna be seven lots of one minus five.

It's two.

We know the coefficient of n is the common difference in the arithmetic sequence, so we're gonna add seven from term to term.

Add seven to two, we get to nine, add another seven, we get to 16, add another seven, add another seven, and we've found the first five terms really efficiently.

For some sequences, this common difference method can be simpler than multiple substitutions.

Seven minus 12 n.

The first term is gonna be negative five.

What's the coefficient of n on this occasion? I do hope you said negative 12, not 12, it's negative 12, so we need to subtract 12 each time.

This is gonna be a decreasing sequence.

Negative five subtract 12 is negative 17, and keep subtracting 12 as we go.

We can use substitution to check that we're right.

We think the fifth term is negative 53.

If we're right, then when we substitute n equals five into our position-to-term rule, we should get negative 53.

We do! Joy of joys.

We're right.

Quick check you've got that.

Generate the first five terms of these two sequences, the sequence six n minus three.

I'm gonna ask you to use substitution and a table of values.

For nine n minus eight, I'll ask you to use that second method I showed you.

I'd like you to find the first term, then use the common difference and count on.

Pause.

Do these two problems now.

welcome back.

Let's see how we did.

Substitution for six n minus three.

We should find the first term is three, the second term is nine, the third term is 15, the fourth term is 21, and the fifth term is 27, and your working out should have looked like that.

There's the first five terms of that sequence.

For using the common difference in counting on for nine n minus eight, we need the first term nine lots of one minus eight, that's one.

And then we know we've got a common difference of positive nine.

The coefficient of n is positive nine, so we're adding nine each time to find the remaining terms in the sequence.

We'll add nine to get to 10, add another nine, add another nine, add another nine.

What do you think we're gonna do now? Absolutely, we're gonna check that's right.

The fifth term is 37.

Is it? When n equals five, term is nine lots of five minus eight is 37.

We know we're right.

As mathematicians we like to be efficient, so we would not use the counting on method if we're trying to find the 50th term of an arithmetic sequence.

For example, find the 50th term of four n minus three.

Let's start on one, and then it goes to five and then nine and then 13 and then 17 and then on and on and on.

We could keep counting on for a long, long time and I think, I think, I think 197 is the one that we're looking for, but there were 49 additions to get to that point.

How can I be sure I didn't make an error? Also, did I count correctly? Perhaps I miscounted and the 50th term is actually next to 197.

It might be 193 or 201.

I could have miscounted.

The good news is we can use substitution to check that we're correct.

When n equals 50, the term is four lots of 50 minus three.

Wonderful.

It is 197.

But look at how efficient that substitution is on its own.

Well, wasn't that just a far quicker and easier method than doing all that counting on? The usefulness of an nth term rule is that it allows us to generate any term efficiently and accurately.

We should use this method to find a term when n is large.

Here's three examples I'm gonna show you now where n is large.

I don't wanna use counting on to find the 50th term.

I certainly don't wanna use counting on to find the 200th term.

So I'll substitute n equals 50 into my position-to-term rule eight n minus 21 and I'll find that 50th term is 379.

How about the 100th term of 21 minus eight n? That would be a lot of counting on.

I think I'll do this instead.

When n equals 100, the term is gonna be 21 minus eight lots of 100.

That's negative 779.

How about that third problem? 200th term of 1/4 n plus 1/2.

That's a lot of counting on, counting on in quarters.

It could take me a while.

Alternatively, when n equals 200, the term will be 1/4 of 200 plus 1/2.

I can write that incredibly succinctly by writing a mixed number, 50 1/2.

Three problems similar to those for you to practise now.

So let's check.

You can do this.

Pause, give those three a go.

Welcome back.

Hopefully you found when n equal 50, the term in 16 n minus 45 is 755.

How about the 100th term of 13 minus five n? You should have found the 100th term of that sequence to be negative 487.

The 200th term of one over eight n plus one over three.

That's gonna be 25 1/3.

The nature of the nth term and the term number we are seeking might dictate which method we use.

For example, find the first five terms and the 10th term of three n minus 12.

Well, the first five terms I can find by substituting into my table of values.

Once I know those first five terms, I can see that counting on is gonna be pretty quick and easy for finding the 10th term.

I'm only counting on in threes.

Count on in threes and I get to 18.

That's the 10th term.

That was nice and easy.

But how about this problem? Find the first five terms and the 10th term of 27 n minus 12.

The first five terms are like so.

So if the sequence starts like that, I could count on, but it's not quite so easy because counting on in 27s is just a little trickier than counting on in threes.

So is it more sensible to use substitution here? Absolutely, when you see that substitution.

When n equals 10, the term is 27 lots of 10 minus 12.

270 minus 12, thank you very much.

That's 258.

The 10th term is 258.

When we're presented with problems in maths, before starting, we should give careful consideration to what's going to be the most efficient and accurate method to execute that problem.

Quick check, you've got that.

I'd like you to find the first five terms and the 10th term of 17 n plus three.

Then I'd like you to find the first five terms and the 10th term of two n plus 19.

For each one of those problems, I'd like you to choose the most appropriate method for that nth term rule.

Pause and give this a go now.

Welcome back.

Let's see how we did.

The first five terms of 17 n plus three, we should have found that, and then counting on will be acceptable, but we're counting on in seventeens.

That's not quite so simple.

So for this one, substitution is really quick and simple.

17 lots of 10 plus three, that's 173.

Wonderful.

That was nice and easy for finding that 10th term.

The first five terms of two n plus 19 would look like so.

And then you could use substitution, but we're counting on in twos.

Counting on in twos is really quick and simple, so we might just use that method here.

When we continue to count on in twos we 39 is the 10th term.

For some problems, a blend of those two methods works well.

For example, find the 102nd term of the sequence 16 n minus five.

Using substitution, we need to do 16 lots of 102 minus five, which we absolutely can do.

If we've got a calculator, we can do that in seconds.

If we've just got a pen and paper, we can still work it out, but 16 lots of 102 is not as easy as something else I'm about to show you.

We could calculate this as it is, but we can think about an alternative way to find the 102nd term.

What's the 100th term? Why do I wanna know the 100th term? Well, 16 lots of 100 minus five, I can do that in seconds.

It's 1,595, and wonderful.

If the 100th term is 1,595, I just need to count on two terms and I'll have the 102nd.

The 100th term is 1,595, add another 16, add another 16, 1,627 is the 102nd term.

One for you to practise.

I'd like to find the 97th term of 23 n plus 70.

Would you like a hint? Find the 100th term first.

Pause and give this problem a go.

Welcome back.

Hopefully you found the 100th term first.

When n equals 100, the term is 2,370.

From there we can count back.

We know the 100th term.

We're gonna count back in 23s and find that the 97th term is 2,301.

Practise time now.

Question one.

I'd like you to generate the first five terms of these two sequences.

For part a, eight n minus 10, I'd like you to use substitution and a table of values.

Show me that you can write that method correctly and populate a table of values.

For b we're gonna be a little quicker, more efficient.

For four n minus 15, I'm gonna ask you to find the first term and then use the common difference to count on.

I need to know you can do both those methods.

So pause and try these now.

Question two.

Andeep has counted on to find the 30th term of 81 minus seven n and he says the 30th term is negative 135.

For part a, there's two errors in Andeep's work.

I'd like you to find them.

For part b I'd like you to show Andeep a more efficient method.

Pause and do this now.

Question three, part a.

I'd like you to use substitutions, meet to complete the table of values.

List the first five terms of the sequence, 37 n minus 25.

For part b I'd then like you to use two different methods to find the 10th term.

And for part c, I'd like to write a comment to explain which method was the most efficient and effective when you did part b.

Pause.

Try these things now.

Question four.

Part a, you're gonna find the 50th term of seven n minus 85.

For part b, you're finding the 100th term of 100 minus nine n.

Part c, find the 80th term of that sequence.

Part d, find the 250th term of that one.

Pause and do these now.

Feedback time.

Let's see how we did.

First five terms of eight n minus 10 using substitution and a table of values.

The first term is negative two, the second term is six, the third term is 14, the fourth term is 22, the fifth term is 30.

Your table of values should be populated with those term values and you're working out should look something like mine.

Part b, find the first term counting on.

The first term, four lots of one minus 15, that's negative 11, and then we're counting on in four, positive four each time.

We'll get those five terms. We'll of course check we're correct.

We think the fifth term is five.

When n equals five, the term is four lots of five minus 15, which is five.

We know we're right.

Question two, I asked you to find two errors in Andeep's work.

Did you notice negative 135 was the 31st term in Andeep's list, not the 30th.

Secondly, did you notice there was a miscount between four and negative two? 81 minus seven n.

There should be a common additive difference of negative seven between each term.

There wasn't between four and negative two, and the annoyance in the counting on method is once you've made one error, it compounds for the rest of the sequence.

For part b, showing Andeep a more efficient method, we should have suggested substitution.

81 minus seven lots of 30 will tell us that 30th term, and 81 minus 210 is negative 129.

Question three, using substitution to complete the table of values for 37 n minus 25, substitutions should have looked like so, and your table of values should be populated with those values.

For part b, two different methods to find the 10th term.

One method is counting on, counting on in 37s.

By the time we count on to the 10th term, we think it's 345.

Substitution confirms that it is.

When n equals 10, the term is 37 lots of 10 minus 25.

It is indeed 345.

For part c, your comment to explain which method was the most effective, you might have written substitution was more efficient and less prone to error.

Question four, 50th term of seven n minus 85 is 265.

For part b, the 100th term of one minus nine n is negative 899.

Part c, the 80th term of 0.

6 n plus 2.

4 is 50.

4.

Part d, the 250th term of this sequence is 149 16/17.

Using a mixed number was a really efficient way to communicate that answer.

On to the second half of our learning now.

We're going to look at identifying the term number.

A position-to-term rule allows us to find any term in a sequence.

For example, find the 50th term of six n minus 11.

When n equals 50, the term is six lots of 50 minus 11, that's 289.

We've moved from the position 50 to the term 289.

A position-to-term rule also allows us to find the term number if we're given a term.

An example of a problem like that would be the term 217 is in the sequence six n minus 11.

What is its position? One approach we can take to this problem is trial and error.

When n equals 10, the term of this sequence is 49.

The 10th term is 49.

What about the 20th term? That's 109.

The 30th term, 169.

The 40th term, 229.

Why are we doing this? Well, at this point we should have spotted 217 is in between the 30th and 40th terms. Furthermore, it's closer to the 40th term.

We can now count back from 229 our 40th term, and we'll find the one that we're looking for.

We know that the sequence six n minus 11 has a common additive difference of positive six, so we need to count down in sixes.

There we go.

217, that's the one we're looking for, but what position is it? We know that 229 is the 40th term, so 223 must be the 39th term and 217 must be the 38th term.

We've cracked it.

217 is the 38th term of the sequence six n minus 11.

We can improve our efficiency with this method by selecting an appropriate starting point.

For example, the term 307 is in the sequence four n plus 15, but where is it? What is its position? When we start by looking at what's in the 10th position, we find that term is 55.

So the 10th term is not a sensible starting point.

It's way too low.

So let's do some estimating here.

Four lots of 70 is 280, and that's getting us close to our target.

So let's start by finding the 70th term.

It's 295.

Wonderful.

We're really close.

The 70th term is a far more efficient starting point.

From the 70th term we'll count up to the 71st, 72nd, 73rd term.

There it is.

307 is the 73rd term.

Quick check.

You can do that.

The term 481 is in the sequence five n plus 71.

I'd like you to use trial and error to find its position.

Pause.

Give this a go.

Welcome back.

You know the 10th term is way too low and you knew the 100th term would be way too high.

Did you spot something like five lots of 90 is 450? That's close to the target figure.

Let's find the 90th term.

It's 521.

It's a bit high.

What's the 80th term? 471.

We're getting really close now.

From 471 we can count up in fives.

476, 481, we found it.

481 is the 82nd term.

We can identify the term number more efficiently by forming and solving an equation.

The term 342 is in the sequence seven n minus 85.

What is its position? Our position-to-term rule looks like that, and we know the term.

The term that we're looking for is 342.

Oh look, we formed an equation which we can go about solving.

Let's add 85 to both sides of that equation.

Let's divide both sides of the equation by seven, and we find n equals 61, but let's not stop there.

Let's interpret that answer.

It means that 342 is the 61st term.

Now I'm gonna do an example like that and then ask you to repeat that skill.

The term 503 is in the sequence eight n plus 23, but where is it? What's its position? That's my position-to-term rule, and I know the term, 503.

I just need to find n, my position.

So I'll add negative 23 to both sides of the equation.

I'll divide both sides by eight and I find n equals 60, and you know I don't stop there.

I'm gonna interpret that answer.

It means 503 is the 60th term.

Your turn.

Pause.

See you in a moment.

Welcome back.

The term 242 is in the sequence seven n minus 17.

What is its position? Hopefully you set up this equation.

Let's go about solving that, and we find that n equals 37.

I do hope you didn't stop there.

I hope you wrote 242 is the 37th term.

Sometimes when we form and solve an equation, we get an unusual result.

Alex says, "I'm gonna find the position of 350 in the sequence eight n plus two." And Alex goes about doing this mathematics.

n equals 43.

5.

Andeep says, "Your working is right, but n equals 43.

5 is not a valid result." What does Andeep mean? Pause and have a think.

Welcome back.

I wonder what you said.

We can find the 43rd term of this sequence by substituting n equals 43 into our position-to-term rule.

We can find the 44th term of the sequence by doing something similar.

Alex has realised something.

"Of course! 346 is the 43rd term and 354th the 44th.

There is no term in between." Andeep says, "That's it! If a term is in the sequence, then n is a positive integer." Quick take.

You've got that.

I'd like you to show that 3,000 is not in the sequence 24 n plus six.

My hint is form and solve an equation.

Pause and give this problem a go.

Welcome back.

Hopefully you formed an equation and set about solving it and found that n equals 124.

75.

In order for a term to be in the sequence, n must be a positive integer value.

In this case n equals 124.

75.

So we know that 3,000 is between the 124th and the 125th terms, but is not itself a term in this sequence.

Andeep and Lucas are looking at the sequence 71 minus three n.

Lucas says, "I've done the maths.

n is an integer, so 80 must be in the sequence." Can you see the problem with Lucas's reasoning? Pause.

Tell a person next to you or tell me aloud on the screen.

I'll see you in a moment.

Welcome back.

Andeep's got this.

Andeep says, "This sequence 71 minus three n starts at 68 and counts down, so 80 cannot be in it.

n must be a positive integer." Anddeep and Lucas now look at this problem.

What position is 15.

5 in the sequence 0.

9 n plus 0.

2? And Lucas says, "I've got this one! 15.

5 is not an integer, so it can't be in the sequence." Do you agree with Lucas? Pause and have a think.

Welcome back.

Hopefully you don't agree with Lucas.

There's our position-to-term rule, and when we put 15.

5 in as a term, we can solve this equation and find n equals 17.

Andeep helps Lucas out.

Andeep says, "T is the term value and it does not have to be an integer.

Provided n is a positive integer, the value will be in the sequence.

In this case 15.

5 is the 17th term." Quick check.

You've got all that.

Which of these statements are true for the sequence T equals 2.

5 n plus seven? Four statements there.

Some are true, some are false.

Pause and decide which are which.

Welcome back.

Hopefully you said a is false.

The sequence starts 9.

5, 12, 14.

5, 17, 19.

5.

We have non-integer values in that sequence and that's fine.

B, it's true.

n will always be an integer value for all terms in the sequence.

C was not true.

In order for a term to be in the sequence, n must be a positive integer.

D was true.

59.

5 will be a term in the sequence.

Practise time now.

Question one I'd like you to use trial and error to find the position of 913 in the sequence six n minus 71.

Pause and do that now.

Question two, parts a, b, and c.

I'm asking you to find the position of some terms in their respective sequences.

Part d is slightly different.

Part d I'd like you to show that that value is not in that sequence.

Pause and try these four problems now.

Feedback time now.

Question one, trial and error to find the position of 913 in this sequence.

A copy of estimation to start will be to recognise that 150 times six is 900, and that's really close to our target figure.

The 150th term is 829.

We can edge a little bit closer by finding the 160th term.

From there we can count on and we spot that 913 is the 164th term.

Question two part a, the position of 142 in that sequence, that's the equation we should have set up, and we should have gone about solving it to find that n equals 38 and then interpreting that answer.

142 is the 38th term.

Part b, there's the equation we should have set up, there's the solution we should have found, and there's our concluding statement, an interpretation of n equals 29.

We should have said negative 216 is the 29th term.

Part C.

We should have started with that equation, and we'll rearrange, like any other equation, and we find that n equals 81, so 995.

9 is the 81st term.

Part d, showing that 1,048 is not in this sequence.

We still start by setting up an equation.

When we solve it, we find that n is not a positive integer value.

So you would write, n is not a positive integer, so 1,048 is not in the sequence.

You must include that sentence in your answer.

We're at the end of the lesson now, sadly.

What have we learned? We've learned that we can use an nth term rule to generate the terms of a sequence or find any term in a sequence.

We also learned that we can identify by using the nth term whether a value appears in a given sequence.

Hope you enjoyed this lesson as much as I enjoyed it, and I look forward to seeing you again soon for more maths.

Goodbye for now.