Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Lovely of you to join me today.

Especially lovely for you because we're looking at arithmetic sequences and they are pretty cool.

So let's get started.

Learning outcome is I'll be able to identify the term-to-term rule and the position-to-term rule and express the latter algebraically.

Some key words, some key phrases that you're gonna hear today, arithmetic, linear, successive terms, constant, nth term, position.

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.

There's a constant difference.

Whereas 20, 40, 80, 160 is not arithmetic.

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.

Look out for those words and phrases throughout.

Two parts to our learning today.

Let's begin by looking at the term-to-term rule.

Let's get you thinking to start.

Can you find the next term in these two sequences? Pause.

See if you can spot them.

Welcome back.

The first sequence is easy to spot as a common, additive difference between each term.

We're adding six each time.

Let's add another six to 43, we get to 49, it's the next term in that sequence.

The second sequence is impossible to predict.

There's no common difference or pattern of any sort between successive terms. So what's that next term? Well, we're absolutely guessing.

We don't know, So let's pay attention to that first sequence where we did know, a sequence which had a pattern to it, which enabled us to make predictions.

A sequence with a difference between successive terms is a constant, in this case, a constant positive six.

It's called an arithmetic sequence.

You may also see them called linear sequences.

You hear the words arithmetic sequence, a linear sequence, they're the same thing.

The common additive difference is the term-to-term rule.

A term-to-term rule describes how to calculate the next term in the sequence from the previous term.

We use the term-to-term rule add six here to find that next term 49.

Quick check you've got that.

What I'd like you to do here is match these arithmetic sequences with their term-to-term rule.

Welcome back.

Hopefully, you paired the first sequence with add three, the second sequence with subtract three, the third sequence with add four, the fourth sequence with add two, the fifth sequence with add seven, and the last sequence with subtract four as it's term-to-term rule.

Another quick check, find the next term in these arithmetic sequences.

You're going to be looking to spot that term-to-term rule.

Remembering that an arithmetic sequence has a constant common, additive difference and using that information to find the next term in each case.

Pause, give these a go now.

Welcome back.

Hopefully, you spotted a term-to-term rule of add three each time in that first sequence.

Add another three, we get to 17.

For the second one, a constant additive difference of negative eight or the term-to-term rule of subtract eight.

The common additive difference can be negative.

It'll give us a decreasing sequence when that happens.

Let's subtract eight from 13 and our next term is five.

For third sequence, a common additive difference of 1.

5.

Be aware the common additive difference may not be an integer that makes the next term in that sequence 7.

3.

Hmm, the fourth one was a tricky problem, but you know there's an additive difference of positive 64 from 153 to 217, and you know it's an arithmetic sequence.

It looked unpredictable because we only knew two terms. However, crucially, you were told it's an arithmetic sequence, so it must have a common additive difference.

Let's keep that common additive difference going.

We add 64 to 217, the next term is 281.

Jacob is considering these sequences.

In that first sequence, he says, "There's a constant plus two, so I think this sequence is arithmetic." The second sequence, "There's a constant change between successive terms, so I think this one is arithmetic," and of that third sequence he says, "This one also goes plus two, plus two, plus two.

So I think it's arithmetic." Do you agree with Jacob with his thoughts on those three sequences? Pause, talk to the person next to you.

Have a good think to yourself.

I'll see you in a moment to see whether we should agree with Jacob or not.

Welcome back.

Hopefully, you absolutely agreed with Jacob on the first one, but disagreed on the second sequence and the third sequence.

The first one is absolutely, definitely arithmetic.

It's a constant additive difference, positive two between successive terms, it's an arithmetic sequence.

That second sequence is not arithmetic.

That third sequence is not arithmetic.

So we'll have a closer look at why.

For the second sequence, 1, 2, 4, 8, 16, it's not arithmetic because in order to be an arithmetic sequence, it must be a constant additive difference between successive terms. What we see in this sequence is a constant multiplicative difference, that makes it something else, but it's not an arithmetic sequence.

For that third sequence, in order to be an arithmetic sequence, the constant additive difference must be the first difference.

If I say to you what's the first difference in that sequence, you would say, well done, add three, add five, add seven, add nine.

We didn't have a constant additive first difference.

That's why that last sequence was not arithmetic.

Quick check you've got that.

What I'd like you to do is complete this sentence.

What word is going in that space there? Pause and answer this now.

Welcome back.

I hope you said it's B, additive.

The sentence should read, "In order to be an arithmetic sequence, there must be a constant additive difference between successive terms." That's an important sentence.

You probably wanna pause here and write it down.

Next, Jacob's considering these sequences, "I understand now.

An arithmetic sequence has a constant additive difference between successive terms, so these two are definitely both arithmetic sequences." Do you agree with Jacob? Pause, have a good think about these two.

You in agreement with Jacob or is there a disagreement here? See you in a moment.

Welcome back, wonder what you thought.

The third sequence, absolutely, definitely, arithmetic.

In the second sequence, absolutely, definitely not arithmetic.

So what's going on here? What we can do is subtract the previous term from the current term in an arithmetic sequence to reveal the term-to-term rule.

When we did that for the first sequence, the second term, subtract the first term, we get four quarters or one whole.

The third term, subtract the second term, the same thing, four quarters, one whole.

The fourth term, subtract the third term, we get four quarters, one whole.

We got a common constant additive difference of one whole from term-to-term.

That's an arithmetic sequence therefore.

The second sequence, 4/7 minus 4/3 is negative 16/21, but the third term minus the second term, 4/11 minus 4/7, that's negative 16/77.

What about the fourth term? Subtract the third term, that's negative 16/165.

We have not got a constant additive difference, so that's not an arithmetic sequence.

You may have even fall into saying it's arithmetic because in 3, 7, 11, 15, there is a pattern to those denominators, but crucially, and you see a sequence of fractions.

If there's an arithmetic pattern to the numerator with no change in the denominator, it will be an arithmetic sequence, but there's a pattern in that denominator.

You have to pay really close attention.

Subtract one term from the previous term to just check, is that arithmetic or not? Quick check you've got that.

Which of these are arithmetic sequences? Three sequences there.

Some are arithmetic and some may not be.

Can you spot which ones are arithmetic? Pause and have a think now.

Welcome back.

Hopefully, you said yes to A, that's arithmetic.

Why? It's common additive difference of positive 3/5 from term-to-term.

B, absolutely was an arithmetic sequence.

You might find it easier to write that sequence with a common denominator for each term.

You can then see adding 1/16 from term-to-term, that's a common additive difference.

It's an arithmetic sequence.

C was not.

You may be drawn into saying it is because you see 3, 6, 9, 12, but they are the denominators.

We're not adding 3/5 or adding 5/3.

If you subtract one term from the previous term, you will find there is no common additive difference in that sequence.

Jacob's now considering these strange looking sequences and Jacob says, "I may have been wrong before, but these ones are definitely all arithmetic sequences!" What do you think? Is it time to start agreeing with Jacob or is there an error in his thinking? Pause.

Have a think about these three.

Welcome back.

Wonder what you thought? In the first case, absolutely, definitely, an arithmetic sequencer, common constant additive difference, positive five, which funnily enough is exactly what we saw in the second sequence, to get from term-to-term, we're just adding five.

That makes that an arithmetic sequence.

What about the third one? Add 5x, add 5x, add 5x.

Well, that's a common additive difference from term-to-term.

It's an arithmetic sequence.

Well done, Jacob.

We will see algebraic terms and sequences as long as there is a common additive difference, they are still arithmetic.

Quick check you've got that.

Which of these are arithmetic sequences? Four to ponder.

See if you can spot the arithmetic ones.

See you in a moment.

Welcome back.

Hopefully, you said the first one is absolutely arithmetic.

That one has a common additive difference of 6/y or positive 6/y.

B, absolutely arithmetic similar to one we saw earlier.

Common additive difference of positive five from term-to-term.

C, absolutely arithmetic.

Why? Because we've got a common additive difference of X plus five where every time from term-to-term we're adding x and adding positive five.

D was not.

For D, I expanded each of those brackets and at that point I can see there's no common additive difference.

Practise time now.

Question one.

I'd like to find the term-to-term rule for each of these arithmetic sequences.

I just want the term-to-term rule.

How are we getting from one term to the next? I don't wanna know any future terms. I just wanna know that term-to-term rule in each case.

Pause.

Do this now.

Question two, I'd like you to find the next term in each of these arithmetic sequences.

You'll want to find that term-to-term rule, but you're then gonna use that term-to-term rule to find the next term in each sequence.

Pause and do this now.

Question three.

Oh, these sequences are beautiful.

What I'd like you to do is find the next term in each of them.

They're tricky, but be fearless, give it a go.

If we make any errors, I'm gonna give you the answer in a few moments, so go for it.

Welcome back.

Feedback time now.

Question one, find the term-to-term rule for each of these arithmetic sequences.

For A, term-to-term rule was add eight.

For B, it was add four.

For C, it was subtract four.

For D, subtract four, for E, add seven.

For F, add 0.

4, and for G, add 0.

04.

Question two, next term in each of these arithmetic sequences for A, it was 21, for B, negative 33, for C, 21, D, negative 44, for E, negative three, for F, 10.

7, and for G, 4.

73.

Question three, these ones were tricky.

We will make some errors, but that's not a problem in maths.

Every error is an opportunity to learn, so let's see what the right answers are so we can compare ours to these.

For A, negative 23/11, for B, negative 3x, for C, 5x plus three, for D, there's 15 lots of x plus six, for E, three quarters.

For E, you probably want to make that a sequence with a common denominator, turn 3/10 into 6/20, 3/5 into 12/20, et cetera.

You can see the next term is 15/20th, which cancels to three quarters.

For F, the next term is six x plus 20 over y.

Onto the second half of our learning now.

What we're going to look at the position-to-term rule.

A term-to-term rule describes how to calculate the next term in the sequence from the previous term.

We know the term-to-term rule.

This sequence is add two each time, so we can find the next term and the next term and the next term.

But what if I said, what is the 250th term of this sequence? Well, that's easy.

We just continue the sequence.

17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, I'm exhausted already.

This is clearly a hugely inefficient method.

105, which is where I ran out of the energy to write anymore.

That was just the 50th term.

The rest will take time and we might err, I, we might make an error, so there's got to be a better way to do this.

For such a question as what is the 250th term, we're better using a position-to-term rule.

I'm gonna give you that position-to-term rule for this sequence.

It's 2n plus five.

You might see it written as T equals 2n plus five, where T is the term value and n is the term number, or n is the position in the sequence.

Well, that means when we apply it, if I want the fourth term in that sequence, I say n equals four, and I substitute n equals four into our position-to-term rule, two lots of four plus five equals 13.

Oh, look, 13 is the term in the fourth position or when n equals four, T equals 13.

Another example, when n equals seven, the term value is two lots of seven plus five, that's 19.

The seventh term has a value of 19 when n equals seven, T equals 19.

Let's see if you can do that.

Can you answer this problem now? What is the 250th term of this sequence? Give it a go.

I'll see you in a moment.

Go through the answer.

Welcome back.

Hopefully, you substituted 250 into a position-to-term rule.

You thought n equals 250.

Put that into that expression.

Two lots of n plus five becomes two lots of 250 plus five, which is 505.

The 250th term, the 250th position, we'll see the term value, 505.

We need to know how to write an expression for this position-to-term rule.

I gave you the position-to-term rule for that example.

You need to know how to write them.

I'm gonna help you do that by giving you the first five terms of some arithmetic sequences.

The sequence 2n goes 2, 4, 6, 8, 10.

The sequence 3n goes, 3, 6, 9, 12, 15.

The sequence 4n goes 4, 8, 12, 16, 20.

What do you think the sequence 5n looks like? Pause, tell the person next to you or say it aloud to me at the screen.

See you in a moment.

Welcome back, well done, I heard you all screaming.

5, 10, 15, 20, 25, that's the sequence 5n.

Why? Well, the sequence 5n when n equals one, the term is five lots of one, five.

When n equals two, the term is five lots of two, 10.

When n equals three, the term is five lots of three, 15, and we can keep going and keep going, that's why we get the sequence 5, 10, 15, 20, 25.

If you've successfully spotted a pattern in all this, you're going to be able to complete this task.

I'd like you to pause, copy this down, and fill in all the blanks for these arithmetic sequences.

See you in a moment.

Welcome back.

The arithmetic sequences, so a sequence starting 6, 12 should continue 18, 24, 30.

The 6n sequence goes 6, 12, 18, 24, 30, so what does the 7n sequence do? Well done, it goes 7, 14, 21, 28, 35.

Look, a 7n sequence has a constant additive difference of positive seven.

What's the next sequence? 8, 16, that's an additive difference of positive eight, so we should continue 24, 32, 40.

A constant common additive difference of eight.

That's the sequence 8n.

Did you get that? If you did, you just found an nth term rule or a position-to-term rule for that arithmetic sequence.

20 something, 40.

Well, in the middle of there, it's gonna be 30, 10, 20, 30, 40, 50, common constant additive difference of 10.

What's the position-to-term rule? Well done, 10n.

The last sequence, 45, 60.

The next term must be 75, and the previous terms must be 15 and 30.

15, 30, 45, 60, 75.

That's a common constant additive difference of 15.

What's the nth term? Well done, 15n.

To get the next bit of the position-to-term rule, we're gonna go back to something that we're very familiar with.

What happens when you add two to five? I know that you know this, we get seven, but how I want you to think of it is like this.

There's a number line.

How did we get from five to seven? We added two.

What happened when we added two to five? We saw from the position five, a translation of two in the positive direction.

We had five and we added two.

We saw a translation of positive two.

Make a mental picture of this because you're going to need again shortly.

My next key question for you, what is the same and what is different about the sequences 5n and 5n plus two? There's the first five terms of the sequence 5n.

There's the first five terms of the sequence, 5n plus two.

I'd like you to pause, I wanna see you again.

I wanna know one thing that's the same about those sequences and one thing that's different.

See you in a moment.

Welcome back.

Hopefully, for same, you said both sequences have a common additive difference of positive five.

A 5n sequence has a common additive difference of five.

What about different? Lots of things you could have said here.

You might have said one starts at five and one starts at seven.

Look at this difference between the two sequences.

When I compare term-to-term, five to seven, 10 to 12, 15 to 17, did you notice this? We're adding two each time.

Why do we see this difference between 5n and 5n plus two? Well, the sequence 5n was generated like this by taking n and multiplying it by five.

Two lots of five, three lots of five, four lots of five, five lots of five.

But when I generated the sequence 5n plus two, I had to take n for the first term, n equals one, multiply it by five and add two.

That's how I got seven for that first term.

For the second term, I had to take n equals two, multiply it by five and add two, that's why I got the second term, 12, and this pattern continues.

So what do you notice? Well done.

The plus two in our position-to-term rule translated all the term values, we added two to each of those 5n term values.

Here is the translation of plus two shown on the number line.

The sequence 5n would look like that on a number line, bouncing from 5 to 10 to 15 to 20.

The sequence 5n plus two would be there on number line 5n plus two is a translation of 5n by two in the positive direction.

Quick check you've got that.

What I'd like you to do is find the first five terms of each sequence, 3n, 3n plus one, 3n plus two.

Draw them on the number lines and state what they all have in common.

Pause, get working out those first five terms and get drawing on those number lines.

Welcome back.

Hopefully, for 3n, you had those terms, on your number line they looked like so.

For 3n plus one, those terms and that on your number line.

3n plus two and the number line.

3n plus three and the number line.

What did they all have in common? Hopefully, you said they have a common difference term-to-term of positive three.

Can you see why they had a common difference of positive three? It's a 3n sequence in each case.

Next check.

For each sequence, describe the translation from 3n.

That first sequence we see there, that first number line, that's the sequence, 3n.

How are the other sequences translated from there? Pause, think about this now.

Welcome back.

Hopefully, you said there's no translation from 3n for the sequence 3n.

You might call it 3n plus zero, but we like to be efficient.

It's just 3n, there's no translation.

For 3n plus one, there was a translation of 3n by positive one.

For 3n plus two, there was a translation of positive two from 3n.

We can see that positive two in our position-to-term rule.

How about that final one 3n plus three? It was a translation of positive three, a translation of 3n by positive three.

Again, that positive three we see in the position-to-term rule.

If we've got that, we should be able to find the nth term of this arithmetic sequence.

We need to know two things.

We need to know the common difference and the translation.

The common difference is positive four.

That means it's a 4n sequence.

What's the translation? The translation from the sequence 4n.

Well, it's one greater or a translation by positive one.

That's why this position-to-term rule, this nth term rule is 4n plus one.

We don't need a number line to find an nth term expression.

It's useful.

Visual representations are really powerful in maths, but we don't need it.

We can compare respective positions in the sequence.

Once we know we've got a common additive difference of four, we know it's a 4n sequence.

We then compare term-to-term, compare the first term to the first term, the second term to the second term.

We see as a translation of positive one each time.

Hence, we can declare the nth term of 5, 9, 13, 17, 21, to be 4n plus one.

Now, I'm gonna do an example and then ask you to have a go at similar problem.

Find the nth term of the sequence, 11, 17, 23, 29, 35.

I've got common additive difference of six.

It must be a 6n sequence.

I'm now gonna compare respective terms, the first term to the first term, the second term to the second term.

When I compare six to 11, it's at five, 12 to 17 at five, 18 to 23 at five, and so on and so on.

I know my nth term is 6n plus five.

You might think of that as six being the common difference, five being the translation.

Your turn, I'd like to pause and find the nth term of this sequence.

Welcome back, let's see how we did.

Hopefully, you spot a common additive difference of positive nine, so it's a 9n sequence.

So it's a translation from the sequence 9n comparing term-to-term, you see negative eight is the translation from term-to-term each time.

So nth term is 9n minus eight.

Common difference of nine and a translation of negative eight.

Aisha and Sophia are discussing the nth term of this sequence, 100, 90, 80, 70, 60.

Aisha says, "With a common difference of negative 10.

I think it's a negative 10n sequence, possibly a 100 minus 10n." Sophia says, "I'd like your thinking, Aisha, but if it is 100 minus 10n, the fifth term would be 100 minus 10 lots of five 50 and it is not." Can you help them? Can you spot the problem here? Pause and have a think.

Welcome back.

Aisha and Sophia got something right, negative 10.

The common difference is negative 10, so it is a negative 10n sequence, but they got the translation wrong.

If you think of on a number line and then compare it to the sequence negative 10n, you can see the translation is indeed positive 110, not positive 100.

So the sequence is 110 minus 10n, not 100 minus 10n.

Whilst the number line helps us visualise a translation, we don't need to use one.

We see a common additive difference in negative 10.

We know it's a negative 10n sequence.

That's the common difference, negative 10.

What's the translation? Compare the first term to the first term, second term to the second term, third term to the third term, and so on.

You can see that translation, positive 110.

This must be the sequence, 110 minus 10n.

Quick check you got that.

What's the nth term of this arithmetic sequence? I've given you three possible position-to-term rules.

Any one of them is right.

Can you spot it? Welcome back, hopefully you said B.

Common difference of negative nine and a translation by positive 59.

We can check that's true by substituting in n equals five.

If we're right, the fifth term should be 59 minus nine lots of five 14.

It is, we know we're right.

Practise time now.

Question one, I'd like to do some matching.

In each row, there's a sequence and then I've got selection of nth term rules.

I'd like you to match up the nth term rule with the right sequence.

Pause and do this now.

Question two, I'd like you to find the nth term of these sequences.

Pause and give these a go now.

Question three, some tricky but lovely problems here.

Can you find the nth term of these sequences? Feedback time now.

Matching nth terms with their sequences.

We should have matched them like so.

You wanna pause and check that your answers match mine.

Question two, the nth term of these sequences, we should have got 8n plus one, 2n plus three, eight minus 2n, negative 5n minus 45 and 0.

0n plus 0.

01.

And for question three, we should have got half n plus a quarter, 8/32n minus 3/32, which you might have cancelled to a quarter n minus 3/32.

And for the last one, I wrote a common denominator to spot this sequence.

It's 13/40n minus 12/40, which you might have cancelled to 13/40n minus 3/10.

We're at the end of the lesson now.

What have we learned? We've learned that we can identify the term-to-term rule of an arithmetic sequence, and we know that the sequence is only arithmetic if the term-to-term rule is a common additive difference.

We've learned that we can algebraically express the position-to-term nth term rule of an arithmetic sequence by finding the common difference and its translation.

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

Goodbye for now.