Lesson video

Hello, Mr. Robson here.

Great decision.

Always a great decision to join me for mathematics.

Especially, when we're doing sequences.

Our learning outcome to begin with.

Our learning outcome is that we'll be able to calculate any term using the nth term.

There are some keywords in today's lesson, which I'd like to remind you of now.

Arithmetic sequences, also known as linear sequence, an arithmetic sequence is a sequence where the difference between successive terms is a constant.

Nth term is something I'll be saying frequently.

Just to remind you, an nth term is a position of a term in a sequence where n stands for the term number.

For example, n equals 10.

That means the 10th term in the sequence.

Because this nth term can be used to calculate any term in a sequence, it's also a formula for finding the nth term.

Three learning cycles in today's lesson.

We're gonna begin by generating sequences using nth terms. Lucas and Alex are discussing the arithmetic sequence with the nth term 3n plus 2.

Lucas says, "I think it starts at three and adds two each time.

The first five terms will be 3, 5, 7, 9, and 11." Alex says, "I think it starts at two and adds three each time.

The first five terms will be 2, 5, 8, 11, and 14." These are perfectly natural things to say.

I see a three and a two in that nth term expression, so it must be something to do with three and two, right? So Lucas, Alex have given us good suggestions? But how could we possibly know who's right? The answer to that is in substitution, a really important skill in algebra.

Substitution will help us to find the first five terms of the sequence with the nth term 3n plus 2.

We'll start with a table whereby we map out the first five terms. That's when n equals one, n equals two, n equals three, n equals four, n equals five.

They're gonna be the first five terms. The relationship between the term number and the term itself is this nth term 3n plus 2.

I'll take the term number, I'll multiply it by three and I'll add two.

So when I want to find the first term number, I substitute into that nth term expression.

Instead of 3n plus 2, I'll do three lots of one plus two.

That will give me the value of five.

That's our first term.

We need to do that for all the terms now.

So 3n plus 2 when n equals two, that'll look like this, three lots of two plus two, giving us a total of eight.

I think you can predict what's coming next.

I want the third term, n equals 3.

So that must be three lots of three plus two, which will give us nine plus two being 11.

The fourth term, n is gonna equal four.

Substitute that is the expression, we'll get a term value of 14.

And when n equals five, substitute that in three lots of five plus two gives us 17.

We've used substitution to complete this table, more importantly, we found the first five terms, 5, 8, 11, 14, and 17.

Something nice you might notice there, three plus two, six plus two, nine plus two, 12 plus 2, 15 plus 2.

Hmm, what changed there? That might be really useful and might come back to be super useful in later learning.

So the terms were 5, 8, 11, 14, and 17.

The first five terms is equal to 3n plus 2 so, were we right? 3, 5, 7, 9, and 11.

Lucas wasn't right.

2, 5, 8, 11, 14.

Alex wasn't right.

It's this table and the substitutions which allowed us to accurately find the five terms. Just wanna check you can do that now.

I'd like to find the first five terms of the sequence 5n plus 1.

I'll do the first one for you.

Term number one, n equals one.

We substitute that into 5n plus 1.

So five lots of one plus one.

That'll give us a term value of six.

Can you do the next four terms for me, please? Pause this video and work those out.

For n equals two, five lots of two plus one, gives us 11.

For n equals three, five lots of three plus one, gives us 16.

For n equals four, five lots of four plus one, gave us 21, for n equals five, five lots of five plus one, gave us 26.

We can use this skill to find the first five terms of any sequence.

0.

3n minus 0.

2.

Okay, this feels like a calculator moment to me.

We need to take the term number, multiply by 0.

3 and subtract 0.

2.

So 0.

3, lots of 1 minus 0.

2.

That'll give us 0.

1.

When n equals two, 0.

3 lots of 2 minus 0.

2, gives us 0.

4.

For n equals three, 0.

3 lots of 3 minus 0.

2, gives us 0.

7, and so on for n equals four, and so on for n equals five, giving us the terms 0.

1, 0.

2, 0.

7 and 1, 1.

3.

Just wanna check that you've got that now.

What are the first five terms of the arithmetic sequence with the nth term rule 1.

5n plus 0.

2? That's right, they're 1.

7, 3.

2, 4.

7, 6.

2, and 7.

7.

We find the first term by substituting in a value of n equals one, 1.

5 lots of 1 plus 0.

2, giving us 1.

7.

For the second term, n equals two, 1.

5 lots of 2 plus 0.

2, giving us a total of 3.

2 and the sequence goes on.

What is the first term, just the first term of the sequence 6.

5n minus 3.

8? Pause this video, find out which one it is.

It's 2.

7.

When n equals one, 6.

5 lots of 1 minus 3.

8 equals 2.

7.

What is the third term of the sequence 6.

5n minus 3.

8? Or more to the point, how would we work it out? Would we use 3 plus 6.

5 minus 3.

8? Would we use 3 lots of 6.

5 minus 3.

8? Would we use 6.

5 minus 3 lots of 3.

8? Which one is it? Tell the person next to you or say it aloud to the screen.

It was option b, 3 lots of 6.

5 minus 3.

8, giving us a third term with a value of 15.

7, that's substitution, n equals three, 6.

5 lots of 3 minus 3.

8.

What's the first term of the arithmetic sequence with the nth term 6.

5n minus 20? Is it 13.

5, negative 13.

5, negative 26.

5 or negative 20? Pause this video, work out which one it is.

It was negative 13.

5.

We want the first term, n has a value of one, 6.

5 lots of 1 minus 20, gives us negative 13.

5.

A point to note, we don't just have positive terms in sequences, we can have negative terms too.

Negative 13.

5 is in this sequence.

Next, let's find the first five terms of the sequence, 100 plus 10n.

Okay, the relationship is, I start with the 100, and I add 10 lots of the term number.

So when n equals one, I'll get a term of 110.

When n equals two, I'll get term of 120.

And this beautiful pattern continues.

So, 110, 120, 130, 140, 150, that's going up in 10s.

Hmm-mm.

What about the sequence 100 minus 10n? What would be the difference? 100 plus 10n.

The terms are 110, 120, 130, 140, 150.

How is this one going to be different? I asked the same question to Aisha and Sophia.

Aisha said, "I think it's the same as 100 plus 10n, but it's in the negatives." So instead of being 110, 120, 130, it's negative 110, negative 120, negative 130, et cetera.

Sophia said, "I think it starts at 100, and goes down by 10 each time.

The first five terms will be 100, 90, 80, 70 and 60." I wonder who's correct.

We'll use substitution to find out.

We won't do any differently just because it's minus 10n, instead of positive 10n, we're gonna substitute in n equals one for the first term.

100 minus 10 lots of 1, gives us 90, 100 minus 10 lots of 2, gives us 80, 100 minus 10 lots of 3, gives us 70, 100 minus 10 lots of 4, 100 minus 10 lots of 5.

We end up with a sequence, 90, 80, 70, 60, 50.

So neither Aisha nor Sophia were correct.

However, we do have a decreasing sequence and Sophia almost had it decreasing by 10 every time.

In fact, Aisha's sequence is going down by 10 every time.

But we didn't start at negative 110 and we didn't start at 100.

Laura writes the first five terms, the arithmetic sequence, 8 minus 1.

1n in this table and she says, "I don't need to do any working out.

It's an arithmetic sequence, so I can just start at eight and subtract 1.

1 each time." That makes sense.

And minus 10n sequence seemed to go down by 10 every time.

So a minus 1.

1n sequence goes down by 1.

1 every time? Explain to Laura what she's done wrong and what she could do better.

Pause this video and say your explanation to yourself.

Your explanation might have included: The sequence does go in steps of negative 1.

1, but the first term is not eight.

Use substitution to accurately find terms. For the first term, when n equals one, 8 minus 1.

1 lots of n will be 8 minus 1.

1 lots of 1, 8 minus 1.

1 being 6.

9.

Your first term will be 6.

9.

And Laura's got 6.

9 down as the second term.

She should have had 6.

9, 5.

8, 4.

7, 3.

6, 2.

5.

So she did spot, it's arithmetic and it's changing by negative 1.

1 each time.

She just started in the wrong position.

Substitution would've helped her start at an accurate position.

Practise time now.

I'd like you to use substitution to find the first five terms of these arithmetic sequences.

Pause this video and give that a go.

Part c and d of question one.

Again, use substitution to find the first five terms of these arithmetic sequences.

These sequences look slightly different.

Pause this video, give it a go.

And for question two, June says, "The sequence n plus three must go up in three's because it says plus three." Use substitution to find the first five terms and show June that he is mistaken.

Pause this video, populate the table, and explain to June why he's mistaken.

Feedback now.

The first five terms of the arithmetic sequence 5n plus 3 would be 8, 13, 18, 23 and 28, and there's the substitutions if you need to check your working out against mine.

For b, trickier sequence, 0.

05n plus 0.

3, we get the terms 0.

35, 0.

4, 0.

45, 0.

5, 0.

55, and here are substitutions if you need to check your working out against mine.

For part c, the terms were 2.

3, 1.

6, 0.

9, 0.

2 and negative 0.

5 and here's your working out.

For part d, the terms were negative 3.

7, negative 4.

4, negative 5.

1, negative 5.

8, negative 6.

5.

And again, there's the working out if you need to check yours against mine.

For question two, n plus three must go up in three's because it says plus three.

That is a perfectly natural thing to think and it's a common misconception that we see when we're learning about arithmetic sequences in nth terms, n plus three, it must go up in three's, it doesn't.

If we use substitution, we'll see where plus three is, but it's not in the difference between the terms. When n equals one, the term is four.

When n equals two, the term is five.

When n equals three, the term is six and so on.

The plus three is the difference between the term number and the term, one plus three, two plus three, three plus three, four plus three.

That's what's generating our terms, but the terms themselves, four, five, six, seven, eight, they just call one each time.

Next, using nth terms to find any term.

The first five terms, the arithmetic sequence, 4n plus 5 are below.

What's the 100th term? Okay, so this sequence goes 9, 13, 17, 21, 25.

Oh, what's that Andeep? You've got this? Super.

You can work out there, I don't have to.

Okay.

I see.

Yes.

Hmm.

Right.

Yeah, I'm not sure, Andeep.

Andeep stopped at 173 whilst counting up in this sequence and asked, "Am I near yet?" I don't know, I really don't know.

I could stare at that for quite a while and I still won't know.

What do you think about Andeep's method? Pause this video, tell the person next to you, You might have said something like it's not very efficient and we might lack accuracy.

If I'm adding four's all those times, I'm pretty sure I'm gonna go wrong at some point.

We like to be accurate in mathematics.

A better way to do it would be to just consider what the nth term is.

How do we find any term in this sequence? Well, we're gonna take that sequence, that term number, multiply it by four and add five.

That's our nth term.

So we can use this nth term to find any term.

We're not limited to finding the first five terms of a sequence.

We just need to consider which term we're after.

In this case, we're after the 100th term, so we're gonna have an n value of 100.

4 lots of a 100 plus 5 is 405.

And we can find that term in a split second rather than having to write down the first 100 terms of this sequence.

Just like to check you've got that now.

Can you use substitution to find the 50th, 100th, and 1,000th terms of the sequence 12n minus 75? Pause this video and work those out.

For the first one, the 50th term number, n equals 50, 12 lots of 50 minus 75, gives us 525.

And then you'll do the same substitution, but 100 this time.

When n equals 100, 12 lots of 100 minus 75, gives us 1,125.

And for the 1,000th term, they're just bigger numbers.

It's no different at skill, it's just substitution 12 lots of 1,000 minus 75 is 11,925.

We can find any term in any sequence, provided we've got an nth term rule for that sequence.

I'm gonna do these two unusual sequences and I'm gonna give you two to practise.

Find the 250th term of the below sequences.

The first sequence I've got is a half n minus three quarters and that is a beautiful sequence.

To find the 250th term, I just need to substitute 250 into that n position.

Half of 250 minus three quarters will be 125 minus three quarters, give me 124 and a quarter.

That second sequence, 3n squared minus 5n, I've got to substitute n in twice here, I'm gonna need 3 lots of 250 squared minus 5 lots of 250.

I'm good, but I'm not that good that I can work out 250 squared multiplied by three without a calculator, hmm, and maybe I could, but I can do it a lot quicker with a calculator.

So I'll just type that into my calculator, and out comes 186,250.

That is the 250th term in the sequence, 3n squared minus 5n.

Over to you.

And you're probably gonna want a calculator, the 250th term of those sequences please.

Pause this video and work those out.

The 250th term in the sequence, 1/5n minus 3/10, (clears throat) we need to substitute n, n equals 250, 1/5 of 250 minus 3/10 will give you 49 and 7/10.

Your calculator might have told you 49.

7, 49.

7, 49 and 7/10 are the same thing.

Either will be acceptable.

For 8n squared minus 3n and your substituting an n twice, 8 lots of 250 squared minus 3 lots of 250, would've given you 499,250.

Next, the first five terms of the sequence, 5n plus one are.

6, 11, 16, 21, 26.

The fifth term is 26, so the 10th term will be 52 because that's double.

I mean, I just sound so logical.

If the fifth term is 26 and I want the 10th term, double five to get 10, double the term 26 to get 52.

That feels pretty logical.

Do you agree? Pause this video, have a conversation with the person next to you.

Do you agree or not? You might have said something along the lines of, "His reason seems sound, but I know he's wrong." You know he's wrong because you could write down the first 10 terms and you don't hit 52, you hit 51.

So you know Alex is wrong, but why? Because it seems so logical.

Double the fifth terms, get to the 10th terms. Double 26 to get to 52.

Why does that not work? Substitution will help us again.

(clears throat) The 10th term, n equals 10, 5 lots of 10 plus 1, that's 51, so that's the 10th term.

We use substitution to accurately find any term in the sequence, 5n plus 1.

So what's the difference between when n equals 10 and n equals five when we use substitution? Does that help us to understand why it's gone wrong? The 10th term, 5 lots of 10 plus 1.

The fifth term five lots of five plus one.

Ah, therein lies the difference.

For the 10th term, we've got 5 lots of 10 or 10 lots of 5.

As for the fifth term, we've only got five lots of five.

The number of five is doubled, but note, the one's did not double.

So we can't just double the fifth term to find the 10th term, double the 10th term to find the 20th.

Substitution again helped us, but looking at that subtle little difference, what is changing here? It's how many five's we've got.

So true or false, the fifth term of 8n plus 5 is 45.

So the 10th term must be 90.

Is that true? Is it false? And how might you justify your answer? It's false.

And the reason it's false is, we can use substitution to show that it's false.

When n equals 10, the term is 8 lots of 10 plus 5, which is 85, not 90.

We can't just double the fifth term, the value of the fifth term to find the value of the 10th term.

Practise time now.

For question one, the first five terms, the arithmetic sequence, 42 minus 5n are 37, 32, 27, 22 and 17.

Jacob wants to find the 20th term.

Is he right? No.

Jacobs declares that the 20th term is negative 57, it's not.

I'd like you to spot Jacob's error and explain a better method to him.

Pause this video, spot the error, and write down an explanation.

Question two, I'd like to use substitution to find the 20th, 100th, and 200th terms of these sequences.

We should have spotted that the error was here.

We have a sequence which is decreasing by five every time, but two minus five does not get us to negative two.

It gets us to negative three.

Once that one's gone wrong, the rest of them are all gonna be wrong in the sequence.

Hence, negative 57 was not the 20th term.

You might have said, "Your method is inefficient.

If you know the nth term, just use substitution." 42 minus 5 lots of 20 gives us 42 minus 100, which is negative 58.

Question two, substitution to find the 20th, 100th, and 1,000th term of these sequences would look like this.

When n equals 20, 7n minus 6 is a term value of 134.

The 100th term was 694, the 1,000th term was 6,994.

For the sequence 6 minus 7n, the 20th term was negative 134.

The 100th term was negative 694.

The 1,000th term was negative 6,994.

For the last, for sequence c, the 20th term was negative 13.

94.

The 100th term was negative 69.

94 and the 1,000th term was negative 699.

94.

The term values were on the screen.

You might wanna pause and just check that you're working out is the same as mine.

Your answer, the same as mine.

For sequences d and e, sequence d is beautiful, n squared plus 3n, the 20th term with 20 squared plus 3 lots of 20, 460, the 100th term, 10,300, the 1,000th term that's, that big number, that's 1,003,000, 1,000 squared plus 3 lots of 1,000, that's 1,003,000.

For sequence e, 3n plus 15 minus 4n squared.

Again, we're substituting n in twice, 3 lots of 20 plus 15 minus 4, lots of 20 squared, gives us negative 1,525.

The 100th term was negative 39,685.

The 1,000th term was negative 3,996,985.

Again, you might wanna pause and just check that your method is the same as mine and your term values are the same as mine.

Finally, comparing the nth terms of different sequences.

Now that we can find the terms for any given sequence, we can use that skill to compare sequences.

The sequences n plus two and n plus three, for example.

Find the first five terms and then compare these sequences.

Give that a go.

Pause this video.

You should have got these term values and notice something going on here between the two sequences.

N plus two gave the term values of three, four, five, six, seven.

N plus three was four, five, six, seven, eight.

And when you come to compare the sequences, you'll notice that they're both arithmetic.

They've both got a common difference of one each time.

But n plus two starts at three, whereas n plus three starts one higher at four and it's always one higher, three to four, four to five, five to six, six to seven.

The sequence n plus three is always one higher than the sequence n plus two.

How interesting.

Find the first five terms of these two sequences and compare them.

Pause this video, give that a go.

So we should have got term values of 2, 4, 6, 8, 10 and term values of 3, 6, 9, 12, 15.

Both sequences are arithmetic.

They have a common difference.

2n goes up in steps of two, whereas 3n goes up in steps of three.

They start at different numbers as well.

2n starts at two, 3n starts at three.

And 3n goes up quicker.

It increases in greater steps.

I'd just like to check that you can do that.

You're independently now.

Find the first five terms of these two sequences and make some comparisons.

The sequences are 5n plus 2 and 5n minus 2.

Right, a table of values will help us find the first five terms of those sequences.

For 5n plus 2, we get 7, 12, 17, 22, 27, 5n minus 2, we get 3, 8, 13, 18 and 23.

You might have said in comparison of those two sequences, they're both arithmetic with a common difference of five.

You might have said they start at different numbers.

5n plus 2 starts at five plus two, seven, whereas 5n minus 2 starts at five minus two, that's three.

You might have said 5n plus 2 is always two above our five times table.

It's not 5, 10, 15, it's 7, 12, 17.

Whereas 5n minus 2 is always two below the five times table.

What if I said find the first five terms and compare these sequences, sequence 2n and the sequence n squared.

How are they different? Find the first five terms. Make some comparisons.

You should have got the terms 2, 4, 6, 8, 10 for 2n and the terms 1, 4, 9, 16 and 25 for n squared.

How are they different? The sequence 2n is arithmetic.

It's going up by common difference of positive two every time.

N squared is not.

It's step is changing each time, it changes by three, then it changes by positive five, then it changes by positive seven and then by positive nine.

2n is increasing slower but consistently, n squared is increasing really quickly and increasing even more with each term.

I'd like to find the first five terms of each of these sequences and then compare them.

All right, pause this video and give that a go.

All right, your last feedback of the day, the sequence 3n went 3, 6, 9, 12, 15, 3n plus 2 looked like that.

3n minus 1 looked like that.

3n plus 0.

7 looked like that.

3n plus a 7th looked like that.

Negative 3n plus 7 looked like that.

And 2 minus 3n looks like that.

So making comparisons, sequences a to g were all arithmetic.

They all had a constant common difference.

That means they're all in that family of arithmetic, also known as linear sequences, a to e, all had a common difference of positive three.

They were all increasing sequences, f and g were a little bit different you might have noticed.

There's a difference of three, but it's negative three, not positive three, giving us decreasing sequences four, one, negative two, negative five decreasing.

In directly comparing those sequences, 3n was the three times table, 3n plus 2 was two above the three times table 3n minus 1 is one less than the three times table 3n plus 0.

7 is 0.

7 more than the three times table and 1/7 more than the three times table in the case of 3n plus 1/7.

Starting points are interesting.

That starts at three plus nothing, three plus two, three minus one, three plus 0.

7, three under seventh.

Yep, they start at 3, 5, 4, 3.

7, 3 and 1/7.

What about f and g? Minus three plus seven, that's four.

Oh, starts at four.

So that would mean that this final one starts at two minus three, pickles.

It starts at negative one.

How did a to g compared to sequence h, which was n cubed? N cubed went 1, 8, 27, 64, 125.

They're your cube numbers.

And you'll notice that that was the odd one out and that it was the only one that wasn't arithmetic.

So to summarise, we can use substitution to find any term in a sequence when we're given the nth term.

For example, the 50th term of the sequence, 12n squared minus 71 is found by substituting in n equals 50.

I hope you've enjoyed today's lesson, I have.

I will see you again soon for more mathematics.