Lesson video

Hello, Mr. Robson here.

Superb choice to join me for math today, especially because we're problem solving and I love problem solving.

Our learning outcome is gonna be that we'll be able to use our knowledge of sequences to solve problems. A nice variety of problems coming your way today.

Keywords that we'll need.

Arithmetic sequences, also known as linear sequences.

They're sequences where the difference between successive terms is a constant.

Five, nine, 13, 17 is arithmetic.

A constant difference of positive four.

One, two, four, eight not arithmetic 'cause it doesn't have a constant difference.

We'll be referring to those words throughout.

Also nth term.

The nth term of a sequence.

The position of a term in a sequence where n stands for the term number, n equals 10 means the 10th term in a sequence.

The nth term can also be used to calculate any term in a sequence.

So there's also a formula for finding any term.

I'll be referring to nth term throughout this lesson.

Three stages to our lesson on problem solving.

We're gonna start with generalising from non-sequential terms. What does non-sequential mean? Well, they're not one after the other.

Find the missing terms in these arithmetic sequences.

We're told the sequences are arithmetic means they have a constant difference.

That first one is easy.

I can see that they're moving up in steps of seven.

I've got a constant difference of positive seven.

So it must continue like that and give me the terms 34 and 41, but the next two are not so easy.

Why are the second and third one's harder? Exactly.

We haven't got consecutive terms, so I need to look at them differently.

Hmm.

The common difference is a little tougher to spot.

13 to 23 in two steps.

Ah, two steps making a difference of 10 that must make each difference positive five therefore the sequence must go 13, 18, 23.

That positive five constant difference is gonna continue because we're told it's an arithmetic sequence so it must go 23, 28, 33.

The bottom one, one, two, three steps to make a difference of nine steps of positive three each time.

So must get 13, 16, 19, 22.

That constant difference continues giving us the final term of 25.

Your turn, just to check that you've got that.

Find the missing terms in these arithmetic sequences.

Pause, figure these out.

The first one we should have spotted.

There's two steps to make negative five.

So those steps must be negative 2.

5, giving us a term in the middle there of 17.

5.

And then if that pattern of negative 2.

5 continues throughout and it does because we're told it's an arithmetic sequences, we can find the other two terms. 22.

5 and 12.

5.

The second one, that was one, two, three, four steps to make one.

Those steps are positive 0.

25.

So we should have got the terms 1.

45, 1.

7 and 1.

95.

Next up, we'll see problems like this in a real life context.

I love it when the maths we see in the classroom is applied in the real world around us.

Buses depart Croydon for Lewisham at regular intervals.

The third bus of the day left at 6:56 AM, the sixth bus is leaving at 8:08 AM.

What time did the first bus leave? Wow, there's a lot in that question.

I'd like you to pause and just give it a go.

I wonder what you came up with.

I don't mind confessing, I find this problem difficult to work with unless I can see what's going on.

What I've shown there is no first bus, no second bus.

The third bus leaving at 6:56 AM, no fourth bus, no fifth bus, but the sixth bus leaving at 8:08 AM and we are told crucially they're regular intervals.

So one, two, three steps to get from our third bus to our sixth bus and those three steps take 72 minutes 'cause there's 72 minutes between those two times.

So share that 72 equally across those three steps because we're told they're regular intervals we get 24 minutes between each bus.

So the fourth bust must be at 7:20 AM and another 24 minutes to 7:44 AM and I know it works because when I add another 24 minutes, I get to 8:00 AM which is the time we're told that the sixth bus was leaving.

What time did the first bus leave? Well, I need to work backwards in those 24 minutes steps and I get to 6:32 AM and then 6:08 AM.

Lovely.

That little visual really helped me out there.

Maybe they'll help you too in the future.

They'll check it out.

The third bus of the day from Lewisham to Greenwich left at 6:50 AM, the fifth bus left at 7:40 AM which is true? Buses leave every 50 minutes.

Buses leave every 45 minutes, buses leave every 25 minutes, which of those is true? Pause and tell the person next to you.

It was C.

Buses leave every 25 minutes.

There's 50 minutes between 6:50 AM and 7:40 AM and bus four would be in the middle of that gap.

50 minutes divided by two, there every 25 minutes.

Next up, the first term of an arithmetic sequence is 11, the fifth term is 31.

Lucas and Izzy are discussing this sequence.

Lucas says, well 31 minus 11 is 20 and it's the fifth term, 31.

So 20 divided by five is four is a common difference of four.

Izzy suggests, so this sequence goes like this.

11, add four to 15, add four to 19, add four to 23, add four to 27.

Wait, that's wrong.

Well done Izzy.

You know it's wrong, but how does Izzy know it's wrong and what has gone wrong in Lucas's mathematics? Pause this video and have a little think about that.

So we're told that the fifth term is 31.

It says that in the question, it's not 27 as Izzy has worked out when she's gone up in steps of positive four.

What's gone wrong is that there's four steps to get from the first term to the fifth.

Lucas has taken the difference of 20 and divided it across five steps because he knew that 31 was the fifth term.

Whereas the difference between the first term and the fifth term is only four steps.

Seeing this table will help us to see those steps.

When I put that information into that table, the first term is 11, the fifth term is 31.

I can see the four steps that get me from the first term to the fifth term.

From the position n equals one to the position n equals five there are five minus one steps, four steps.

So Lucas should have taken that difference of 20 between the terms and divided it across the four steps, not dividing it across five steps, there's only four.

That gives us a common difference of five, which would give us those terms, 16, 21, 26.

Let's check you've got that.

How many steps are there between the first and the 10th terms in an arithmetic sequence? Is it nine, 10, or 11? Pause and tell the person next to you.

It's option A.

Nine steps from n equals one to n equals 10 there are 10 minus one steps.

How many steps are there between the 71st and the 95th terms in an arithmetic sequence? Is it 25, 24, or 23? Pause, tell the person next to you.

It's B.

24 steps from n equals 71 to n equals 95, there's 95 minus 71 steps.

The 71st term of an arithmetic sequence is 579.

The 95th term is 771.

Find the nth term of the sequence.

Are you kidding? That's impossible.

This is way too hard.

It's not, we can do it.

We just need to break this problem down.

The 71st term, the 95th term.

Okay, to get from the 71st position to the 95th position, I need to take 24 steps.

In this 24 steps, how much did we go up? 771 minus 579 means we've gone up 192 over 24 steps that makes each step eight.

It's got a common difference of eight.

This must be an eight n sequence.

It's not exactly eight n though, but we can compare eight n to this sequence.

What do we know? We know the 71st term, so let's make a table about that 71st term and then let's compare eight n to this sequence.

If the sequence were eight n, eight lots of 71, 568 but we're told the 71st term of our sequence is 579, we need the translation from the eight n sequence to our sequence.

To get from 568 to 579, we need to translate by positive 11.

So our sequence must be eight n plus 11.

Wow, just two terms and their positions in the sequence and we can figure out the nth term of an arithmetic sequence.

How awesome is that? All right, I'll do one example and I'll hand over to you to try an example.

We've got the 25th term of an arithmetic sequence, 18.

45 and the 32nd term 20.

9, and I'd like to find the nth term.

So from n equals 25 to n equals 32, that's seven steps.

How far have I gone up? I've gone up 2.

45 over those seven steps.

That gives me a constant difference each time of 0.

35.

So I have a 0.

35 n sequence, but it's not exactly 0.

35 n.

So I'm gonna set my 25th term, my given bit of knowledge and see if I can figure my translation from there.

0.

35 multiplied by 25 will give me 8.

75.

So the 25th term in 0.

35 n is 8.

75.

The 25th term in my sequence is 18.

45.

I need to figure out the translation from 8.

75 to 18.

45.

It's a translation of 9.

7.

So this sequence must be 0.

35 n plus 9.

7.

Your turn.

There's your question, pause this video, see if you can replicate what I've done.

Good luck.

All right, well done for giving that a go.

Did we start with there are 19 steps from the 18th term to the 37th term.

We went up by 45.

6 over 19 steps, steps of 2.

4.

It's a 2.

4 n sequence, but what's the translation from the sequence 2.

4 n to what we have? We didn't have 43.

2 as an 18th term, we had 40.

2, which is a translation of negative three.

So this nth term would be.

Lovely.

2.

4 n minus three.

Practise time now.

Question one, I'd like to fill in the missing terms in these arithmetic sequences.

If you look at row A blank, blank, blank, one, 10, blank.

Can you spot the constant difference in that arithmetic sequence and then tell me the missing sixth term, the missing third term, second term, first term.

Do that for each row please.

You'll notice something unusual about row E and row F panic not, you've got this.

Question two, what happens when you try to find the missing terms in this sequence and what does that tell you about this sequence? Two, blank, blank, 32, blank, 98.

Hmm, try and find the missing terms and then you'll notice something.

Question three, buses leave Chorley for Bolton at regular intervals.

The second bus of the day left at 6:35 AM, the sixth bus left at 8:03 AM and you just missed it.

Oh no.

What time is the next bus? That is exactly the kind of thing we need to work out in the world around us.

Pause and give those two questions a go.

Question four, the 23rd term of an arithmetic sequence is 187.

The 45th term is 363.

Find the 100th term of the sequence and part B, will 3,634 be a term in this sequence? Delightful.

Pause, give it a go.

Feedback time now.

Missing terms in the sequences.

It helps to understand the term-to-term rule in order to spot the missing terms and the term-to-term rule in the first row, A is positive nine, a constant difference is positive nine, which is gonna give you those terms. In the second row, we've got one, blank, blank, 10, that's three steps to make a difference of nine.

That's steps of positive three each time our missing terms must be those.

In the third row, the term-to-term rule was plus 4.

5.

Two steps to get from one to 10.

We're going up in 4.

5 giving you those terms. and then d, a decreasing sequence.

One, two, three, four steps to go down by nine.

That's negative 2.

25 as a constant difference giving you those terms. You wanna pause this video now and just check that your terms match mine.

Did you spot the right term-to-term rules.

Right.

Part E and part F of that question were a little bit trickier on account of there's fractions in there.

Three steps from one-tenth to one.

If I change that one to 10 tenths, it's easier to see that there's a term-to-term rule of positive three-tenths between the terms. That gives you those terms and there's a reason why I've left two rows in this table because the sequence starts negative five-tenths, negative two-tenths.

One-tenth, one-tenth is the only one I've said that's in its simplest form.

You would cancel down negative five-tenths to negative a half, negative two-tenths to negative one-fifth.

So you should have written negative half, negative-fifth, one-tenth, two-fifths, seven-tenths, one.

The same thing for f.

I'm gonna turn that from one-tenth and one to two-20th and 20-20th because that enables me to see that we're going up by nine-20th each time.

If I write that for you in 20th, it looks like that.

When I cancel down, it looks like that.

Pause and just check that you've got what I've got.

Question two, what happens? Well, we've got three steps to get from two to 32 and we've got two steps to get from 32 to 98.

Therein lies the problem.

We're not going to have a constant difference if we don't get a common constant difference between the terms, it's not an arithmetic sequence.

All right, our real life problem now.

I'm going to find this easiest to just map out those buses.

The first bus, we don't know the second bus we do third, fourth, fifth, we don't, the sixth we do.

What time's the next bus? I need to know when the seventh bus is.

So one, two, three, four steps to make 88 minutes between those two times.

We're told that they're regular intervals.

So I can split those 88 minutes equally across the four steps and we'll get a 6:57 7:19 AM, 7:41 AM and I'm gonna add the different 22 minutes.

I know I'm right because I get to 8:03 AM so if the next bus is leaving at another regular interval of 22 minutes, it's on its way at 8:25 AM.

I've just gotta wait until that time to get my next bus.

Do I need to know the first bus time? No, no.

We were asked what time is the next bus.

The next bus is 8:25 AM, we've answered that question however, I kind of don't wanna leave it unfinished.

It's 6:13 AM if we go back 22 minutes, and I know for next time when I'm in Chorley looking to go to Bolton, the earliest I can go is 6:13 AM.

The fourth one we can deduce an awful lot.

You give me two terms and their positions in the sequence tell me it's an arithmetic sequence.

We can do all sorts.

Find the 100th term.

Well, We've jumped by 176 numerically in 22 steps.

Each step is of eight, therefore it's an eight n sequence.

That 23rd term, the eight n sequence will be 184.

Ours isn't, ours is 187 giving us a translation of plus three.

So we've got the nth term of eight n plus three.

I wanna find the 100th term, so I substitute in n equals 100 and I find a 100th term of 803.

Will 3,634 be a term in this sequence? Nope, we have odd terms and an even common difference, therefore 3,634 an even number will not be in the sequence.

We deduced all that just from two terms and their positions.

How nice.

Right, fractional sequences.

These are lovely.

Is this an arithmetic sequence? Hmm.

Look, one, two, three, four, five, six.

It looks like one, but it's not one.

To get from one to a half, we need to minus a half.

To get from a half to a third, minus a sixth, minus a 12th, minus a 20th, minus one-30th.

We've not got a constant common difference.

It's not an arithmetic sequence, but there is a pattern going on.

How do we define this pattern? If I put it into a table like that, the numerator's constant, it's always one, but there's clearly a relationship between the term number and the denominator.

Have you spotted it? Yes.

For any term in this sequence, I can tell you the term value is one over n, the fifth term, one over five, the sixth term, one over six.

The nth term, one over n.

The sequences might not be arithmetic, but we are able to define an nth term.

We're now defining nth terms of non-arithmetic sequences.

Well, that's pretty awesome maths.

If it's not arithmetic, it's not linear.

What does it look like? You know that arithmetic sequences, when we plot them, the points align.

They make a straight line.

I wonder what this one looks like.

Wow, when we plot one, a half, a third, a quarter, a fifth as a sequence, it looks like that.

How cool is that graph? You ever seen a graph like that before? Maybe, maybe not.

It is delightful.

You don't need to know about those graphs just yet.

I just wanted to show you because I think it's beautiful.

Quick check.

What is the nth term for this non-arithmetic sequence? Two thirds, two quarters, two fifths, two sixths, two sevenths.

Is it two over n plus two, two over n plus three or two over three n? Pause this video.

See if you spot which one it is.

It was the top one, two over n plus two.

The numerators are constant, two, the denominator is the sequence, n plus 2, a translation from the sequence n by positive two.

And if you're interested, the graph of this sequence would look like that.

It's not linear, not arithmetic, but it's beautiful nonetheless.

What's the nth term for this non-arithmetic sequence? Is it two n over three n, two n over n plus two or n plus two over n plus three? Pause so you can figure out which one it is.

It was b, two n over n plus two.

The numerator two, four, six, eight, that's a sequence, two n.

The denominator was a sequence n plus two.

A translation positive two from the sequence n.

And again, I can graph that for you.

That graph's different, but equally as beautiful.

Finding the nth term of non-arithmetic sequences, how nice.

What's the nth term for this sequence? Tricky looking one.

Is it three n minus 12, one over one minus three n, is it three n minus nine over one minus three n? Or is it three n minus 12 over minus two minus three n, negative two minus three n, which one is it? Pause this video.

See if you can spot it.

It was a, three n minus 12 over one minus three n.

The numerator is three n minus 12.

The denominator is one minus three n.

There's the graph of that one, which is different and interesting again.

As he says, "I've never seen a graph like this before." You don't need to know about those yet.

I'm just showing you because they are absolutely lovely.

Practise time now.

Jun and Laura are looking at a sequence, one over five, one over 10, one over 15, one over 20, one over 25.

Jun says "five, 10, 15, 20, 25.

This is an increasing sequence." Laura says, "five, 10, 15, 20, 25, that's five n.

This is an arithmetic sequence." I'd like you to write a sentence to justify why Jun is wrong and write a sentence to justify why Laura is wrong.

Pause and give that a go.

Next up, I'd like you to write an nth term for these fractional sequences.

Five sequences there.

Pause this video, see if you can find those nth terms. So Jun and Laura, five, 10, 15, 20 is an increasing sequence.

It looks like it when you just look at those numbers, but these fractions tell a different story.

One-tenth is less than a fifth, one-15th is less than a tenth, one- 20th is less than the 15th.

One-25th is less than one 20th.

The sequence is decreasing.

We could plot those points to show that it's decreasing.

Another beautiful graph and a visual aid to see and to show Jun why it's not an increasing sequence.

Onto Laura's comment.

Five, 10, 15, 20, 25, that five n is an arithmetic.

Okay, that is the arithmetic sequence, five n, but that's not what we've got here.

We've got a non-arithmetic sequence because to get from one-fifth to one-tenth, we need to minus the 10th.

And then the next step is minus the 30th, minus the 60th, minus 100th.

That's a non-arithmetic sequence.

There's not a constant common difference there.

But Laura has spotted something really useful by identifying that the denominators for the pattern are five n.

We can write an nth term for non-arithmetic sequence one over five n.

Say thank you for that Laura Nth term for these fractional sequences.

The first one was quite kind, one over n plus one.

Second one's a little bit trickier.

N over n plus one.

N over two n plus one.

Next one's trickier again, five n minus one over two n plus one.

The bottom one, a constant numerator of one and then one, four, nine, 16.

Did you spot real square numbers? That gives us an nth term of one over n squared.

Last learning cycle coming up now, pattern spotting.

15 people meet for the first time and shake hands with each other just once.

How many handshakes will occur? Okay, Laura considers the problem and says for three people it's three handshakes.

So for 15 people there'll be 15 handshakes.

Very logical.

Alex says for three people it is three handshakes, but for four people, it's not four handshakes, it's bigger than four.

So maybe our solution is bigger.

Maybe there's 15 times 15, 225 handshakes.

Interesting, nice hypothesis.

Aisha says, I'm just drawing pictures.

Hmm, no one's correct yet, but have our Oak students done anything useful? What's your intuition? Anything useful happening here? Pause this video.

Tell the person next to you.

So no one's right yet, but there's some useful strategies going on here.

A strategy we mathematicians employ to solve big problems like this and our Oak students have done some really useful things.

What Aisha has done, drawing pictures really, really, really useful in mathematics.

Can we visualise this problem? Aisha's, right to try and draw this one out.

The kind of thing she's drawing, 15 people.

Let's imagine they're stood in a circle like that and they shake hands with each other like that with the person next to them.

That's 15 handshakes.

But then that person has to lean across and shake that hand and that hand and that hand and every hand.

So that person ends up doing all those handshakes and the person to their left does all those handshakes.

There's a lot of handshaking going on here.

Aisha visuals are gonna be really useful to help us to solve this problem, but it's becoming difficult to see exactly how many.

If I finish that pattern, it will look beautiful, but I've gotta count all those handshakes.

That's gonna be difficult.

So Laura said for three people it's three.

So for 15 people it'll be 15.

Laura did something really useful too.

It's hard to start thinking immediately about 15 people.

She broke it down into smaller steps.

What if it were three people, two people, one person breaking big problems down into small steps is hugely useful when solving big problems in mathematics.

So for three people, it's three handshakes.

There we go, three people, three handshakes.

There's no other handshakes to happen.

Three people really is three handshakes.

And then Alex tested this and this was useful.

Three people, three handshakes, four people is not four handshakes.

I can see six handshakes going on between those four people.

So Alex tested Laura's hypothesis, which is another really important math skill.

We must make suggestions, what do we think it's going to be and then test the truth in those suggestions.

That's what Alex has done here.

Between three of them, we had some really nice ideas.

So let's combine those ideas and solve this problem.

Break the problem down into smaller steps.

Let's say one person, two people, three people.

Let's visualise the problem as Aisha did when she started drawing.

That's two people, three people, four people, five people.

That gives us those handshakes.

One person, there's no handshakes.

Two people, one handshake, three people, I can see the three handshakes and so on.

Ah, Aisha has now spotted a pattern.

I didn't need to draw the sixth one.

I spotted a pattern in the numbers.

Can you see Aisha's pattern? Look at those numbers.

Zero, one, three, six, 10.

What do you see going on? Did you see difference of one, difference of two, difference of three, difference of four, difference of five? Yes.

If you draw out six people and test it, you'll find there's 15 handshakes.

That pattern does continue.

If we continue it all the way to having 15 people, it will look like this going up by six and then seven and then eight and then nine, et cetera.

Until you reach 105.

There will be 105 handshakes when 15 people meet and shake each hand, shake the hands of each other just once.

One, three, six, 10, 15.

Did they look familiar? If they did, you might have recognised 'em as your triangular numbers.

They were behind the solution to this problem.

105 handshakes.

Over to you now.

When faced with a complex problem, what useful approaches can be taken? Try to visualise the problem, break it down into smaller steps, hypothesise and test the hypothesis.

Look for patterns.

Which of those are useful strategies? Pause, tell a person next to you.

Visualise the problems, absolutely.

Break the problem down into smaller steps, absolutely.

Hypothesise and test the hypothesis.

Yes, please.

Look for patterns, always.

Practise time now.

How many squares are there on a chessboard? That's a chessboard.

It's eight by eight broken down into squares.

This is easy, right? Don't forget to consider all the squares, not just those squares.

Those squares and those same squares in different positions and those squares.

Don't forget those squares.

I'd like you to find all the squares.

This might take you a while, so pause.

You might not even all come up with the same answer.

That will be exciting.

I'll be back with a solution in just a moment.

Okay, break the problem down.

One square, an eight by eight square, there's one of them.

A seven by seven square, did you see them? Four of them.

A six by six square.

Were you counting? I hope so.

There were nine of them.

Five by five.

Do I need to draw those or have you spotted a pattern? One, four, nine.

What's next in that sequence? You should have noticed they are square numbers.

So five by five, 16 of them.

That's what I think it's gonna be.

You would test that.

If you drew five by five squares and tested them, you'd find 16 of them.

So with that pattern in mind, one, four, nine, 16, 25, 36, 49, 64, yes, I can see there's 64 one by one squares.

So I just need the sum of the first eight squares and I can see how many squares are on a chess board.

Add those square numbers together.

204.

That was your answer.

That's the end of the lesson now.

To summarise, we can use our knowledge of sequences such as pattern spotting and finding nth terms to solve a variety of problems. I hope you've enjoyed this lesson a lot, I certainly have.

I'll see you again soon for more mathematics.