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.

A 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.

5, 9, 13, 17 is arithmetic, a constant difference of +4.

1, 2, 4, 8, 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 is 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 is 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 7.

I've got a constant difference of +7, so it must continue like that and give me the terms 34 and 41.

But the next two are not so easy.

Why 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 +5.

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, two, three steps, to make a difference of 9, that's steps of +3 each time.

So I must 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 -5, 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 asthmatic 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, not 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 seen 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 you a go in.

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 the regular intervals, we get 24 minutes between each bus.

So the fourth bus must be at 7:20 am, then another 24 minutes to 7:44 am.

And I know it works because when I add another 24 minutes, I get to 8:08 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 up.

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 2 they're 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 5 is 4, there's a common difference of 4." Izzy suggests, "So this sequence goes like this, 11, add 4 to 15, add 4 to 19, add 4 to 23, add 4 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 on the question.

It's not 27 as Izzy has worked out when she's gone up in steps of +4.

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

From the position n equals 1 to the position n equals 5, there are 5 minus 1 steps, a 4 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 9, 10, or 11? Pause and tell a person next to you.

It's option a, nine steps.

From n equals 1 to n equals 10, there are 10 minus 1 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 set 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 those 24 steps, how much did we go up? 771 minus 5 79 means we've gone up 192 over 24 steps.

That makes each step 8.

It's got a common difference of 8, this must be an 8n sequence.

It's not exactly 8n though, but we can compare 8n 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 8n to this sequence.

If the sequence were 8n, 8 lots of 71, 568.

But we're told the 71st term of our sequence is 579.

We need the translation from the 8n sequence to our sequence.

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

So our sequence must be 8n + 11.

Wow, just two terms and they're 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.

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

35n.

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 multiply by 25 will give me 8.

75.

So the 25th term in 0.

35n 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.

35n 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 to term to the 37th term.

We went up by 45.

6 over 19 steps, steps of 2.

4.

It's a 2.

4n sequence.

But what's the translation from the sequence 2.

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

2 as an 18th term.

We had 40.

2, which is a translation of -3.

So this nth term would be, lovely, 2.

4n minus 3.

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, 1, 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.

Question two, what happens when you try to find the missing terms in this sequence and what does that tell you about this sequence, 2, blank, blank, 32, blank, 98? Hmm, try and find the missing terms and then you'll notice something.

Question three, 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 +9, a constant difference is positive nine, which is gonna give you those terms. In the second row, we've got 1, blank, blank, 10, that's three steps to make a difference of 9.

That's steps of +3 each time.

Our missing terms must be those.

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

5.

Two steps to get from 1 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? Question two, what happens? Well, we got three steps to get from 2 to 32 and we 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.

We can deduce an awful lot.

You give me two terms and their positions in the sequence.

Time is 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 8n sequence.

That 23rd term in the 8n sequence will be 184 ask isn't ask 187 giving us a translation of +3, so we've got an nth term of 8n +3.

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

Will 3,634 be a term in their 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, 1, 2, 3, 4, 5, 6.

It looks like 1, but it's not 1.

To get from 1 to a 1/2, we need to -1/2.

To get from 1/2 to 1/3, -1/6, -1/12, -1/20th, -1/30th.

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? Well, if I put it into a table like that, the numerator's constant, it's always one, but there's clearly 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 1 over n.

The fifth term, 1 over 5, the sixth term, 1 over 6, the nth term, 1 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, pretty awesome maths.

If it's not arithmetic, i.

e.

, 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 1, 1/2, 1/3, 1/4, 1/5 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 want to show you because I think it's beautiful.

Quick check.

What is the nth term for this non-arithmetic sequence, 2/3, 2/4, 2/5, 2/6, 2/7? Is it 2 over n plus 2? 2 over n plus 3? Or 2 over 3n? Pause this video, see if you can spot which one it is.

It was the tough one.

2 over n plus 2.

The numerator is a constant 2.

The denominator is a sequence n plus 2, a translation from the sequence n by +2.

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 2n over 3n? 2n over n plus 2? Or n plus 2 over n plus 3? Pause so you can figure out which one it is.

It was b, 2n over n plus 2.

The numerator 2, 4, 6, 8, that's a sequence 2n.

The denominator was a sequence n plus 2, a translation +2 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 3n minus 12 over 1 minus 3n.

Is it 3n minus 9 over 1 minus 3n? Or is it 3n minus 12 over minus 2 minus 3n? Minus 2, -2 minus 3n, which one is it? Pause this video, see if you can spot it.

It was a, 3n minus 12 over 1 minus 3n.

The numerator is 3n minus 12, the denominator is 1 minus 3n.

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

Izzy 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, 1 over 5, 1 over 10, 1 over 15, 1 over 20, 1 over 25.

June says, "5, 10, 15, 20, 25, this is an increasing sequence." Laura says, "5, 10, 15, 20, 25, that's 5n.

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, 5, 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.

1/10 is less than 1/5, 1/15 is less than 1/10, 1/20 is less than 1/15, 1/25 is less than 1/20.

The sequence is decreasing.

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

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

On to Laura's comment, 5, 10, 15, 20, that 5n is not okay.

That is the arithmetic sequence, 5n, but that's not what we've got here.

We've got a non-arithmetic sequence, because to get from 1/5 to 1/10, we need to minus the 1/10, and then the next step is minus the 1/30, minus the 1/60, minus the -1/100.

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 of 5n, we can write an nth term for non-arithmetic sequence 1/5n.

So thank you for that Laura.

nth term for these fractional sequences.

The first one was quite kind 1 over n plus 1.

Second one's a little bit trickier, n over n plus 1.

n over 2n plus 1.

Next one's trickier again, 5n minus 1 over 2n plus 1.

The bottom one, a constant numerator of 1 and then 1, 4, 9, 60.

Did you spot a real square numbers? That gives us an nth term of 1 over n squared.

Last learning cycle coming up now, pattern spotting.

15 people meet for the 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." Pretty 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 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.

There are strategies 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 things 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.

Hm-mm.

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's 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 3, 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.

And 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, 0, 1, 3, 6, 10.

What do you see going on? Did you see difference of 1, difference of 2, difference of 3, difference of 4, difference of 5? Yes.

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

That pattern does continue.

If we continue all the way to having 15 people, it will look like this, going up by 6 and then 7 and then 8 and then 9, 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.

1, 3, 6, 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 the 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 on a chess board? That's a chess board.

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, and 8x8 square is one of them.

A 7x7 square.

Did you see them? Four of them.

A 6x6 square.

Were you counting? I hope so.

There were nine of them.

5x5, do I need to draw those or have you spotted a pattern? 1, 4, 9, what's next in that sequence? You should have noticed they are square numbers.

So 5x5, 16 of them.

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

You would test that.

If you drew 5x5 squares and tested them, you'd find 16 of them.

So with that pattern in mind, 1, 4, 9, 16, 25, 36, 49, 64, yes, I can see there's 64 1x1 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 there.

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.