Lesson video

Hello, Mr. Robson here, welcome back for more maths, securing understanding of arithmetic sequences.

I know I'm gonna enjoy this one, I hope you will too.

Our learning outcome is that we'll begin to generalise a sequence, and we all have generalisations.

They're such an awesome part of mathematics.

Key words that you'll hear me using throughout the lesson, arithmetic or linear sequence.

Three parts to today's lesson and we're gonna start by reviewing the nth term.

Time tables can be described in nth term form.

For example, two, four, six, eight, 10.

We could put it in a table, the first term of our two times table is two, the second term is four, the third term is six, and you'll notice that the term value is always two times the term number.

This is written more simply as an nth term of 2n.

Any times table can be described this way.

For example, term number times five, giving us the term, well, one times five, two times five, three times five.

The sequence 5n is five, 10, 15, 20, 25, our five times table.

The sequence 2n is our two times table, the sequence 5n is our five times table.

Notice the term-to-term difference.

Positive five, sequence 2n went up by two.

The sequence 5n goes up by five.

Let's check you've got that.

Fill in the below blanks for this times table.

Pause and have a go at this now.

I hope you found the terms 14, 21, 28, 35, two times seven, three times seven, four times seven, five times seven, a common difference of positive seven between those terms. So we can say the sequence 7n is the seven times table and the term-to-term difference is positive seven.

If we've noticed that about the sequence is 2n, 5n and 7n, then we should be able to fill in these blanks now for these arithmetic sequences.

Pause, get this written down, fill in the blanks.

See you in a moment.

Welcome back, 3n, it's your three times table, three, six, nine, 12, 15.

The sequence 3n has got a common difference of positive three.

9n, well, that must be your nine times table.

Let's just check.

If we count back, 45, 36, 27, 18, nine, yep, that worked, sequence 9n.

It's got a common difference of positive nine between terms. So four, eight, 12, 16, 20, can we write an nth term for that sequence? Absolutely, it's your four times table.

It's a sequence 4n.

Common difference, positive four between terms. I gave you very little information for this bottom one, but you can see a common difference of six between the two terms that we've given.

You know it's an arithmetic sequence so that common difference repeats.

So counting backwards, 30, 24, 18, 12, six, you can see it's your six times table, which we call 6n, sequence 6n, common difference of plus six.

Sometimes it won't be that simple.

Sometimes the relationship, our nth term, might be two step.

What if I said the term number multiplies by four and adds one to become the term? In this case, the term value is four times the term number plus another one.

We'd call that the nth term 4n plus one.

We can work out each term though.

When n is one, we substitute that into 4n plus one, four lots of one plus one, that's five.

The first term is five.

When n is two, four lots two plus one is nine, and then you can do four lots of three plus one, four lots of four plus one, four lots of five plus one, and you'll get the terms 13, 17, and 21.

So we got the sequence 4n and we got the sequence 4n plus one.

Jacob and Jun are now discussing the link between the terms of these sequences.

Jacob says, "The key is to spot the shift between the two sequences, the shift between the two sequences." Jun says, "I think I see it." Do you, can you spot what Jacob and Jun have spotted? Pause, have a conversation with the person next to you or have a good think to yourself.

I'll see you in a moment.

Welcome back.

"The terms of 4n plus one are always plus one away from the terms of 4n," says Jun.

Did you see that, did you spot that? Take the first term of 4n four, add one, we get the first term of 4n add one.

Take the second term of 4n eight, add one, and we get the second term of 4n add one, and the pattern repeats throughout.

This is the shift that Jacob was on about.

The sequence 4n shifted by positive one each time becomes a sequence 4n plus one.

The sequence 4n plus one, I hope you noted, still has a common difference of positive four, but the terms have shifted.

We can see this shift as a translation on a number line.

The sequence 4n, our four times table, you would've seen that mapped on a number line before.

So when we map this 4n plus one, look where it is.

4n plus one is a translation of the sequence 4n by positive one.

Can you see in each case the terms in the sequence 4n are one to the right to reveal the sequence 4n plus one.

Let's check you've got that.

The sequence 5n is shown above this number line, where? There, that's the sequence 5n, your five times table.

So what is this sequence? Seven, 12, 17, 22, 27.

Compare your 5n above the line to this one below the line.

Can you spot the shift, the translation? Can you give me an nth term? Pause, talk to the person next to you, have a good think.

I'll reveal the answer in a moment.

Welcome back.

Did you say that must be the sequence, 5n plus two? Common difference of positive five, just as a sequence 5n has a common difference of positive five, 5n plus two has a common difference of positive five, but the terms are translated by positive two along the number line each time.

So what's this sequence? Pause, have a good think, have a conversation with the person next to you.

I'll see you in a moment.

Welcome back.

How did we get on? Any good suggestions? Did you suggest it's 5n minus four? Well done, why? Well, it's still got common difference positive five, but the terms are translated by negative four.

Positive five difference in the sequence 5n, positive five difference in this sequence, but all of the terms are translated by negative four to go from the sequence 5n to our sequence.

So we call our sequence one, six, 11, 16, 5n minus four.

Another useful observation about this when we compare 5n to 5n plus two, the five is our common difference.

We'd call this the coefficient of n.

Five was a common difference, the two term was the shift or the translation.

You are going to need this information, so you might wanna pause and copy it down now.

Let's check you've got that, true or false.

The sequence 7n plus three has a common difference of positive three.

Is that true or is it false? Once you've decided, I'd like you to use one of the two statements at the bottom to justify your decision.

Is it statement A that justifies your answer or is it statement B? Pause, have a good think, I'll see you in a moment.

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

There's no more instruction than that.

Five sequences, all arithmetic.

I'd like an enter term, please.

Pause, give it a go.

Finding the nth term of these sequences.

I quite like thinking of it in terms of common difference and translation.

So common difference of positive eight.

So think about the sequence 8n, eight, 16, 24.

That's a shift of positive three to get to our sequence 11, 19, 27.

Common difference of positive eight, shift of positive three, that must be 8n plus three.

For part B, same common difference, positive eight, but it's a translation by negative four.

So this is the nth term, 8n minus four.

C was different, a decreasing sequence, common difference of negative four.

So we're thinking about the translation from the negative 4n sequence.

How different are these terms to the terms in your negative four times table? Well, there was a translation of positive 15, so you would get the nth term negative 4n plus 15, which you might see written as 15 minus 4n.

We like efficiency and 15 minus 4n is quicker to write than minus 4n plus 15.

Both answers are right, but you are more likely to see this referred to as 15 minus 4n.

D was different or was it? Not really, common difference, positive 0.

8, just 'cause it's a decimal, didn't behave any different.

Common difference of 0.

8 or positive 0.

8 and a translation from the sequence 0.

8n by positive 0.

3.

So it's 0.

8n plus 0.

3.

For part E, again, no different, common difference of positive three 11ths.

And then it's not our sequence, three over 11n, it's always one 11th above it.

So that's the sequence, three 11ths n plus one 11th.

Onto the second part of the lesson now, using the nth term.

Defining the nth term of a sequence has many uses.

For example, what's the 150th term in this sequence? It'd be highly inefficient to write the next 145 terms out.

I've started and it took me a long while, but I'm not even at the 50th term yet.

Do I want to write out another hundred? Not really, it'll take me ages and I might well make an error.

More efficiently, and we'd say this arithmetic sequence is to defined by the nth term 12n plus five, so what's the 150th term? Well, that's when n equals 150.

Substitute it in, 12 times 150 plus five is 1,805.

So what's the 150th term? It's 1,805.

We like that, that's way more efficient.

We can also work backwards, solving an equation to find a term's position in a sequence.

Same, same sequence, is 2,585 in the sequence and what position is it? I don't wanna write out the rest of that sequence until I hit 2,585, and then count how many terms I wrote to get there.

That would take me a while.

S0 it's 12n plus five, the nth term hasn't changed.

Will the nth term ever hit 2,585? Well, we set up an equation, 12n plus five, when does it hit 2,585? We add negative five to both sides of that equation, we divide both sides by 12, n equals 215, same.

2,585 is in the sequence.

It's the 215th position in the sequence.

What about when a term is not in a sequence? Again, same sequence, the sequence, 12n plus five.

And if we know that 2,585 is in the sequence and there's a common difference of positive 12 between terms, we know that 2,590 is not in this sequence.

So if we tried to solve this equation, when does our sequence, 12n plus five, hit 2,590? We solve it just as we'd solve any equation.

Add negative five to both sides, divide through by 12.

Something different about the answer now, n equals 215.

417.

I've rounded that answer to three decimal places.

But importantly, the position n has to be a positive integer for this sequence.

This equation does not have a whole number solution.

So 2,590 is not in the sequence.

Quick check you've got that.

True or false, we can solve the equation 4n minus 11 equals 307, therefore 307 must be in the sequence.

Is that true or is it false? Once you decided, can you justify your answer with if you can solve the equation using the nth term, then it must be in the sequence or we can solve equations to find the position of a term, but the solution must be a positive integer.

So true or false and pick a statement to justify your decision.

Pause and I'll see you in a moment.

Welcome back, I hope you said false, and I hope you justified it with, we can solve equations to find the position of a term, but the solution must be a positive integer.

For example, if we try to solve this, let's add positive 11 to both sides, let's divide through by four, n equals 79.

5.

What's wrong with that answer? There is no 79.

5th term, therefore 307 is not in this sequence.

Practise time now, for question one, there's three sequences there, sequence A, sequence B, sequence C.

For each of those, I'd like you to do three things.

I'd like you to find the nth term, use your nth term to find the 100th term, and then I'd like you to use that nth term to identify if 500 is in the sequence, and if so, what's its position? Question two, assuming each pattern continues to grow the same way, will we see a pattern with exactly 250 circles? Pause, give this a go.

Feedback time, for question one, I asked you to do three things with each sequence, the first of which was find the nth term.

For part A, the nth term was 3n minus 13.

For part two, using the nth term, the 100th term is three lots of 100 minus 13, it's 287.

For part three, is 500 in the sequence? Yes, it's in the sequence.

We can say 500 is the 171st term.

For part B, the nth term was 1,002 minus 3n, a decreasing arithmetic sequence.

The coefficient of n is negative.

For the 100th term, it's 1,002 minus three lots of 100, giving us 100th term of 702.

Is 500 in the sequence? If we know it's not a positive integer solution, we know that 500 is not a term in the sequence.

For part C, nth term of 0.

4n plus eight, giving us a 100th term of 48.

Is 500 in this sequence? 500 is in this sequence, it's the 1,230th term.

For question two, my patterns.

If they continue to grow in the same way, will we see a pattern with 250 circles? Well, turning it from a pattern into an a numerical sequence was one way to do this.

The first pattern goes five circles, eight circles, 11 circles.

That's an nth term of 3n plus two.

Can we solve the equation 3n plus two equals 250, IE, will that sequence ever have 250 circles? We can solve it, but it's not a positive integer solution, it's 82.

6 recurring, so no, that pattern will not have 250 circles.

For part B, the circles went eight, 10, 12.

That's the N term 2n plus six.

Can we solve the equation 2n plus six equals 250? We can, n equals 122.

So we can say yes, the 122nd pattern will have exactly 250 circles, but you could have spotted eight, 10, 12.

Well, we're definitely gonna hit 250 because we're gonna hit every even greater than or equal to eight, so you might have come to that solution in a slightly more efficient way.

For part C, the circles went 10, 14, 18, that's an nth term of 4n plus six.

Can we solve 4n plus six equals 250? We can, n equals 61.

So yes, in the 61st pattern, there will be exactly 250 circles.

On to the third part of the lesson now, generalising finding the nth term.

There's more than one way to generalise an arithmetic sequence and it's useful for disproving a common misconception, the sequence 3n plus two, it can cause confusion.

For example, our pupils, Sam and Alex, they say respectively, "I think 3n plus two starts at three and goes up by two each time," and, "I think 3n plus two starts at two and goes up by three each time." These are common misconceptions.

Both pupils are wrong, but can you see where they got their ideas from? Pause this video, tell the person next to you where Sam and Alex got their misconceptions from.

See you in a moment.

Welcome back.

You can see where they got those misconceptions from.

There's a three, there's a two.

It must start at three and go up by two or start at two and go up by three.

Sam sensibly says, "Let's work out the first few terms to see who is right," and they work out when n equals one, there's a term value of five, when n equals two, there's a term value of eight, when n equals three, there's a term value of 11.

They get the sequence five, eight, 11, 14, 17, and Alex says, "We are both wrong." So let's look at the sequence in a different way.

If we see it building, five, eight, 11, 14, 17.

Can you see it building, can you see what it's adding? If we look at this in a table, when n equals one, there's a term of five, when n equals two, there's a term of eight.

How were that five, that eight, that 11 constructed? Well, five is just five.

That's the first time in the sequence.

The eight is made of five and three.

The 11 is made of five and two threes, and then it's five and three threes, and then it's five and four threes.

If you can see a building like that, it's going to lead us to a different generalisation.

Can you see how many threes we're adding each time? When n equals one, we don't add n threes.

When n equals two, we add one three.

When n equals three, we add two threes.

When n equals four, we add three threes.

When n equals five, we add four threes.

Can you generalise? What if I said find any term in this sequence? How many threes will you add? Well done, you'll add n minus one threes.

So we can express this sequence as five plus how many threes? N minus one threes.

As a generalisation, we call this the first term plus the difference multiplied by n minus one.

You'll see it written algebraically as a plus d, bracket n minus one.

It looks a little scary, but it is in fact remarkably simple, a is the first term and d is the difference or the common difference between terms. You're going to need this information, so you might want to pause and copy that down, a plus d bracket n minus one, whereby a is the first term and d is the difference between terms. Right, I'm gonna do two questions now and then ask you to repeat the same skill.

So using the generalisation, a plus d bracket n minus one.

Write the first five terms of the sequence where a equals 11 and d equals four.

But when we use this generalisation, it's remarkably simple, a is 11, so my first term is 11, my d equals four, my difference is positive four.

So I have a sequence which goes up by four every time, 11, 15, 19, 23, 27, done.

There's the first five terms. Can I go backwards? Can I take the sequence 12, 17, 22, 27, 32, and write that sequence using the generalised form a plus d bracket n minus one? I absolutely can, a, the first term, is 12, d, the difference, is positive five.

So I can express this sequence as 12 plus five bracket n minus one.

It's really useful to check your work in mathematics, so I'm just gonna check my generalisation works.

I'm gonna substitute in n equals five, and if I've done it correctly, I should get 32 'cause that's the fifth term.

Oh, I do.

Your turn now, pause, give this a go.

I'll see you in a moment.

Right, the first five terms of sequence where a equals negative eight and d equals five.

We start on negative eight.

We have a common difference of positive five, so it goes negative eight, negative three, two, seven, 12.

Can you write this generalised form for the sequence one, eight, 15, 22, 29? Well, we've got a first term of one and we've got a difference of positive seven, so we could call that sequence one plus seven bracket n minus one, and I hope once you wrote your generalisation, you just checked that it worked.

If I substitute in n equals five, it tells me that the fifth term is 29, which is what I hoped it would.

That generalisation worked.

Let's compare this form, a plus d bracket n minus one to the nth term form we've seen previously.

We might have said the sequence one, eight, 15, 22, 29 has an nth term of seven n minus six.

If I ask you to write in this form, you tell me it's one plus seven lots of bracket n minus one.

Expand those brackets and simplify.

Oh look, 7n minus six, they're one in the same thing, just generalised slightly differently.

Let's check you've got that now.

I'd like you to write the first five terms, the arithmetic sequence 9n minus one, and then write it in the generalised form, a plus d bracket n minus one.

Pause, give that task a go, I'll see you in a moment.

I hope you generated the terms eight, 17, 26, 35, 44, and then expressed it in the form a plus d bracket n minus one by saying eight plus nine bracket n minus one, and then did you check that you were right? If you substitute in n equals five, you should get the term 44 or you could expand those brackets and simplify and get back to your original nth term, 9n minus one.

Either of those checks would've proof that you were right.

Practise time now.

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

They're written in the form a plus d bracket n minus one where a is the first term and d is the common difference.

You might want to know that.

Pause, give these a go.

For question two, I'd like you to write two different generalisations for each of these arithmetic sequences.

Pause and give these four a go.

Question three, for the sequence 24 plus seven bracket n minus one, I'd like you to tell me the first five terms, the 100th term, and I'd like you to find the position of the term 388 in that sequence.

Pause and try that now.

Feedback time, question one, part a, the first five terms of that sequence, a is five, d is four, so we start on five and we have a common difference of positive four.

For part b, a, the first term is four, d, the common difference is positive five, so we start on four and have a common difference, positive five.

For part c, a, the first term is four, d, the common difference is negative five, so we get those terms. Just pause and check that yours match mine.

For question two, two different generalisations for each of these arithmetic sequences.

You might have called that top one 3n plus four, or you might have said, well, it starts on seven and has a common difference of three.

For the second one, 17 minus 10n or it starts on seven and has a common difference of negative 10 in that form.

Part C, 0.

25n plus 0.

45 or it starts on 0.

7 and has a difference of positive 0.

25 if you'd written it in that form.

For part d, four over 7n minus three sevenths or one seventh plus four seventh bracket n minus one.

For question three, the first five terms went 24, 31, 38, 45, 52.

The 100th term, substitute in n equals 100, and you get the term 717.

And then the position of 388 in that sequence, we need to solve this equation.

Subtract 24 from both sides, divide three by seven, n equals 53, it's the 53rd position.

I'll end the lesson now.

In summary, using nth terms expressions, we can generalise arithmetic sequences in a number of ways.

Another way we've learned to generalise arithmetic sequences today is to use the form a plus d bracket n minus one, where a is the first term and d is the common difference between terms. We can use these generalisations to justify if a term is part of a sequence or not.

Hope you've enjoyed our lesson today, I certainly have, and I look forward to seeing you again soon for more mathematics.

Goodbye for now.