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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.
Keywords that you'll hear me using throughout the lesson, arithmetic or linear sequence.
Three parts to today's lesson.
And I'm gonna start by reviewing the nth term.
Time tables can be described in nth term form.
For example, 2, 4, 6, 8, 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 5, 10, 15, 20, 25, a 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 if 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 sequences: 2n, 5n and 7n, then you 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.
3, 6, 9, 12, 15.
The sequence 3n, it's 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, 9.
Yep, that worked.
Sequence 9n, they've got a common difference of positive nine between terms. So for a 12, 16, 20, can we write an nth term for that sequence? Absolutely, it's your four times stable, it's a sequence 4n.
The 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, 6, you can see it's a six times table, which we call 6n, it's sequence 6n, common difference of plus six.
Sometimes it won't be that simple.
Sometimes the relationship or 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, owe a good thing 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 June.
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 all 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 are four times table, you would've seen that mapped on a number line before.
So when we mapped 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.
I think 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? 7, 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 the 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 do we get on? Any good suggestions? Did you suggest it's 5n minus four? Well done! Why? Well, it's still got common difference of 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: 1, 6, 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.
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.
Welcome back.
I hope you said false, and I hope you justified it with the coefficient of n in the expression.
7n plus three tells us the common difference of the sequence.
If we worked out the first five terms of the sequence, seven lots of one plus three is 10, seven lots of two plus three is 17, etc.
, we'd get this sequence or these terms. So a common difference of positive seven, and it's a translation of positive three from the sequence 7n.
There is 7n mapped against our 7n plus three, still got a common difference of seven, but our translation of positive three is what takes us from the sequence 7n to 7n plus three.
Practise time now.
For question one, I'd like to match the nth term with the sequences in the table.
By sequences in the table, I mean, row by row.
The top row reads 3, 6, 9, 12, 15.
That sequence matches with which nth term from the left-hand side.
You'll notice there's seven rows, therefore seven sequences, but I've given you 10 nth terms. So three of those nth terms have no partner in the table.
Can you spot which three they are? And can you match up the other seven? Pause, give this a go now.
Question two, 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 nth term, please.
Pause, give it a go.
Feedback time.
Matching the nth term with the sequences in the table.
The top one is 3n, it's a three times table.
The one below it: 5, 14, 23, that's a common difference of nine, but it's not the sequence 9n, it's shifted by negative four.
Oh, it's the sequence 9n minus four.
In the third row, 3n plus two.
Notice the shift from 3n to this one.
It's plus two every time from that sequence, hence it's 3n plus 2.
The fourth row, 18 minus 3n.
What was different about this sequence? It was a decreasing sequence: 15, 12, 9, it's going down by three each time, or it's got a common difference of negative three, hence we've got a negative coefficient of n on that occasion.
For the fifth row, that's a sequence 9n plus one, one above your nine times table.
The sixth one, 3n plus 50, 50 above your three times table.
For the last one, 9n minus 22.
Tricky 'cause it starts on negative terms, but it's just a shift of negative 22 from your nine times table.
So these three had no partner.
Some people would have said the sequence: 5, 8, 11, 14, 17, that's 3n plus five 'cause it starts on five and goes up by three each time.
That's a common misconception.
So how do we undo this misconception? Well, we could work out the first three terms of 3n plus five.
Three lots of one plus five, three lots of two plus five, three lots of three plus five, the sequence 3n plus five starts 8, 11, 14, which is not how that row in the table starts, so you can't match that.
9n minus 13.
Some people would have told me that's the bottom row.
It starts on negative 13, common difference of positive nine.
That must be the sequence 9n minus 13.
But it's not.
How do we know? Well, 9n is your nine times table: 9, 18 27.
If we needed to shift it, so the first term wasn't nine, the first term was negative 13.
Shifting by negative 13 won't do it, we need a far bigger shift of negative 22 to get all the way from the sequence 9n down to this one.
So we need a translation of negative 22, not negative 13.
That's why it's not 9n minus 13.
9n minus 13 had no match.
Similar thing for 3n plus 53, people might have told me that belongs there.
It does not, it's not a big enough shift, or it's too big a shift.
You needed a shift of 50 to go from the sequence 3n to the sequence: 53, 56, 59.
So 3n plus 53 had no match.
Question two, 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: 8, 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, they are 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 doesn't behave any different.
Common difference of 0.
8 or positive 0.
8, and the translation from the sequence 0.
8 and by positive 0.
3, so it's 0.
8n plus 0.
3.
For part E, again, no different.
Common difference of positive 3/11, and then it's not our sequence three over 11n, it's always one 1/11 above it.
So that's the sequence, 3/11n plus 1/11.
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 100? Not really.
It'll take me ages, and I might well make an error.
More efficiently, and we'd say this arithmetic sequence is 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 of 150 plus five is 1805.
So what's 150th term? It's 1805.
We like that, that's way more efficient.
We can also work backwards, solving an equation, to find the term's position in a sequence.
So, same sequence.
It's 2585 in the sequence.
And what position is it? I don't wanna write out the rest of that sequence until I hit 2585, and then count how many terms I wrote to get there.
It would take me awhile.
Say it's 12n plus five, the nth term hasn't changed.
Will the nth term ever hit 2585? Well, we set up an equation: 12n plus five.
When does it hit 2585? We add negative five to both sides of that equation, we divide both sides by 12, and equals 215.
Same.
2585 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 2585 is in the sequence, and there's a common difference of positive 12 between terms, we know that 2590 is not in this sequence.
So if we tried to solve this equation, when does our sequence 12n plus five hit 2590? 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 2590 is not in the sequence.
Quick check if 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, you can justify your answer with: If you can solve the equation using 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 4, n equals 79.
5.
Strong 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 is 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.
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 100 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 1002 minus 3n decreasing arithmetic sequence, the coefficient of n is negative.
For the 100th term, it's a 1002 minus three lots of 100 giving us a 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, a 48.
Is 500 in this sequence? 500 is in this sequence, it's the 1230th 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 a miracle 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, i.
e.
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 8, 10, 12, that's the nth 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 8, 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.
Onto 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: 5, 8, 11, 14, 17.
And Alex says, "We were both wrong!" So let's look at the sequence in a different way.
If we see a building: 5, 8, 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 did that five, that eight, that 11 constructed? Well, five is just five, it'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 it 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 any 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. All right, I'm gonna do two questions now, and then ask you to repeat the same scale.
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.
Well, 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 could express this sequence as 12 plus five brackets 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.
All right, in the first five terms of the sequence where A equals negative eight and D equals five, we start on negative eight, and we have a common difference of positive five, so it goes negative eight, negative 3, 2, 7, 12.
Can you write this generalised form for the sequence: 1, 8, 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: 1, 8, 15, 22, 29 has an nth term of 7n minus six.
If I ask you to write it in this form, you'd 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 and the same thing, just generalised slightly differently.
Let's check if you've got that now.
I'd later 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: 8, 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 proved 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 it 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 seven n minus 3/7, or 1/7 plus 4/7 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, to five three by seven, n equals 53, it's the 53rd position.
That's the end of the lesson now.
In summary, using nth term's expressions, we can generalise arithmetic sequences in a number of ways.
Another way we've learned to generalise nth term 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.
I 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.
