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Hello, Mr. Robson here.
Welcome to Maths.
Well done for joining me today, always a good choice.
We're reasoning in today's lesson, which is a super important skill in maths.
So let's find out what it's all about.
Our learning outcome is I'll be able to reason whether a value appears in a given sequence.
Lots of interesting words we're gonna hear throughout today's learning.
An arithmetic or linear sequence is a sequence where the difference between successive terms is a constant.
For example, 20, 30, 40, 50 is an arithmetic sequence, whereas 20, 40, 80, 160 is not.
The nth term of a sequence is the position of a term in a sequence where N stands for the term number.
For example, if N equals 10, this means the 10th term in the sequence.
Two parts to our learning today.
Let's begin by using the nth term to justify.
In some cases it is easy to spot if a number is in a given sequence.
For example, is 80 in the arithmetic sequence starting -10, -1, 8, 17, 26? In such a case, it's easy enough to count on, there's a start of our sequence.
We notice a common additive difference from term to term of positive nine, and we keep counting on in positive nines, and we see, yes, 80 is indeed in the sequence.
In some cases though, counting on is not an appropriate method.
Is 309 in the arithmetic sequence starting 5, 9, 13, 17, 21? Let's have a look at counting on.
There's the first five terms, the first 10 terms, 15 terms, and on and on and on.
Hmm, this method doesn't feel very efficient.
It is neither efficient nor is it simple.
Additionally, the chance of me reaching 309 without making an error is pretty low.
In such a case, we can find and use an nth term rule.
Is 309 in the arithmetic sequence starting 5, 9, 13, 17, 21? Let's answer the same problem using an nth term rule.
In order to find an nth term rule, we need two things.
We need the common difference of the arithmetic sequence and the translation.
So in 5, 9, 13, 17, 21, we have a common additive difference of positive four, that tells us it's a 4n sequence.
The sequence 4n goes 4, 8, 12, 16, 20 at its start.
What is the shift? The translation between successive terms. Between the first term of 4n and the first term of our sequence, we got a shift of plus one.
Between the second terms, a shift of plus one, between the third terms, the fourth terms, the fifth terms, and all terms has a shift of positive one from the sequence 4n.
So what do you think we're going to call our sequence? Well done, 4n + 1.
In 4n + 1, we found our nth term rule, but it's sensible to test that this is the nth term rule that it works.
We know the fifth term of this sequence is five.
When n equals five, the term is 21.
Let's substitute that into our position to term rule and just check that we're right when we say this sequence is 4n + 1.
Substitute in N equals five and T equals 21, and we get an equation that balances.
Now that we know our nth term rule works, we can use it to justify if 309 is in the sequence.
In order to do this, we need to form and solve an equation, there's our position to term rule again.
And we're looking for the term 309, so we substitute 309 into the T term position.
From here we'll resolve.
Let's add a negative one to both sides, let's divide both sides by four, and we find that n equals 77.
What does that mean? It means that 309 is in the sequence and it is the 77th term.
Quick check you've got all this so far.
If we're faced with the question is 1005 in the arithmetic sequence starting 5, 13, 21, 29, 37? Can you explain why counting on would not be suitable for this question and can you suggest a better method? Pause, make that explanation and suggestion now.
Welcome back.
Hopefully you said something along the lines of counting on would be inefficient and prone to error.
Can you imagine how long it would take us to get to 1005? You should also have said something along the lines of a more suitable method would be to find an nth term rule and form and solve an equation.
So let's set about doing that.
Let's start to find an nth term rule.
What is the common difference in the arithmetic sequence? That's the same sequence you saw a moment ago.
What I'd like to know, which one of those three options is the common difference? Pause, see if you can spot it.
Welcome back.
Of course you can spot it, that's a positive eight from term to term, that's our common difference in this aromatic sequence.
Same sequence, next question though.
What is the translation of the arithmetic sequence from 8n? If we know there's a common difference of positive eight, it's an 8n sequence, but I'd like you to compare the successive terms of 8n and our sequence to tell me what that translation that shift is.
Pause and take a pick from those three options.
Welcome back, hopefully you said option C.
When we compare prospective terms, the two sequences, we can see that our sequence is a shift of negative three from 8n.
Same sequence, now that we're armed with all that information, we're ready to answer the key question, what is the nth term of this arithmetic sequence? Three options to choose from.
Pause, take your pick.
Welcome back.
Hopefully you said option A.
There's a common difference of positive eight and a translation of negative three from 8n, hence this sequence is 8n - 3.
Now that we know that nth term rule, I'd like you to use it to justify if 1005 is in the arithmetic sequence starting 5, 13, 21, 29, 37.
My hint for this is form and solve an equation.
Pause and give it a go.
Welcome back.
Hopefully you started by writing out our position to term rule and then substituted in the term we are seeking, 1005 goes in the T position.
From here, we can start to resolve by adding positive three to both sides, dividing both sides by eight and finding that n equals 126.
Now I do hope you didn't stop there, you went on to say, yes, 1005 is in the sequence, it is the 126th term.
It's useful to include that sentence to interpret that answer, n equals 126.
This method can provide other useful information.
Jacob and Aisha are faced with a problem, is 234 in the arithmetic sequence starting -15, -9, -3, 3, 9? Jacob says, "I'm great at algebra, I've solved this one," and that's Jacob's solution.
Aisha says, "Your working is correct, but n equals 42.
5 is not a valid result." Can you see what Aisha means? Pause and have a think about this.
Welcome back.
I wonder what you said n equals 42.
5.
Well, n is the position in the sequence.
What's the 42nd term in the sequence? Well, when N equals 42, the term is 231.
What's the 43rd term? When N equals 43, the term is 237.
Upon realising this, Jacob says, "Of course, 231 is the 42nd term and 237 is the 43rd term.
There is no term in between, so 234 can't be in that sequence." Aisha says, "That's it.
If a term is in the sequence, then N has to be a positive integer." So look out for that when you're forming and solving equations to justify if a term is in a sequence, n has to be a positive integer.
I'm gonna do an example just like the one Jacob and Aisha did, then ask you to repeat that same skill.
I'm gonna show that 168 is not in the arithmetic sequence starting 21, 26, 31, 36, 41.
I've got a common difference of positive five and a translation of positive 16 from 5n.
So my position to term rule is T equals 5n + 16, I'm now looking for 168.
Is that a term? I can add -16 to both sides of that equation, divide both sides by five, n equals 30.
4.
I don't stop there.
I write n is not a positive integer, therefore 168 is not in the sequence, we must include that sentence in our answer.
Your turn has a problem just like that one, pause and give it a go.
Welcome back.
Hopefully you spotted a common difference of positive four and a translation of negative three from 4n, so your position to terminal was 4n - 3.
From there, seeking the term 216, and you solve that equation like so, and you find that N equals 54.
75.
What is significant about that result? Well, I hope you wrote n is not a positive integer, therefore 216 is not in the sequence.
Well done.
Practise time now.
Question one, which of the below numbers are terms in the arithmetic sequence starting 11, 19, 27, 35, 43? In each case, write a sentence to justify your answer.
Pause, do these four problems now.
For question two, an arithmetic sequence starts 17, 14, 11, 8, 5.
Part A, I'd like you to show that negative 289 is in the sequence.
Part B, I'd like you to show that negative 131 is not in the sequence.
Pause and do those problems now.
Question three, an arithmetic sequence starts 17, 17.
4, 17.
8, 18.
2, 18.
6, 19.
Jun says 42.
2 is not an integer, therefore it cannot be a term in the sequence.
What I'd like you to do is show that Jun is incorrect and write a sentence correcting his misunderstanding.
When I say show that Jun is incorrect, you're going to need to use some mathematics to show that.
You'll then add to your mathematics a sentence, explain it to Jun, what's gone wrong in his thinking.
Pause and do that now.
Feedback time.
Let's see how we did it.
Question one was spotting which terms are in this arithmetic sequence and following our answers with a sentence to justify them.
A, 83, is that in this sequence? Lots of ways we could have tackled this, but 83 wasn't that higher term value.
So by counting on in eights, we can see that 83 is in the sequence.
You might have done that another way as long as you justify that 83 is in the sequence, your answer is fine.
For 283, there's a little bit too high to count on, so we'll find a position to term rule 8n + 3.
Is 283 a term? We can solve that equation like so, n equals 35.
We don't stop there, we write a sentence.
Yes, 283 is in the sequence, it is the 35th term.
Part C, 511, again, way too far in the sequence to be counting on, so we'll use our nth term rule, we'll solve that equation like so.
N equals 63.
5, it's not enough to stop there, we must say, n is not a positive integer, therefore 511 is not in the sequence.
Part D, 2651.
Definitely not counting on for that one.
We'll solve this equation, n equals 331, yes, 2651 is in the sequence, it is the 331st term.
Question two, part A, I asked you to show that negative 289 is in the sequence.
For that sequence, we've got common difference of negative three and a translation of positive 20 from -3n.
So opposition's term rule is T equals 20 minus 3n.
Let's look for that term negative 289, solve our equation, n equals 103.
We say that n is a positive integer, therefore 289 is in the sequence, in fact it's the 103rd term.
For part B, we were showing that value is not in the sequence.
Again, we're setting up and solving an equation, and when we solve this one, we find that N is not a positive integer, therefore negative 131 is not in the sequence.
Question three, we were showing that Jun is incorrect and writing a sentence to correct his misunderstanding.
In the sequence, we've got a common difference of 0.
4 and a translation of 16.
6 or positive 16.
6 from 0.
4n.
So our position to term rule, our nth term rule is 0.
4n + 16.
6.
We can show that 42.
2 is indeed in the sequence.
We substitute it into opposition to term rule, solve our equation, we find that n equals 64.
So you might have written, n is a positive integer, therefore 42.
2 is in the sequence.
Whilst n must be a positive integer, T the term value can be a non integer.
Onto the second half of our learning now, where we're gonna look at multiple methods for reasoning.
I love it in maths when we come across multiple ways to solve the same problem.
It is very powerful to be able to solve the same problem in multiple different ways.
Reasoning, if a term is in a given sequence, may be more efficient and less work for certain sequences.
Is 191 in the arithmetic sequence starting 20, 28, 36, 44? I've got Jun, Isha, and Jacob working on this problem.
Jun says, "I'm going to count on," and proceeds to count on and concludes, "No, it is not." That's the sequence, June did not hit the term 191, so it's not.
Well done, Jun.
Aisha said, "I shall use the nth term" and found an nth term rule, solved an equation, and said, "No, it is not." Well done, Aisha.
Jacob's got a different approach.
Jacob says, "I will use reasoning," and then proceeds to say "The first term is even and there is an even common difference, therefore all terms in the sequence will be even.
191 is odd, so cannot be the sequence." All three pupils reached the correct answer.
What do you think of each method? Which one do you think you would've used? Pause.
Have a think about that.
Welcome back.
Let's have a look at each of these three methods.
When looking at Jun's method, I hope you said something along the lines of counting on was really inefficient in this case, and when counting on so far we may well make a mistake and reach the wrong result.
So was there a better method? Of course, there was.
Aisha's method was way better.
Using the nth term was more efficient, way more efficient, and it enabled us to see not only is 191 not in the sequence, but we can say that 191 is somewhere between the 22nd and 23rd terms. That's useful information.
Great method, Aisha.
How about Jacob's method? What did you make of this? In this case, Jacob's reasoning was a simple and efficient way to show that 191 was not in the sequence.
If you could justify that the sequence will only ever be populated by even numbers, saying that 191 is an odd number is wonderfully simple and efficient, isn't it? And we like efficiency.
Great, Jacob, you've got that.
Which method would you use to solve this problem? Is 6728 in the arithmetic sequence starting 1, 12, 25, 37? Would you be counting on for that one? Would you use an nth term, form and solve an equation? Or could you use reasoning? Pause, tell a person next to you or say it aloud to me at screen.
See you in a moment.
Welcome back.
Hopefully, we said we're not counting on.
This will be very inefficient and will be definitely prone to error.
Trying to count on in twelves that far, not sure I could do that accurately.
The nth term, while it would be efficient, it would require us finding an nth term rule and then forming and solving an equation and doing all those things correctly.
C is a winner.
It's so efficient and so simple.
Saying something along the lines of the first terms are odd, an even difference means all terms in this sequence will be odd too, 6728 is even therefore will not be in this sequence.
We're able to solve this problem with just a couple of sentences of reasoning.
Jun and Jacob are looking at another problem.
Is 19281 in the arithmetic sequence starting 22, 30, 38, 46? Jun says, I like your method Jacob, I think I will use it for this problem.
Remember Jacob's method was reasoning.
Jacob says, "Great, give it a go" and Jun says, "No, because odd." What do you think? Would you accept Jun's answer if you were the teacher? Pause, tell the person next to you or say it aloud to meet at screen.
Welcome back.
Are you accepting that answer? Jacob certainly isn't.
Jacob says, "That is not enough, Jun.
When reasoning you need to use full sentences and be sure to include all the necessary details that justify your decision." This is good feedback by Jacob, inspires Jun to have another go at the problem, and Jun says, "The first four terms are even and there is an even common difference, therefore all terms in the sequence will be even.
19281 is odd, so cannot be in this sequence." Jacob says, "That is much better, Jun." I'm inclined to agree, that's a fabulous second attempt, Jun.
Quick check you've got that.
I'd like you to improve Alex's answer to this problem.
Pause, have a read and then improve that answer.
Welcome back.
You might have said the first terms are odd and even difference means all terms in this sequence will be odd too.
374 is even, therefore will not be in this sequence.
If you wrote something like that to improve Alex's answer, that's super, importantly be aware, Alex's answer did not have enough detail to justify whether 374 was in that sequence or not.
Our reasoning will not always be a comment on whether terms are odd or even.
For example, is 78365 in the arithmetics sequence starting 53, 73, 93, 113, 133? Can you see how we might reason this one and do it differently to the odd even explanations we saw earlier.
Pause, have a quick think.
Welcome back and well done, I can tell that you are itching to tell me the answer.
The common difference is positive 20.
Therefore, all terms in this sequence will end with a ones digit of three.
The ones digit in 78365 is five, therefore it will not be in this sequence.
That would be a lovely piece of reasoning.
Let's see if you can reason this one.
Is 617 in the arithmetic sequence starting 24, 29, 34, 39? Pause, reason this one out.
Welcome back, let's have a look at your reasoning.
You might have said there is a common difference of positive five between successive terms, so they will always end with a ones digit of four or nine in this sequence.
The ones digit in 617 is seven, therefore it is not in this sequence.
If you wrote something like that, that's a marvellous piece of reasoning.
Jun, Aisha, and Jacob are now looking at this problem.
Is 44 in the arithmetic sequence starting -26, -19, -12, -5, 2? Jun says, "I'm gonna count on." Aisha says, "I shall use the nth term." Jacob says, "I will use reasoning." Which method would you use in this case? Pause, tell the person next to you or me aloud on screen.
Welcome back.
There's Jun's counting on, there's Aisha's nth term.
When we come to reasoning, it's tricky.
There's no obvious pattern in this sequence on which to base simple reasoning.
So in this case, counting on was the most efficient and simple method.
In maths, we must be fluent in our use of methods, and in each case ask ourselves, what is the most simple way we can solve this problem? Truth be told, we, mathematicians, are a little bit lazy.
We're always looking for the laziest way to solve problems. Quick check you've got that.
Which method would you use for each of these problems? In each case, justify your selection.
Pause and have a think about those three.
See you in a moment.
Welcome back.
Hopefully you opt for A, you said reasoning.
There's an obvious pattern in the term values making reasoning as a method really efficient and simple to say that negative 662 would not be in that sequence.
For B, nth term, there's no obvious pattern with which to reason, so we can't use that method.
It's too far to count on, so we can't use that method, but we know we can use an nth term efficiently and accurately.
For C, counting on would work, there's no obvious pattern with which to reason.
We don't need to do the nth term and form and solvent equation because it's close enough as a value to quickly and simply count on.
Practise time now.
Question one, I'd like you to solve this problem using three different methods.
You're gonna do that same problem three times.
Once you're finished, you're gonna write a sentence to comment on the efficiency of each method.
Pause and do that now.
Question two.
For this one, you're gonna select one method and use it to solve this problem.
I'm not asking you to do this problem three times with all three methods, I want you to do it once with the best method you think for this example.
Once you've solved the problem using your preferred method, I'd like you to, for the two methods you did not use, write a sentence explaining why you didn't choose those methods.
Pause and do this now.
Question three.
I'd like to explain what is wrong with Jun's reasoning in this problem.
There's the problem.
You can read Jun's reasoning and I'd like at least one sentence explaining what is wrong.
Pause and do that now.
Question four.
Initially this one looks tricky, but trust me you can do this.
Be fearless, give a go.
355 is the 60th term in an asthmatic sequence, 361 is the 61st term.
That's all the information that we're given, but that is all you need in order to solve these problems. In each case, be sure to select an efficient and accurate method.
Pause, be fearless, get stuck in to these three problems. Welcome back.
Feedback time.
Question one, I asked you to solve this problem using three different methods and comment on their efficiency.
For counting on it would've looked like so, and you would've said, no, 351 is not in the sequence.
Counting on was very time consuming and had high potential for error is the comment you might have made on efficiency.
For nth term, there's our nth term rule, there's the equation formed and solved, with a statement n is not a positive integer, therefore 351 is not in the sequence.
In terms of efficiency, you might have said using the nth term was efficient and accurate.
For reasoning, you might have said there's a common difference of positive 15 between successive terms. They will always end with ones digit of two or seven, the ones digit and 351 is one, therefore it is not in this sequence.
And your comment on efficiency, you should have said using reasoning was easily the most efficient method, in this case end result is accurate.
For question two, I asked you to select one method, the best method, hopefully you selected the nth term.
There's our position to term rule or nth term rule.
Here we are forming and solving an equation to find that n equals 811.
714 and so on, with a subsequent statement, n is not a positive integer, therefore 570.
65 is not in the sequence.
Why we wouldn't have used counting on and I asked you for a sentence explaining why you didn't use the other two methods.
You could have written, counting on would've been very time consuming as well as being prone to error.
For reasoning, you might have said, the noticeable characteristic of the known terms is a five in the hundredth position.
However, 570.
65 also has a five in that position.
We can neither rule it out nor confirm it is in the sequence with reasoning.
Question three, I asked you to explain what's wrong with Jun's reasoning.
You might have written,, the common difference is positive four.
So whilst all terms will be even, this sequence will only include every second even number greater than or equal to 10, we therefore cannot confirm without further reasoning whether 876 is or is not in the sequence.
You might have written further reasoning, you may have formed and solved an equation to show that n is not a positive integer value, therefore 876 is not in the sequence.
Question four, this was a lovely problem and a tricky looking one, but it's achievable.
Is four in that sequence? You might have resolved this one with reasoning.
Our two known terms are odd and there's an even common difference, therefore all terms in the sequence will be odd.
Four is even, therefore it cannot be in the sequence.
Is 393 in the sequence? Counting on was really efficient and simple.
In this case, 355 is the 60th term.
361 is the 61st term.
If we keep counting on, we can see that, no, 393 is not in this sequence.
For C, 1393.
Well, counting on and reasoning are gonna be inefficient and ineffective in this case, we need the nth term.
We know there's a common difference of six, and we know our position to term rule will look something like that with B being a constant.
Let's substitute in something we know, the 60th term is 355, we can find that constant negative five.
So this sequence is 6n - 5.
Let's form and solve an equation.
And we find that n is a positive integer of value, therefore 1393 is in the sequence, it's the 233rd term.
That's the end of lesson now, sadly, but what have we learned? We've learned that we can sometimes reason whether a value appears in a given arithmetic sequence.
We've learned that there are multiple methods that can be used to justify whether a value is in a given sequence and selecting the most appropriate method reduces the time and work required, and we do like efficiency as mathematicians, don't we? Well, I hope you've enjoyed this lesson as much as I have.
I look forward to seeing you again soon for more mathematics.
Goodbye for now.