Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Good of you to join me today.

Today we're looking at conditions in arithmetic sequences.

Spotting patterns in sequences and working with conditions in sequences is a really important piece of mathematics.

So you are going to enjoy this lesson.

Let's get started.

Our learning outcome is that we'll be able to find the first value, bigger or smaller than a given value in the sequence.

Lots of useful language we're gonna hear throughout our lesson today.

Let's have a closer look at some of it.

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.

Nth term.

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.

Look out for all that language throughout our learning today.

To pass this lesson let's begin with trial and improvement.

It can be useful to know when the term values in a sequence reach or surpass a certain value and there's more than one way to work this out.

For example, what is the first term in the following arithmetic sequence greater than 100? We could use a simple counting on method.

If we continue that sequence and continue that sequence, we reach 101, and that's the first value greater than 100, but which term is it? Is it the 20th, the 30th, the 40th term? I'm not sure.

101 is the 32nd term, but this method was time consuming and prone to error in counting.

A more efficient method is to find the nth term and use that to find the term number.

The same problem, we're looking for the first term over a hundred in that arithmetic sequence, but we can find an nth term rule.

We know the sequence has a common difference, a positive three, so it's a 3n sequence.

When we compare the respective terms of the sequence 3n and our sequence, we find it's a translation or a shift of positive 5.

So our sequence is 3n plus 5.

That's our nth term rule, our position term rule for our arithmetic sequence.

Now that we have an nth term rule, we can use trial and improvement to find the first term over a hundred.

We can find the 10th term by substituting in n equals 10 into our nth term rule.

We find that the 10th term is 35.

What's the 20th term? Well, when n equals 20, the term is three lots of 20 plus 5, it's 65.

The 30th term, that's 95.

From here we know we've got a common difference of three, so we can count on ninety five, ninety eight, one hundred and one.

Wonderful, we found it.

It's the 32nd term.

That's the first one over a hundred.

We'll finish by fully justifying our answer.

Mathematics, it's really important that we're able to justify an answer.

We would say the 32nd term is 101.

The 31st term is 98, therefore the 32nd term is the first term greater than 100.

Quick check you've got that.

What is the first term in the following arithmetic sequence greater than 250.

That's the problem we want to solve.

The first thing we need to do is find the nth term of our sequence.

Our sequence is 15, 24, 33, 42, 51.

Which of those three nth terms represents that sequence? Pause, and take your pick.

Welcome back.

Hopefully you said it's option A, 9n plus 6.

Why? Well, we've got common difference of positive nine and when we compare it to the sequence 9n, we've got a shift or translation of positive six.

When we've got a common difference of positive nine and a translation of positive 6 from 9n, our sequence is 9n plus 6.

Now that we know the nth term is 9n plus 6, I'd like you to use that nth term rule to find the 10th term, the 20th term, the 30th term of the sequence.

Pause and do that now.

Welcome back.

Hopefully you found that the 10th term is 96, the 20th term is 186 and the 30th term is 276.

And I do hope that your method looks a little bit like mine.

Now that we know this, we're ready for the final step.

with that information, the values of the 10th, 20th, and 30th terms we can finally answer the question.

Once you've answered the question, What's the first term greater than 250, I'd like you to write a sentence to fully justify your answer.

You can pause and do this now.

Welcome back.

Hopefully you started with that 30th term, 276 and then counted backwards in steps of nine to find the 29th term, the 28th term, the 27th term, which wonderfully fell below 250 when counting backwards.

So we can fully justify our answer by saying the 28th term is 258.

The 27th term is 249, therefore the 28th term is the first term greater than 250.

Let's look at this problem now.

What's the first term in the following arithmetic sequence greater than 1,000.

For this example, it will be inefficient to find the 10th, 20th, 30th terms and so on.

The nth term is 5n minus 13.

That's a common difference of positive 5 between the terms and a shift of negative 13 from the sequence 5n.

When we substitute n equals 10 into our nth term rule, we find the term is 37, the 20th term is 87, the 30th term is 137.

Thirty terms in and we're still nowhere near a thousand.

So how do we become a little more efficient? In this case, we use estimation to decide a sensible starting point to begin our trial and improvement from.

The nth term is 5n minus 13, and we know that five lots of 200 is a thousand, so let's start by finding the 200th term.

When n equals 200, the term is 987.

987 is really close to our target of 1,000.

From here, we can quickly count on in fives.

We find the 201st term, 202nd term, 203rd term.

We can finish by fully justifying an answer.

We can say the 203rd term is 1,002.

The 202nd term is 997, therefore the 203rd term is the first term greater than 1,000.

Quick check you can operate efficiently like that.

Which term number is the most sensible one to use as a starting point for this problem? You can read that problem, look at that sequence then you tell me that you can start at the 50th term, the 100th term or the 125th.

See you in a moment so you can compare your answer to mine.

Welcome back.

Hopefully you said it was option C when n equals 125 or the 125th term.

The nth term of the sequence is 4n plus 7 and we know that four lots of 125 makes 500.

So 125th term will get us really close to our target of 500.

What I'd like you to do now for the exact same problem is start at n equals 125.

Once you've found that 125th term, you can solve this problem.

Don't forget to write a sentence to fully justify your answer.

Pause and do this now.

Welcome back.

Let's see what we found.

Hopefully you found that the 125th term is 507.

From there we can count back in fours to 503 to 499 and then we can fully justify our answer.

124th term is 503.

The 123rd term is 499.

Therefore the 123rd term is the last term lower than 500.

It's important that we fully justify our answers.

Izzy and Sam are looking at this problem.

Izzy finds that position to term rule finds a 200th term, the 300th term and then the 250th term and then concludes our answer is 1,705 is the first term greater than 1,700 in this sequence.

Sam says that is not a full enough answer, Izzy.

Can you see what Sam is getting at? Pause.

Have a debate with a person next to you or a good think to yourself.

What does Sam mean? Welcome back.

Hope you're on Sam's side in this argument.

Whilst you may know the common difference of seven means the term before the 250th will be smaller than 1,700, you still need to communicate this.

You need to calculate the 249th term and show that it is smaller than 1,700 and also show that the 250th term is greater than 1,700.

When n equals 249, the term is 1,698.

Izzy's back with an improved answer.

So a better answer is the 249th term is 1,698, the 250th term is 1,705, therefore the 250th term is the first term greater than 1,700.

Izzy, that is a fabulous fully justified answer.

Quick check you can do as well as Izzy just did.

Here's a problem that Sofia was working on and she says 645 is the last term smaller than 650 in this sequence.

I'd like you to improve Sofia's answer for this problem.

Pause and write a sentence improving that answer.

Welcome back.

Sofia should have calculated the 111th term and I hope that you did too.

The 110th term is 645.

The 111th term is 651 and seeing that they are either side of 650 is crucial.

Sofia then could have written, the 110th term is 645, the 111th term is 651, therefore the 110th term is the last term smaller than 650.

This mathematics can also be very useful in a practical context.

For example, company A make 200,000 units of product A.

They sell 9,000 units of product A each week.

When will they run out of stock? We can make a position to term rule for this scenario.

The term is 200,000 minus 9,000n.

Now that term is gonna be the amount of stock they've got left and n in this case will be the number of weeks, so the 10th term is 110,000, or after 10 weeks they've got 110,000 units left.

The 20th term is 20,000.

After 20 weeks, there's 20,000 units left.

From here we can count on as the 20th term, 21st term, 22nd term, the 23rd term, they will still have stock of product A after 22 weeks.

They will run out of stock during the 23rd week.

Quick check you can do a similar problem.

Company B makes 15,000 units of product B.

They sell 800 units of product B each week.

When will they run out of stock? Pause.

Give this problem a go.

Welcome back.

A position to term rule or a week's to stock rule.

The 10th term will be 7,000, the 20th term negative 1,000, so let's go to that 20th term and count backwards in weeks.

We can say they will have stock of product B After 18 weeks.

They will run out of stock during the 19th week.

Practise time now.

There's four problems for you to have a go at.

They're quite self-explanatory, so I'll leave you to read them rather than me talk you through them all.

The only thing I do have to say is make sure you include a sentence that fully justifies each answer.

Pause and give these four problems a go now.

Question five, it's practical context.

Company C make 24,000 units of product C.

They sell 775 units each week.

When will they run out of stock? Question six is another practical context but slightly different one.

Izzy opens a savings account with 150 pounds.

She puts an extra 12 pound into the account each week.

When will her savings first exceed 400 pounds? Don't forget both those problems to include a sentence to justify your answer.

Pause, give these a go now.

Feedback time now.

Question one first term in that arithmetic sequence greater than 100.

The nth term is 2n plus 3, 50th term is 103.

From there we can count backwards and you might have written something along the lines of the 48th term is 99, the 49th term is 101, therefore the 49th term is the first term greater than 100.

To question two, the nth term was 9n minus 14.

The 100th term was 886.

Let's move a little further.

The 110th term, 976.

From there we can count on and we can conclude the 112th term is 994.

The 113th term is 1,003.

Therefore the 113th term is the first term greater than 1,000.

Question three, the nth term is 4n minus 3.

The 150th term was 597.

Count on from there and justify with, the 150th term is 597, 151st term, 601, therefore the 150th term is the last term smaller than 600.

For four the nth term 65 minus 4n.

The 75th term is negative 235.

The 80th, negative 255.

Counting backwards from that 80th term and then writing the 78th term is negative 247.

The 79th term is negative 251.

Therefore the 79th term is the first term smaller than negative 250.

Question five, the practical context.

We could have written a position to term rule or a week to stock rule of 24,000 minus 775n.

In the 30th week we'll have 750 units left.

Count on from there and we can conclude they will still have stock of product C after 30 weeks.

They will run out of stock during the 31st week.

For question six, a position to term rule would a read T equals 150 plus 12n.

This is a week's to savings rule in the context of Izzy's savings account.

After 20 weeks, 390 pound and we can count on from there and conclude Izzy will have 390 pound of savings after 20 weeks and reach 402 pounds of savings after 21 weeks.

She exceeds 400 pounds after 21 weeks.

Onto the second half of our learning now.

We're going to be reasoning algebraically.

We do not have to use trial and error to solve such problems. We can reason algebraically using the nth term rule, forming and solving an equation.

For example, what's the first term in the following arithmetic sequence greater than 2,000.

We know a position to term rule will be 8n minus 26.

From there we can set up an equation.

We want to know when the sequence is greater than 2,000, so we put 2,000 in the term position.

From there we can solve that equation and we'll find that n equals 253.

25.

What we need to do next is interpret this answer.

Andeep says it's the 253.

25th term.

Laura says 253.

25 rounds down.

It's the 253rd term.

Lucas says it's the 254th term.

Who do you agree with? Pause and have a think.

Welcome back.

We can immediately rule out Andeep's answer.

There's no 253.

25th term.

There is a 253rd term and a 254th, but there's no term in between, so Andeep's answer's gone.

It's down to Laura and Lucas now.

One of them rounded up, one of them rounded down.

Who is right? Pause.

Have another think.

Welcome back.

It was Lucas's answer that we were looking for.

Rounding up from 253.

25 give us the answer we were looking for.

You know that the nth term is 8n minus 26.

It's an increasing sequence.

Therefore Lucas rounding up gives us the correct answer.

To fully justify we can calculate the 253rd and the 254th terms. There's a 253rd term, the 254th term, and we can say the 253rd term is 1,998.

The 254th is 2006 therefore the 254th term is the first term greater than 2,000.

Quick check you can do that.

What's the first term in the following arithmetic sequence greater than 900.

I'd like you to find the nth term and use this to justify your answer.

Pause, give this a go now.

Welcome back.

Hopefully you found a position to term rule 12n minus 5.

We then substitute 900 into the term position in the equation.

From there we can solve.

We find that n equals 75.

416 and so on.

This is an increasing sequence, therefore we round up to the 76th term, but we need to show that this is the case.

We'll find the 75th term and the 76th term and we'll state the 75th term is 895.

The 76th term is 907.

Therefore, the 76th term is the first term greater than 900.

Crucially, you should have identified that when we put T equals 900 into our equation, n equals 75.

416.

From there you absolutely want to work out the 75th term, the 76th term, and then that sentence at the bottom to fully justify our answer is an important one to include.

We won't always be rounding up to the next term after solving our equation.

Let's contrast these two problems. In the first problem we've got an nth term 244 minus 3n.

We're looking for when the sequence is below 50, so I'll form that equation and solve.

n equals 64.

666 and so on.

We find the 64th term, the 65th term, and we round up to the 65th term in this case.

When is that sequence below 50 when it reaches the 65th term? We had to round up from 64.

666.

Let's have a look at this second problem.

You have the term 244 minus 3n, but we're looking for the last positive term, I.

e, the last one above zero, so we put zero into our equation and we solve and we find that N equals 81.

333, so we just round up here, yeah? We're gonna be saying that the 82nd term is the one we're looking for.

No, be careful.

Find the 81st term, find the 82nd term.

Oh look, we round it down to the 81st term in this case.

So why in the first one did we round up from 64.

666 to find the 65th term, whereas we rounded down from 81.

333 to declare the answer, the 81st term in the second problem? What was the difference? It's useful to think of these two problems in this way.

In the first case, we wanted to know the value beyond a threshold, so we rounded up.

We wanted to know when the sequence had gone beyond 50.

In the second problem, we wanted to know the value before a threshold.

We wanted to know the last value before this sequence passed zero.

That's why we ended up rounding down.

If you can see the difference between these two problems, you have cracked this topic.

Quick check you've got that.

What is the last term in the following arithmetic sequence below 100.

Find the nth term and use this to justify your answer.

Pause, give this a go.

Welcome back.

We have an nth term of 2n minus 41.

We can form that equation and when we solve, we find that N equals 70.

5.

From here we find the 70th term, the 71st term.

In this case, we wanted to know the value before a threshold, so we rounded n down and the 70th term was our answer.

When problems look familiar, we can make mistakes.

Now mistakes are welcome in maths, they're part of the learning process, but sometimes we'll make some very avoidable mistakes, so let's see if you can spot what's going on here and then avoid mistakes like these in the future.

Let's look at this problem.

What's the first value greater than 200 in the following sequence.

I know what you would've done here.

You'd have formed and solved that equation.

And then with that answer N equals 48.

25, you'd have found the 48th and the 49th term.

Then you know the solution is n equals 48.

25 and in this case we're rounding up, so it's 49.

Can you see a problem with that answer? If I leave that as my answer, is there an error there, or am I right? Pause.

Have a conversation with a person next to you, or a good think to yourself.

And we'll go through it in a moment.

Welcome back.

Did you spot it? Let's look back at the question.

What's the first value greater than 200? 49 is not greater than 200.

49 cannot be the answer.

We were asked for the value, not the term number.

Whilst the term number might be 49, that's not the term value.

The term value was 203.

We should have stated that as our answer.

We need to be really careful and make sure we're answering the question that is asked.

Quick check you've got that.

How about this problem? What's the last value greater than zero in the following sequence? I've done some of the work for you and I've left you with three possible answers.

Which one are you going for? Pause and take your pick.

Welcome back.

I do hope you went for option C.

Why don't you go for option A? Well, just because N equals 23.

75 that does not make that our answer.

Why not B? The 23rd term is indeed the last one greater than zero, but that's not our answer.

The 23rd term is three.

That's the last value greater than zero and we were asked for the value.

That's why C is the answer here.

Practise time now.

There's three problems there which are quite self-explanatory, so I'll leave you to have a go.

Pause and do it now.

Questions four and five.

Question four, what's the last positive value in that sequence? Question five.

This one's nice.

What's the term number of the last three digit value in the following arithmetic sequence? Pause and enjoy these two problems. Feedback time now.

Let's see how we did.

Question one, we should have formed and solved that equation, found the 72nd and 73rd term numbers and concluded the 72nd term is 445.

The 73rd term is 451.

Therefore, the 73rd term is the first term greater than 450.

Question two, form and solve that equation, find the 167th, 168th terms and conclude the 167th term is 2492, 168th term is 2,507, so the 168th term is the first term greater than 2,500.

Question three, form and solve that equation.

Find the 99th and the 100th terms and conclude the 99th term is negative 49, the 100th is negative 51.

Therefore, the hundredth term is the first term below negative 50.

For four, what's the last positive value in that sequence? We'll form and solve that equation.

Find the 28th term, the 29th term, and then be very careful with this conclusion.

Look at the question, what's the last positive value? The 28th term is one.

The 29th term is negative six, therefore one is the last positive value in the sequence.

Question five asked for a term number.

We'll form and solve that equation.

We'll find the 90th term and the 91st term and we'll conclude the 90th term is 992, the 91st term is 1,003, therefore, the 90th term is the last three digit value.

We're at the end of the lesson now, sadly, but we have learned that we can find the first value bigger or smaller than a given value in a sequence.

This can be done using trial and improvement or by using the nth term to form and solve an equation.

When using the latter, we can identify given the context, when to round up and when to round down to reach a final value of n.

I hope you've even enjoyed this lesson as much as I've enjoyed it and I look forward to seeing you again soon for more mathematics.

Goodbye for now.