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### Arithmetic Sequences & Series

• A sequence is arithmetic if the difference between any two consecutive terms is constant:
 an−an−1 = d,
where d is a constant. We call d the common difference. Every "step" in the sequence has the same change. The difference can be positive or negative, so long as it's always the same.
• The formula for the nth term (general term) of an arithmetic sequence is
 an = a1 + (n−1)d.
• To find the formula for the general term of an arithmetic sequence, we only need to figure out its first term (a1) and the common difference (d).
• We can use the following formula to calculate the value of an arithmetic series. Given any arithmetic sequence a1, a2, a3, …, the sum of the first n terms (the nth partial sum) is
 n 2 ·(a1 + an).
• We can find the sum by only knowing the first term (a1), the last term (an), and the total number of terms (n). [Caution: Be careful to pay attention to how many terms there are in the series. It can be easy to get the value of n confused and accidentally think it is 1 higher or 1 lower than it really is.]

### Arithmetic Sequences & Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:05
• Definition: Arithmetic Sequence 0:47
• Common Difference
• Two Examples
• Form for the nth Term 2:14
• Recursive Relation
• Towards an Arithmetic Series Formula 5:12
• Creating a General Formula 10:09
• General Formula for Arithmetic Series 14:23
• Example 1 15:46
• Example 2 17:37
• Example 3 22:21
• Example 4 24:09
• Example 5 27:14

### Transcription: Arithmetic Sequences & Series

Hi--welcome back to Educator.com.0000

Today, we are going to talk about arithmetic sequences and series.0002

Now that we have an understanding of sequences and series, we are ready to look at specific kinds of sequences.0006

The first that we will consider is an arithmetic sequence, a sequence where we add a constant number each step.0011

We will add some number, and we keep adding the same number every time we go forward a term.0018

Sequences of this form pop up all the time in real life, and we often need to add up their terms.0022

We will explore the creation of a formula for arithmetic series that will allow us to quickly and easily add up those terms.0027

Plus, by the end of this lesson, we will be able to add all of the numbers between 1 and 1000 in less time than it takes to put on a pair of shoes.0033

I think that is pretty cool; we will be able to do something really, really fast that seems like it would take a long time, just like that.0041

All right, a sequence is arithmetic if the difference between any two consecutive terms is constant.0048

We can show that with the recursive relationship an - an - 1 = d.0055

Notice: the nth term minus the n - 1 term (that is, the term one before the nth term) equals d.0061

Some term minus the term before it equals d.0069

And d is just some constant number; we call d the common difference.0073

Here are two examples of arithmetic sequences: they are arithmetic because every step in the sequence has the same change.0079

For example, here, 1 to 4, 4 to 7, 7 to 10, 10 to 13...it is + 3, + 3, + 3, + 3.0087

It is always the same amount that we change; it is always adding a constant number.0099

Over here, it is 5 to 3, 3 to 1, 1 to -1, -1 to -3...it is - 2, - 2, - 2, - 2; and that pattern would continue, as well.0103

In this case, our common difference is -2; we are adding -2 each time; or we can think of it as subtracting 2 each time.0116

The important thing is that it is always the same; the difference can be positive; the difference can be negative;0124

but it always has to be the same for each one--that is what makes it an arithmetic sequence.0128

How can we find the nth term? The definition for an arithmetic sequence is based on a recursive relation.0135

It is based on an = an - 1 + d, that some term is equal to its previous term, with d added to it.0141

So, how can we turn this formula into something for the general term--how can we get a general term formula out of this?0148

Remember: a recursive relation needs an initial term--we have to have some starting place.0154

There is nothing before our starting place to refer back to, so we actually have to be given the initial term directly.0158

Now, we don't know its value yet; so we will just call it a1; we will call it the first term--we will leave it as that.0164

Now, from an = an - 1 + d, we see that a1 relates to later terms.0171

At the most basic level, we have that a2 is equal to a1 + d.0177

The second term is equal to the first term, adding d onto it; that is what it means for it to be an arithmetic sequence.0183

We can take this out and continue looking at later terms.0188

a3 would be equal to a2 + d, based on this recurrence relation.0191

But we just figured out that a2 is equal to a1 + d, so we can swap out for a2:0196

a1 + d, which now gets us a1 + 2d when we add this d to that one there.0203

So, we have a1 + 2d for a3; when we work on a4,0210

well, a4 is going to be a3 + d, the previous term, adding d.0215

But once again, we have just figured out that a1 + 2d is what a3 is.0220

So, we can plug in for a3; we have a1 + 2d, and now we can add that d onto the 2d.0225

So, we end up getting that a1 + 3d is what a4 is equal to.0231

So, we notice that this pattern is going to keep going; we are just going to keep adding on more and more d's to our number.0238

So, a1 = a1; a2 = a1 + d; a3 = a1 + 2d;0245

a4 = a1 + 3d, and this pattern will continue down.0253

We see that the nth term is n - 1 steps from a1.0257

The first term is at 1, and the nth term is at n; so to get from the first term to the n term, we have to go forward n - 1 steps.0262

Since it is n - 1 steps away from a1, we will have added d for each of those steps;0270

so we will have added d that many times, or n - 1 times d.0276

That means that the an term is equal to a1 plus all of those steps times d.0280

So, the nth term, an, equals a1, our first term, plus n - 1 times d.0288

Thus, to find the formula for the general term of an arithmetic sequence, we only need to figure out the first term, a1, and the common difference, d.0294

With those two pieces of information, we automatically have the general term; we automatically have that nth term formula--pretty great.0306

How about if we want an arithmetic series formula?0314

Consider if we were told to add up all of the integers from 1 to 100; how could we find 1 + 2 + 3 + 4 + 5 + 6...+ 98 + 99 + 100?0316

How could we add that whole thing up? Well, we could do it by brute force, where we would just sit down0327

with a piece of paper or a calculator and just punch the whole thing out.0334

We could do it by hand; but that is going to take a long time.0337

And any time that we end up seeing something that is going to take us forever to do, we want to ask ourselves,0341

"Is there a way to be clever--is there an easier way that I can do this that will be able to take away some, or a lot, of the time and effort?"0345

How are we going to do that? We want to look for some sort of pattern that we can exploit.0353

We want to find a pattern that we can exploit--something that will keep happening--0357

something that we can rely on, that will keep us from having to add up all of these numbers,0361

because we can instead use this pattern to give us a deeper insight to what is going on.0365

So, if we look at this for a while, we might start to realize that there is a pattern in the numbers.0368

But that doesn't help us, because that is just adding the numbers.0374

But is there a way that addition itself has a pattern?0376

There is something that we could match up--something that we could create--and this is where we are getting really clever.0379

This is the hard part, where you really have to sit down and think about it for a long time.0383

And hopefully you just end up getting some "lightning bolt" of insight.0386

And hopefully, at some point, we will notice that here is 100; here is 1; if we add them together, we get 101.0390

But not only that--if we had 99 (let's use a new color)...if we use 99 and 2, we get 101.0396

If we add 98 and 3, we get 101; if we keep doing this, working our way in, we are going to keep adding things up to 101, 101, 101...0404

So, if we notice that we can add 1 and 100 to get 101; 2 and 99 to get 101; 3 and 98...we get 101; and so on and so on and so on...0415

what we can do is pair up each number from 1 to 50 with a number from 51 to 100.0425

And we will always be able to make 101 out of it.0430

We start out at the extremes, 1 and 100, and we work in: 2 and 99; 3 and 98; 4 and 97; until we finally make our way to 50 and 51.0433

So, we were able to figure out this pattern; there is something going on.0446

Now, we are finding something; now we have something that we can pull into a formula that will make this really easy.0449

With this realization in mind, let's look for an easy way to pair up the numbers.0455

The first thing: it is nice to give names to things in algebra--it lets us work with them more easily.0458

So, let us have s denote the sum of 1 to 100; so s is equal to 1 + 2 + 3...+ 99 + 100; it is all of those numbers added up together.0462

Now, notice: we can rewrite the order of those numbers, since order of addition doesn't matter.0472

1 + 2 + ... + 99 + 100, here, is the same thing as 100 + 99 + ... + 2 + 1.0476

We can swap the order, and we still have the same value in the end.0484

That is one of the nice things about the real numbers: order of addition doesn't matter.0489

Furthermore, we can add two equations together--that is elimination.0492

Remember: when we worked on systems of linear equations, if you have an equation, you can just add it0497

to another equation, because they are both working equations.0501

You can add the left sides and the right sides, and you know that everything works out; there is nothing wrong with doing that.0503

What we have is: we can add the top equation there and the bottom equation, the normal order and the reversed order; we can add them together.0508

What do we end up getting? Well, here we have a hundred and one, so we get 101; here we have 99 and 2, so we get 101;0516

here we have 2 and 99, so we get 101; here we have 1 and 100, so we get 101.0523

And we know that we are going to end up having 101 show up for every one of the values inside of here, as well.0527

How many terms are there total? Well, we had 1, 2, 3...99, 100; so we had 100 terms, left to right.0532

So, if we had that many terms total, well, even after we add them up, and each one of them becomes 101, then we have 100 terms total.0544

We have 100 terms on the right side; so if we have the same number appearing 100 times, we can just condense that with multiplication.0553

We can condense all of that addition with multiplication: 3 + 3 + 3 + 3 is the same thing as 3 times 4, 4 times 3.0561

So, if we have 101 appearing 100 times, then we can turn that into 100 times 101.0569

Our left side is still just 2s; so we have 2s = 100(101).0576

What we are looking for is the sum, s = ...up until 100; so we just divide both sides by 2 to get rid of this 2 here.0580

Divide both sides by 2; 100 divided by 2 gets us 50, so 50 times 101 means we have an answer of 5050.0588

So, that probably took about as much time as if we had added up 1 + 2, all the way up to 100.0596

If we had done that whole thing by hand, it would have taken a while.0601

And now, we have the beginning kernel to think, "We can just do this for anything at all, and it will end up working out!"0603

Indeed, that is what will work out.0609

We have this method in mind of being able to string all of the things in our arithmetic sequence together,0610

and then flip it and add them together and see what happens.0616

We can now figure out a general formula for any finite arithmetic series.0619

Let sn denote the nth partial sum--that is, the first n terms of the sequence, added together, of some arithmetic sequence.0623

So, sn = a1 + a2 + a3 + ... up until we get to + an - 1,0632

up until, finally, an is our end, because we have the nth partial sum; great.0639

Earlier, we figured out the general term for any arithmetic sequence is an = a1 + (n - 1)d.0645

So, we can swap out a1 for what it is in the general form, a2 for what is in the general term,0652

an - 1 for what it is in the general term, an for what it is in the general term.0658

This will get everything in terms of a1 and d and that n; great.0662

Thus, we can write out sn if we want to; we can write it out as sn = a1,0667

and then a2 would be a1 + d (2 minus 1, so 1 times d...a1 + d).0673

We work our way out: an - 1 would be a1 + (n - 2)d;0679

we plug in n - 1 for the general n term, so n - 1, minus 1...n minus 2 times d.0683

And finally, the an would be n plus n minus 1 times d.0689

Great; so we have this thing where the only thing showing up there is a1, n, and d.0693

We have far fewer things that we have to worry about getting in our way.0698

Furthermore, we can write sn in the opposite order; we are allowed to flip addition order.0702

So, we write it in the opposite order as sn =...the last thing now goes first...a1 + (n - 1)d.0706

a1 + (n - 2)d goes next; and then finally, we work our way down: a1 + d...a1...0713

so now, we have the equation in its normal order and the equation in its opposite order.0719

We can add these two equations for sn together; they are both equal; they are both fine equations.0723

There is nothing wrong with them, so we are allowed to use elimination to be able to add equations together.0728

We add them together, and we have our normal way of writing it, sn = a1 + ...0732

+ up until our an term, a1 + (n - 1)d.0738

And then, the opposite order is sn = a1 + (n - 1)d + ... up until a1.0741

We add these together; a1 + a1 + (n - 1)d ends up getting us 2a1 + (n - 1)d.0748

Over on the far end, we will end up having the exact same thing: a1 + (n - 1)d + a1 will get us 2a1 + (n - 1)d.0756

And we are going to end up getting the same thing for every term in the middle, as well.0766

All of those dots will end up matching up, as well, for the same reason that we added 1 and 100, then 2 and 99, then 3 and 98.0770

They all ended up matching up together; the same thing happens.0776

We will always end up having that be the value for each of the additions through our elimination.0779

So, notice, at this point, that we can do the following: we can write this 2a1 + (n - 1)d here: 2a1 + (n - 1)d.0785

Well, that is the same thing: we can split the 2a1 into two different parts.0793

So, we have a1 plus...and then we can just put parentheses: a1 + (n - 1)d.0798

Well, we already have a way of writing this out: a1 + (n - 1)d is the an term.0803

So, what we have is a1 + an; so we can write this as a1 + an.0809

We swap each one of them out; we now have that 2sn is equal to a1 + an + ... + a1 + an.0814

How many terms are there total? There are n terms here total, because we started at a1 here,0822

and we worked our way up until we finally got to an here: first term, second term, third term...up until the nth term.0828

The first term to the nth term--that means that we have a total of n terms.0835

So, a1 + an gets added to itself n times (n terms, so n times, since they are all identical).0839

At that point, we have 2sn = n(a1 + an).0845

And since what we wanted on its own was just sn, we divide 2sn by 2 on both sides of our equation.0850

And we get n/2(a1 + an); great.0857

Thus, we now have a formula for the value of any finite arithmetic series.0863

Given any arithmetic sequence, a1, a2, a3...the sum of the first n terms is n/2(a1 + an).0868

This works for any finite arithmetic sequence, starting at the first term and working up to the nth term.0884

So, we can find the sum by only knowing the first term, a1, the last term, an, and the total number of terms, n.0890

That is all we need, and we can just easily, just like that, find out what the value of a finite arithmetic series is--that is pretty great.0904

Before we go on, though, one little thing to be careful about: be careful to pay attention to how many terms are in the series.0911

It can be easy to get the value of n confused and accidentally think it is one higher or one lower than it really is.0918

We will see why that is the case in the examples; so just pay really close attention.0925

If you are working from a1 up until an, then that is easy, because it is 1, 2, 3, 4...up until the n.0928

So, it must be that there are n things there.0934

But it can start getting a little bit more confusing if you start at a number that isn't 1--0935

if you start at 5 and count your way up to 27, how many things did you just say out loud?0939

We will see what we are talking about there as we work through the examples.0943

All right, let's see some examples: Show that the sequence below is arithmetic; then give a formula for the general term, an.0947

First, to show that it is arithmetic, we need to show that it has a constant difference.0953

To get from 2.6 to 3.3, we add 0.7; to get from 3.3 to 4, we add 0.7; to get from 4 to 4.7, we add 0.7;0956

and we can see that this is going to keep going like this, so it checks out.0968

It is an arithmetic sequence, because there is a common difference; its common difference is 0.7.0972

To figure out the general term, an, we want to figure out what our a1 is.0979

a1 is just the first term, which is 2.6; so our general term, an, always ends up working like this.0983

It is the first term, plus (n - 1) times the common difference.0990

So now, we can just plug in our values: an =...we figured out that a1 is 2.6, plus (n - 1)...0995

that is just going in because it is the general term...times our difference of 0.7.1001

And there we are; there is our general term; there is the formula for the nth term.1007

Alternatively, if we wanted to, we could also simplify this a little bit more, so it isn't n - 1 (that part doesn't show up).1011

Sometimes it is useful to have it in this format; but other times we might want to simplify it.1017

So, if we decided to simplify it, we would have an = 2.6 + n(0.7), so 0.7n, minus 1(0.7), so minus 0.7;1021

so the 2.6 and the -0.7 interact, and we have 1.9 + 0.7n.1033

Alternatively, we could write it like this: either of these two ways is perfectly valid.1040

Either one of these two things is a formula for the general term.1045

Sometimes it will be more useful to write it one way, and sometimes it will be more useful to write it the other way.1048

So, don't be scared if you see one written in a different way than the other one; they are both totally acceptable.1051

The second example: Find the value of the arithmetic series below.1058

What is our difference? That will help us understand what is working on here.1062

The difference will not actually be necessary to use our formula for an arithmetic series,1066

but it will help us see what is going on just a little bit on our way to using it.1070

We have a difference of 5 each time; so it is + 5, + 5, + 5...difference = 5.1073

We need three things to know what the series' value is.1080

We need to know the first term; that is easy--we can see it right there: a1 = 7.1086

We need to know the last term; that is easy, as well: an = 107.1091

And we need to know what the number of terms is, n = ?.1096

So, how can we figure out how many terms there are?1101

We might be tempted to do the following: 107 - 7 comes out to be 100; and then we say,1104

"Oh, our difference is 5, so let's divide by 5," and so we get 20; so n must be 20...NO, that is not the case.1114

Now, to understand why this is not the case, we need to look at something.1123

Let's create a little sidebar here to understand what is going on a little better.1128

Look at...if we wanted to talk about the number from 1 to 25, if we wanted to count how many numbers there are between 1 and 25,1133

we count: 1, 2, 3, 4...25--pretty obvious: that means that the number of numbers is 25.1140

There are 25 things there; great--that makes sense.1148

What if we were talking about going from 25 to 50?1150

Well, we might say that we can count by hand...50 - 25...so then, there are a total of 25 terms, because 50 - 25 is 25...No.1155

Wait, what? Well, let's count it by hand: how does this work.1165

25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50; that is 26 things that we just counted out there.1170

So, what is going on? 25 counts as something we have to count.1189

When we count from 1 to 25, if we just subtracted 25 minus 1, if we subtracted 1 from 25, 25 - 1, that is 24.1195

But we don't say, "Oh, from 1 to 25, there must be 24 numbers, because 25 - 1 is 24"--no, we don't think like that.1202

We know that counting from 1 to 25, it is 25 things there.1207

So, counting from 25 to 50 means a difference of 25; that means that there are 25 steps to get to 50 from 25.1211

But we also have to count the first location that we started on.1220

We have to actually count to 25, as well; so that is a total of 26.1223

What we have is: we have 50 - 25 = 25; but then we have to have 25 + 1 = 26.1227

So, our number is 26 for how many things we ended up doing.1240

It is the same thing with 7 to 107; how many steps?1245

If we have a distance of 5 for each step, how many steps do we have to take from the 7?1250

Well, we have to take 20 steps, because 20 times 5 is 100.1256

So, if we take 20 5-distance steps from 7, we will make it to 107; so we have 20 steps that we take.1260

But we also have to count the 7, so we end up actually have n = 21.1268

And so, that is our value for n.1278

It is really easy to just say, "OK, I took that many steps, so that must be my value."1283

No, you have to really pay attention to make sure that you are counting also where you started.1287

But sometimes, you have to pay attention and think about it: "Did I already count where I started?"1291

So, you really have to be careful with this sort of thing.1295

It is easy, as long as you can get a1, an, and the number of steps, n.1296

But sometimes, it is hard to really realize just how many steps you have precisely.1300

So, be careful with that sort of thing.1304

All right, at this point, we are ready to use our formula: n/2(a1 + an) is the value1306

of the sum of all of those numbers, the sum of that finite arithmetic series.1313

Our n was 22, over 2; our a1 was 7; our an was 107, so + 107; so we get 21/2(114).1317

We punch that into a calculator, and we end up getting 1197; 1197 is our answer for adding that all up.1331

All right, the third example: here is my thing that I said at the beginning, when we talked through the introduction.1341

At this point, we are now able to add the numbers from 1 to 1000 in less time than it takes to put on a pair of shoes and tie them up.1346

If you are going to take off your shoes and test if that is really the case, now is the time to do it, before we start looking at this problem.1354

Are you ready? OK, let me read the problem, and then we will have things be a fair challenge between shoe-putting-on and massive addition.1360

Add all of the integers from 1 to 1000 (so 1 + 2 + 3...+ 999 + 1000).1369

The first term is equal to 1; our last term is equal to 1000; the total number of terms we have from 1 to 1000 is simply 1000.1380

So, it is 1000, the number of terms, divided by 2 times the first term, plus the last term...1000, so 500 times 1001...is equal to 500,500; I am done!1388

It's pretty amazing how fast we can end up adding everything from 1 to 1000 in that little time.1401

This is the power of the series formula; this is the power of studying series--1409

the fact that we can add things that would take so long to work out by hand, like that.1413

We can do this stuff really, really quickly, once we work through this.1418

Our first term was 1; our last term was 1000; so a1 = 1; an = 1000.1422

How many things are there from 1 up to 1000? Well, that one is pretty easy; that one is 1000.1428

So, we have n/2, 1000/2, times 1 + 1000, so 500 times (I accidentally made a little bit of a typo as we were writing that out) 1001.1432

Multiply that out, and you get 500 thousand, 500; it is as simple as that.1445

The fourth example: Find the value of the sum below.1449

To do this, let's write out what this sigma notation ends up giving us.1453

i = 4 is our first place, so that is going to be 53 minus...oops, if it is going to be an i here and a k here, they have to agree on that.1458

So, that should actually read as a k, or the thing on the inside should read as an i, for this problem--I'm sorry about that.1468

53 - 4 times 4 is our first one; plus 53 - 4 times 5 (our next step up--our index goes up by one) plus 53 - 4 times 61474

(our index goes up one again); and it keeps doing this, until we get to our last upper limit for our sum, 53 minus 4 times 25; cool.1491

Now, at this point, we think, "OK, how can we add this up?"1503

Well...oh, this is an arithmetic sequence; it is 4 times some steadily-increasing, one-by-one thing.1505

So, it is an arithmetic sequence, an arithmetic series, that is appearing here.1513

If that is the case, what do we need?1517

We need to know the first term, the last term, and the number of terms that there are total.1518

If that is the case, all we really care about is this first term and this last term.1525

All of the stuff in the middle--we don't really need to work with it.1529

53 - 4 times 4, so 53 - 16, plus...up until our last term of 53 - 4 times 25 is 100; 53 - 16 comes out to be 37, plus...plus...53 - 100 comes out to be -47.1532

Our first term is 37; our last term is -47; the only real question that we have now is what is the value for n.1554

How many terms are there total? We are counting from 4 up to 25.1566

So, from 4 up to 25, how many steps do we have to take there?1570

25 - 4 means 21 steps; but notice, it is steps; there are 21 steps, but we also have to count the k = 4.1573

It is 21 steps above 4, so we also have to count the step at 4; so 21 + 1 counts where we start.1584

It is not just how many steps you take forward, but how many stones there are total, so to speak.1592

21 + 1 = 22 for our value of n; so we get n = 22; great.1598

n = 22; we know what the first one is; we know what the last one is; we are ready to work this out.1606

22 is our n, divided by 2; n/2 times the first term, 37, plus the last term, -47...we work this out.1611

22/2 is 11, times 37 + -47 (is -10)...we get -110; that is what that whole series ends up working out to be; cool.1622

All right, and we are ready for our fifth and final example.1635

An amphitheater has 24 seats in the third row, 26 in the fourth.1637

If this pattern of seat increase between rows is the same for any two consecutive rows, and there are 27 rows total, how many seats are there in total?1641

The first thing we want to do is understand how this is working.1650

Well, it is an amphitheater; we can see this picture here to help illustrate what is going on.1652

As we get farther and farther from the stage, it curves out more and more.1657

It is pretty small near the stage; as it gets farther and farther, it expands out and out.1660

So, that means there are more seats in every row, the farther back in the row we go.1665

Later rows will end up having more seats than earlier rows.1670

That is why we have 24 in the third, but 26 in the fourth.1673

We can see this: the early rows have fewer seats than the later rows, from how far they are from the stage.1676

OK, what we are looking for is how many seats there are total.1683

What we can do is talk about the third row having 24 seats; the fourth row has 26 seats; so we could think of this as a sequence,1689

where you know that it has the constant increase; the pattern of seat increase between rows is always the same.1698

What we have here is an arithmetic sequence; it makes sense, since that is what the lesson is about.1703

We can write 24 seats in the third row as a3 = 24.1708

We also know that it is 26 in the fourth row; so a4 = 26.1716

OK, so if that is the case, and the pattern of seat increase is always the same for two consecutive rows,1726

that means...to get from 24 to 26, we added +2; we have a common difference of positive 2.1734

So, if that is the case, what would the second row have to be?1741

Well, it would have to be -2 from the third row, so it would be at 22; 22 + 2 gets us to 24.1744

The same logic works for the first row, so the first row must be at 12 20, 22, 24, 26...that is the number of seats.1750

We see that we have a nice arithmetic sequence here.1758

What we are really looking to do is take a finite arithmetic series.1760

We are looking to figure out what is the 27th partial sum, because what we want to do1763

is add the number of seats in the first, second, third, fourth...up until the twenty-seventh row.1768

And we will be able to figure out all of those.1774

So, what we need to use the formula that we figured out: we need to know how many seats there are in the first row,1776

how many seats there are in the last row, and the total number of rows.1781

How many seats are there in the last row?1784

a27 is going to be our last row, because there are 27 rows total.1787

a27 is going to be the number in the first row, plus...27 - 1 (n - 1 is 27 - 1) times our difference (our difference is 2, so times 2).1791

This makes sense, because what we have here is that the 27th row1803

is going to be equal to our first row, 20, plus...how many steps is it to get from the first to the twenty-seventh row?1806

That is going to be 26 steps, times an increase of 2 for every row we go forward.1812

We work this out; that means that our 27th row is equal to 20 plus 26 times 2 is 52;1818

a27, our 27th row...the number of seats in our 27th row, 20 + 52, is 72 seats total.1825

So, at this point, we have 72 seats for our final row, 20 seats for our first row...1833

how many total rows are there? Well, that is going to be...if we are going from the first row,1839

up until the 27th row, then we can just count: 1, 2, 3, 4...counting up to 27.1843

That is easy; that is 27, so n = 27.1847

So, our formula is the number of terms total, divided by 2, times the first term plus the last term.1851

So, our number of terms total (number of rows total) is 27, divided by 2, times...1859

what is the first term, the first number of seats? 20, plus...what is the last number of seats, our last term? 72.1864

27/2 times 92...we work that out with a calculator, and we end up getting 1242 seats total in the amphitheater.1872

Great; there we are with the answer.1885

All right, in the next lesson, we will end up looking at geometric sequences and series,1886

which give us a way to look at this through multiplying instead of just adding.1890

All right, we will see you at Educator.com later--goodbye!1894