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Arithmetic Sequences & Series
 A sequence is arithmetic if the difference between any two consecutive terms is constant:
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.a_{n}−a_{n−1} = d,  The formula for the n^{th} term (general term) of an arithmetic sequence is
a_{n} = a_{1} + (n−1)d.  To find the formula for the general term of an arithmetic sequence, we only need to figure out its first term (a_{1}) and the common difference (d).
 We can use the following formula to calculate the value of an arithmetic series. Given any arithmetic sequence a_{1}, a_{2}, a_{3}, …, the sum of the first n terms (the n^{th} partial sum) is
n 2
·(a_{1} + a_{n}).  We can find the sum by only knowing the first term (a_{1}), the last term (a_{n}), 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

 A sequence is arithmetic if the difference between any two consecutive terms is constant. In other words, the difference between the first and second terms is the same as the difference between the second and third, the third and fourth, etc.
 To check if a sequence is arithmetic, we simply need to find the difference between any two consecutive terms, then see if that's the difference everywhere else in the sequence.
 Let's start with the first sequence:
Begin by noticing that the exponents of the terms follow an arithmetic pattern: they add 2 each time. However, it doesn't matter if the exponents follow an arithmetic pattern: we need the terms themselves to follow an arithmetic pattern. To see if that is the case, we first need to write each term as a clear number:3^{1}, 3^{3}, 3^{5}, 3^{7}, …
From here, it's pretty easy to see that the sequence is not arithmetic. The difference between any two terms does not remain constant. We can make this very clear by comparing the difference between the first and second with the difference between the second and third:3^{1}, 3^{3}, 3^{5}, 3^{7}, … ⇒ 3, 27, 243, 2187, …
Thus, since the difference does not stay the same for each consecutive pair, the sequence is not arithmetic.3^{3} − 3^{1} = 9 − 3 = 6 ≠ 3^{5} − 3^{3} = 243 − 27 = 216  Now let's consider the second sequence:
At first, it's difficult to tell if this sequence is arithmetic because the format of each term keeps changing. One way to deal with that would be to put them all in a common format (such as putting them all on a common denominator of 9). Alternatively, we can just check by looking at the differences:1, 7 9, 5 9, 1 3, … First to Second: 7 9− 1 = 7 9− 9 9= − 2 9Second to Third: 5 9− 7 9= − 2 9
If we look at the difference between each pair of consecutive terms, we see that the difference is always −[(2)/(9)]. Thus the sequence is arithmetic.Third to Fourth: 1 3− 5 9= 3 9− 5 9= − 2 9

 The number given by a_{1} is the first term of the sequence. The value of d is how much each subsequent term changes by. That is a_{1} + d = a_{2}, and so on for later terms.
 The n^{th} term formula a_{n} is a formula where we can plug in n (the number of the term we want) and get out the value of the term for that n^{th} term. For an arithmetic sequence, in the video lesson we saw that there is a simple formula if you know the first term and the difference. It is simply
a_{n} = a_{1} + (n−1)d.  Since we have a formula and already know a_{1} and d, we can just plug in:
From there, we can easily simplify a bit:a_{n} = 47 + (n−1)·8
Therefore we have the general term formula a_{n} =8n + 39.a_{n} = 47 + (n−1) ·8 = 47 + 8n − 8 = 8n+39
[If we want to check our answer, we know the first term is a_{1}=47 and the difference is d=8, so we can easily write out the first few terms:
Now that we have the first few terms, check to make sure the formula gives the same values:47, 55, 63, … n=1 ⇒ a_{1} = 8(1) + 39 = 47
Great: our formula for a_{n} checks out, so we know our answer is correct.]n=3 ⇒ a_{3} = 8(3) + 39 = 63

 For an arithmetic sequence, in the video lesson we saw that there is a simple formula if you know the first term and the difference. It is simply
a_{n} = a_{1} + (n−1)d.  This means we can easily find the formula if we know the first term and the difference. Looking at the sequence, the first term is clearly there, so we know a_{1} = 103. Now we just need to find the difference. To do that, just find the difference between two terms (remember, we find difference as "later minus earlier", for example, third minus second). To be sure we got it correctly, do it for two different pairs. Below we'll check first and second along with fourth and fifth:
Thus the difference is d=−3.100 − 103 = −3 = 91 − 94  Now that we know a_{1} = 103 and d=−3, we just plug in:
We can easily simplify a bit:a_{n} = a_{1} + (n−1)d = 103 + (n−1)·(−3)
Therefore we have the general term formula a_{n} =−3n + 106.a_{n} = 103 + (n−1)·(−3) = 103 −3n + 3 = −3n + 106
[If we want to check our answer, we already know the first few terms of the sequence, so we can check that our formula gives the same values:n=1 ⇒ a_{1} = −3 ·1 + 106 = 103
Great: our formula for a_{n} checks out, so we know our answer is correct.]n=4 ⇒ a_{4} = −3 ·4 + 106 = 94

 We could find the value of the sum by just getting a calculator (or a piece of scratch paper) and adding everything up. It would take a little bit of time, but it could be done that way. However, instead of that, let's start by noticing that each term in the sum is effectively a term in an arithmetic sequence. Thus, we're working with an arithmetic series: a sum where every subsequent term has a common difference with its preceding term.
 From the video lesson, we learned that the sum of any arithmetic sequence is
where n is the total number of terms in the series, a_{1} is the first term, and a_{n} is the last term.n 2·(a_{1} + a_{n}),  Looking at this problem, the first term and last term are obvious:
We also need to know the number of terms, n. We can approach this in two ways: one, we can simply count the number of terms by hand to find that there are 10 terms, thus n=10. However, for more difficult problems, it will be hard to simply count the terms by hand. Instead, we can find the number of terms by considering the common difference d and how many times that has been added to the starting term of a_{1} to get the final term of a_{n}. Notice, for this problem:a_{1} = 3, a_{n} = 30
Thus, we added 3 to a_{1} a total of 9 times. Thus, that's 9 steps, but we have to count the first location we started at (a_{1}), so we have a total of n=10. Either way we do it, we can now find the sum with the formula:a_{n} − a_{1} = 30 − 3 = 27 ⇒ 27 3= 9 n 2·(a_{1} + a_{n}) = 10 2·(3 + 30) = 5 ·33 = 165
 The question asks for us to sum up a portion of the positive, even integers. These begin as below:
Notice that the even integers make up an arithmetic sequence, since we get each following term by adding 2.2, 4, 6, 8, …  From the video lesson, we learned that the sum of any arithmetic sequence is
where n is the total number of terms in the series, a_{1} is the first term, and a_{n} is the last term. Clearly, we have a_{1}=2 and since we're adding up the first 1000 positive even integers, that means there must be n=1000 of the numbers total. The only potentially difficult part for this problem is figuring out what the last term (a_{n}) is.n 2·(a_{1} + a_{n}),  If you have difficulty finding the last term, try thinking about what the fifth term would be: 2, 4, 6, 8, 10. Thus, for the fifth term, we simply have 2·5 = 10. By this same logic, we can see that the 1000^{th} term will be
Now that we know all the necessary values for the arithmetic series formula, we can just plug in:a_{n} = 2 ·1000 = 2000 n 2·(a_{1} + a_{n}) = 1000 2·(2 + 2000) = 500 ·2002 = 1 001 000

 (Note: If you are unfamiliar with using sigma notation (Σ) for compactly showing a sum/series, make sure to check out the lesson Introduction to Series. How to read and use the notation is carefully explained in that lesson, but it will be assumed you already understand it in the below steps.) Begin by noticing that the notation indicates an arithmetic series. It will have a difference of d=8 for every term because of the 8i in the sigma notation. Thus, we can apply what we know about finding arithmetic series.
 From the video lesson, we learned that the sum of any arithmetic sequence is
where n is the total number of terms in the series, a_{1} is the first term, and a_{n} is the last term. To find the first term, just plug in the lowest value our sigma index can give: i=4.n 2·(a_{1} + a_{n}),
To find the last term, just plug in the upper value our sigma index can give: i=53.a_{1} ⇒ i = 4 ⇒ 8 ·(4) + 2 = 34 a_{n} ⇒ i = 53 ⇒ 8 ·(53) + 2 = 426  The trickiest part is probably figuring out what the the number of terms (n) is. To do this, notice that the first term has an index of i=4, while the last term has an index of i=53. Thus there are 53−4 = 49 steps between the two terms. However!, we must also remember to include the starting location, since it doesn't get counted as a step. Thus there are a total of n=50 numbers (49 steps plus 1 for "home"). Now that we know all the necessary values for the arithmetic series formula, we can just plug in:
n 2·(a_{1} + a_{n}) = 50 2·(34+426) = 25·460 = 11 500

 (Note: If you are unfamiliar with using sigma notation (Σ) for compactly showing a sum/series, make sure to check out the lesson Introduction to Series. How to read and use the notation is carefully explained in that lesson, but it will be assumed you already understand it in the below steps.) Don't let yourself get freaked out by the appearance of two sigma (Σ) signs: it just means that we need to find each of the two sums, then subtract one from the other. Like in the previous problem, notice that each of the series are arithmetic, so we can use the formula for an arithmetic sequence:
where n is the total number of terms in the series, a_{1} is the first term, and a_{n} is the last term.n 2·(a_{1} + a_{n}),  Begin by finding the value of the first series. The first term will occur at the lowest index value of i=1:
The last term occurs at the highest index value of i=40:a_{1} ⇒ i = 1 ⇒ 2+5(1) = 7
Finally, the number of terms is the number of "steps" plus one for "home": 40−1=39 steps, then one more for the first term, giving n=40. Plug in to the formula to find the sum:a_{n} ⇒ i = 40 ⇒ 2+5(40) = 202 40
∑
i=1(2+5i) = n 2·(a_{1} + a_{n}) = 40 2·(7+202) = 20 ·209 = 4180  Next, we find the value of the second series. The first term will occur at the lowest index value of k=4:
The last term occurs at the highest index value of k=22:a_{1} ⇒ k = 4 ⇒ 100−4(4) = 84
Finally, the number of terms is the number of "steps" plus one for "home": 22−4=18 steps, then one more for the first term, giving n=19. Plug in to the formula to find the sum:a_{n} ⇒ k=22 ⇒ 100−4(22) = 12 22
∑
k=4(100−4k) = n 2·(a_{1} + a_{n}) = 19 2·(84+12) = 19 2·96 = 912  From the above work, we now know
Our goal was to find the value of the below40
∑
i=1(2+5i) = 4180 ⎢
⎢22
∑
k=4(100−4k) = 912
so we just plug in the values that we found:40
∑
i=1(2+5i) − 22
∑
k=4(100−4k), 40
∑
i=1(2+5i) − 22
∑
k=4(100−4k) = 4180 − 912 = 3268
 From the problem, we're adding up some number of terms from an arithmetic sequence that begins at a_{1}=3 with a difference of d=4:
We know that if we add up some number of terms, we get the below:3, 7, 11, 15, 19, …
Our goal is to figure out how many terms we have to add up and what the value of that last term will be.3 + 7 + 11 + … + ? = 4095  From the video lesson, we learned that the sum of any arithmetic sequence is
where n is the total number of terms in the series, a_{1} is the first term, and a_{n} is the last term. Thus we can plug in to that and we know that the value it should output is 4095:n 2·(a_{1} + a_{n}),
The only issue is that the above equation has two unknowns: both n and a_{n}. To be able to solve the equation, we need to somehow put it in terms of only a single unknown. We need to somehow express n or a_{n} in terms of the other.n 2·(3 + a_{n}) = 4095  Notice that each subsequent term is based off the previous terms. Consider the below:
We can see that getting to a given term is a case of adding up the difference the correct number of times. To get to the n^{th} term from the first term, it will take (n−1) "steps". Thus, we must add the difference a total of (n−1) times to the starting value. Therefore, to get to our last term a_{n}, we havea_{2} = 7 = 4+3, a_{3} = 11 = 2·4 + 3, a_{4} = 15 = 3·4 + 3
which we can simplify to a_{n} =4n−1.a_{n} = (n−1)·4 + 3,  Now we can plug in to our arithmetic series formula to get an equation with a single unknown:
From here, work towards solving for n:n 2·(3 + a_{n}) = 4095 ⇒ n 2·(3 + [4n−1]) = 4095
We can solve using the quadratic formula, so set it up:n 2·(3 + [4n−1]) = 4095 ⇒ n 2·(4n+2) = 4095 ⇒ 2n^{2}+n = 4095
Working with a calculator, we get2n^{2} + n − 4095 = 0 ⇒ n = −1 ±
√1^{2} − 4·2·(−4095)2·2
which provides us with two values for n:n = −1 ±181 4,
Since n is the number of terms we must add together, it clearly makes no sense to have a negative (or fractional) value for n, so we throw out the second answer, leaving us with n=45.n = −1+181 4= 180 4= 45 ⎢
⎢n = −1−181 4= −182 4= − 91 2  Finally, we can use this value to find the last term. Since
we can plug in n=45 to find the value of a_{n}:a_{n} = (n−1)·4 + 3,
[Since we put so much work into the problem, it would probably be a good idea to check our result since it is easy to check. Just use the formula for arithmetic series with the values of a_{1}, n, and a_{n}:a_{n} = (45−1) ·4 + 3 = 44 ·4 + 3 = 179
Great! Since the formula gives the same sum as the problem told us it should have, we know our results for n and a_{n} are correct.]n 2·(a_{1} + a_{n}) = 45 2·(3 + 179) = 45 2·182 = 4095
 Notice that the prizes are given as an arithmetic sequence, since each prize has a difference of d=−200 between it and the previous one. Because we're dealing with adding up the terms from an arithmetic sequence, we can use our formula:
n 2·(a_{1} + a_{n})  From the problem, it's quite clear there are a total of 20 prizewinners, so n=20, and that first place gets $5000, so a_{1} = 5000. The only difficult part is figuring out the value of the final prize, a_{n}. To figure it out, remember it is the 20^{th} prize so it is 19 steps away from first place (be sure to notice that it is 19 steps, not 20: first place does not count as a step). Thus we apply the difference 19 times and add that to the first term:
Therefore the last prize winner gets a_{n}=1200.a_{n} = a_{20} = 19·(−200) + 5000 = 1200  Now that we know all the pertinent values, we can apply the formula:
n 2·(a_{1} + a_{n}) = 20 2·(5000+1200) = 10 ·6200 = 62 000
 Notice that the number of seats that each row has increases with a common difference of d=5: this means we're working with an arithmetic sequence. We want to add up the number of seats in the first 23 rows, so we can do that by adding up the first 23 terms of the arithmetic sequence.
n 2·(a_{1} + a_{n})  From the problem, we know there are a total of 23 rows, so n=23, and that the first row has 60 seats, so a_{1}=60. The only difficult part is figuring out the number of seats in the final row, the 23^{rd} row, a_{23}. To figure it out, remember it is the 23^{rd} row, so it is 22 steps away from first place (be sure to notice that it is 22 steps, not 23: first place does not count as a step). Thus we apply the difference 22 times and add that to the first term:
Therefore the final row has a_{n}=170 seats.a_{n} = a_{23} = 22·(5) + 60 = 170  Now that we know all the pertinent values, we can apply the formula:
n 2·(a_{1} + a_{n}) = 23 2·(60 + 170) = 23 2·230 = 2645
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Arithmetic Sequences & Series
Lecture Slides are screencaptured 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
Precalculus with Limits Online Course
Transcription: Arithmetic Sequences & Series
Hiwelcome 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 a_{n}  a_{n  1} = d.0055
Notice: the n^{th} term minus the n  1 term (that is, the term one before the n^{th} 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 onethat is what makes it an arithmetic sequence.0128
How can we find the n^{th} term? The definition for an arithmetic sequence is based on a recursive relation.0135
It is based on a_{n} = a_{n  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 termhow can we get a general term formula out of this?0148
Remember: a recursive relation needs an initial termwe 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 a_{1}; we will call it the first termwe will leave it as that.0164
Now, from a_{n} = a_{n  1} + d, we see that a_{1} relates to later terms.0171
At the most basic level, we have that a_{2} is equal to a_{1} + 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
a_{3} would be equal to a_{2} + d, based on this recurrence relation.0191
But we just figured out that a_{2} is equal to a_{1} + d, so we can swap out for a_{2}:0196
a_{1} + d, which now gets us a_{1} + 2d when we add this d to that one there.0203
So, we have a_{1} + 2d for a_{3}; when we work on a_{4},0210
well, a_{4} is going to be a_{3} + d, the previous term, adding d.0215
But once again, we have just figured out that a_{1} + 2d is what a_{3} is.0220
So, we can plug in for a_{3}; we have a_{1} + 2d, and now we can add that d onto the 2d.0225
So, we end up getting that a_{1} + 3d is what a_{4} 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, a_{1} = a_{1}; a_{2} = a_{1} + d; a_{3} = a_{1} + 2d;0245
a_{4} = a_{1} + 3d, and this pattern will continue down.0253
We see that the n^{th} term is n  1 steps from a_{1}.0257
The first term is at 1, and the n^{th} 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 a_{1}, 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 a_{n} term is equal to a_{1} plus all of those steps times d.0280
So, the n^{th} term, a_{n}, equals a_{1}, 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, a_{1}, and the common difference, d.0294
With those two pieces of information, we automatically have the general term; we automatically have that n^{th} term formulapretty 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 cleveris 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 exploitsomething that will keep happening0357
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 upsomething that we could createand 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 thatif 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 algebrait 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 togetherthat 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 s_{n} denote the n^{th} partial sumthat is, the first n terms of the sequence, added together, of some arithmetic sequence.0623
So, s_{n} = a_{1} + a_{2} + a_{3} + ... up until we get to + a_{n  1},0632
up until, finally, a_{n} is our end, because we have the n^{th} partial sum; great.0639
Earlier, we figured out the general term for any arithmetic sequence is a_{n} = a_{1} + (n  1)d.0645
So, we can swap out a_{1} for what it is in the general form, a_{2} for what is in the general term,0652
a_{n  1} for what it is in the general term, a_{n} for what it is in the general term.0658
This will get everything in terms of a_{1} and d and that n; great.0662
Thus, we can write out s_{n} if we want to; we can write it out as s_{n} = a_{1},0667
and then a_{2} would be a_{1} + d (2 minus 1, so 1 times d...a_{1} + d).0673
We work our way out: a_{n  1} would be a_{1} + (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 a_{n} would be n plus n minus 1 times d.0689
Great; so we have this thing where the only thing showing up there is a_{1}, n, and d.0693
We have far fewer things that we have to worry about getting in our way.0698
Furthermore, we can write s_{n} in the opposite order; we are allowed to flip addition order.0702
So, we write it in the opposite order as s_{n} =...the last thing now goes first...a_{1} + (n  1)d.0706
a_{1} + (n  2)d goes next; and then finally, we work our way down: a_{1} + d...a_{1}...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 s_{n} 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, s_{n} = a_{1} + ...0732
+ up until our a_{n} term, a_{1} + (n  1)d.0738
And then, the opposite order is s_{n} = a_{1} + (n  1)d + ... up until a_{1}.0741
We add these together; a_{1} + a_{1} + (n  1)d ends up getting us 2a_{1} + (n  1)d.0748
Over on the far end, we will end up having the exact same thing: a_{1} + (n  1)d + a_{1} will get us 2a_{1} + (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 2a_{1} + (n  1)d here: 2a_{1} + (n  1)d.0785
Well, that is the same thing: we can split the 2a_{1} into two different parts.0793
So, we have a_{1} plus...and then we can just put parentheses: a_{1} + (n  1)d.0798
Well, we already have a way of writing this out: a_{1} + (n  1)d is the a_{n} term.0803
So, what we have is a_{1} + a_{n}; so we can write this as a_{1} + a_{n}.0809
We swap each one of them out; we now have that 2s_{n} is equal to a_{1} + a_{n} + ... + a_{1} + a_{n}.0814
How many terms are there total? There are n terms here total, because we started at a_{1} here,0822
and we worked our way up until we finally got to a_{n} here: first term, second term, third term...up until the n^{th} term.0828
The first term to the n^{th} termthat means that we have a total of n terms.0835
So, a_{1} + a_{n} gets added to itself n times (n terms, so n times, since they are all identical).0839
At that point, we have 2s_{n} = n(a_{1} + a_{n}).0845
And since what we wanted on its own was just s_{n}, we divide 2s_{n} by 2 on both sides of our equation.0850
And we get n/2(a_{1} + a_{n}); great.0857
Thus, we now have a formula for the value of any finite arithmetic series.0863
Given any arithmetic sequence, a_{1}, a_{2}, a_{3}...the sum of the first n terms is n/2(a_{1} + a_{n}).0868
This works for any finite arithmetic sequence, starting at the first term and working up to the n^{th} term.0884
So, we can find the sum by only knowing the first term, a_{1}, the last term, a_{n}, 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 isthat 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 a_{1} up until a_{n}, 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 10935
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, a_{n}.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, a_{n}, we want to figure out what our a_{1} is.0979
a_{1} is just the first term, which is 2.6; so our general term, a_{n}, 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: a_{n} =...we figured out that a_{1} 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 n^{th} 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 a_{n} = 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 easywe can see it right there: a_{1} = 7.1086
We need to know the last term; that is easy, as well: a_{n} = 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...25pretty obvious: that means that the number of numbers is 25.1140
There are 25 things there; greatthat 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 5distance 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
This is why you really have to think about this stuff carefully.1280
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 a_{1}, a_{n}, 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(a_{1} + a_{n}) 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 a_{1} was 7; our a_{n} 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 shoeputtingon and massive addition.1360
Add all of the integers from 1 to 1000 (so 1 + 2 + 3...+ 999 + 1000).1369
All right, are you ready? Ready, set, shoes on now!1376
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 series1409
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 a_{1} = 1; a_{n} = 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 problemI'm sorry about that.1468
53  4 times 4 is our first one; plus 53  4 times 5 (our next step upour 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 steadilyincreasing, onebyone 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 middlewe 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 a_{3} = 24.1708
We also know that it is 26 in the fourth row; so a_{4} = 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 27^{th} partial sum, because what we want to do1763
is add the number of seats in the first, second, third, fourth...up until the twentyseventh 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
a_{27} is going to be our last row, because there are 27 rows total.1787
a_{27} 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 27^{th} row1803
is going to be equal to our first row, 20, plus...how many steps is it to get from the first to the twentyseventh 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 27^{th} row is equal to 20 plus 26 times 2 is 52;1818
a_{27}, our 27^{th} row...the number of seats in our 27^{th} 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 27^{th} 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 latergoodbye!1894
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