Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Pre Calculus
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for Educator.com

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.
  • Learn at your own pace... anytime, anywhere!

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

For each sequence below, decide whether or not it is an arithmetic sequence.
31,  33,   35,   37,   …             
             1,    7

9
,    5

9
,    1

3
,   …
  • 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:
    31,  33,   35,   37,   …
    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:
    31,  33,   35,   37,   …       ⇒       3,   27,   243,   2187,   …
    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:
    33 − 31 = 9 − 3 = 6        ≠        35 − 33 = 243 − 27 = 216
    Thus, since the difference does not stay the same for each consecutive pair, the sequence is not arithmetic.
  • Now let's consider the second sequence:
    1,    7

    9
    ,    5

    9
    ,    1

    3
    ,   …
    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:
    First to Second:       7

    9
    − 1     =     7

    9
    9

    9
        =     − 2

    9

    Second to Third:                                                                                        5

    9
    7

    9
        =     − 2

    9

    Third to Fourth:       1

    3
    5

    9
        =     3

    9
    5

    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.
First sequence is not arithmetic, second sequence is arithmetic.
Give the nth term formula an for the arithmetic sequence that has the first term (a1) and difference (d) given below.
a1 = 47,        d=8
  • The number given by a1 is the first term of the sequence. The value of d is how much each subsequent term changes by. That is a1 + d = a2, and so on for later terms.
  • The nth term formula an 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 nth 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
    an = a1 + (n−1)d.
  • Since we have a formula and already know a1 and d, we can just plug in:
    an = 47 + (n−1)·8
    From there, we can easily simplify a bit:
    an = 47 + (n−1) ·8     =     47 + 8n − 8     =     8n+39
    Therefore we have the general term formula an = 8n + 39.

    [If we want to check our answer, we know the first term is a1=47 and the difference is d=8, so we can easily write out the first few terms:
    47,   55,   63,   …
    Now that we have the first few terms, check to make sure the formula gives the same values:
    n=1     ⇒     a1 = 8(1) + 39     =     47    

    n=3     ⇒     a3 = 8(3) + 39     =     63    
    Great: our formula for an checks out, so we know our answer is correct.]
an = 8n + 39
Give the nth term formula an for the arithmetic sequence written below.
103,  100,   97,   94,   91,   …
  • 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
    an = a1 + (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 a1 = 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:
    100 − 103     =     −3     =     91 − 94
    Thus the difference is d=−3.
  • Now that we know a1 = 103 and d=−3, we just plug in:
    an = a1 + (n−1)d     =     103 + (n−1)·(−3)
    We can easily simplify a bit:
    an = 103 + (n−1)·(−3)     =     103 −3n + 3     =     −3n + 106
    Therefore we have the general term formula an = 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     ⇒     a1 = −3 ·1 + 106     =     103    

    n=4     ⇒     a4 = −3 ·4 + 106     =     94    
    Great: our formula for an checks out, so we know our answer is correct.]
an = −3n + 106
Find the value of the below sum.
3+6+9+12+15+18+21+24+27+30
  • 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
    n

    2
    ·(a1 + an),
    where n is the total number of terms in the series, a1 is the first term, and an is the last term.
  • Looking at this problem, the first term and last term are obvious:
    a1 = 3,        an = 30
    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 a1 to get the final term of an. Notice, for this problem:
    an − a1     =     30 − 3     =     27        ⇒        27

    3
        =     9
    Thus, we added 3 to a1 a total of 9 times. Thus, that's 9 steps, but we have to count the first location we started at (a1), so we have a total of n=10. Either way we do it, we can now find the sum with the formula:
    n

    2
    ·(a1 + an)     =     10

    2
    ·(3 + 30)     =     5 ·33     =     165
165
Find the sum of the first 1000 positive, even integers.
  • The question asks for us to sum up a portion of the positive, even integers. These begin as below:
    2,   4,   6,   8,   …
    Notice that the even integers make up an arithmetic sequence, since we get each following term by adding 2.
  • From the video lesson, we learned that the sum of any arithmetic sequence is
    n

    2
    ·(a1 + an),
    where n is the total number of terms in the series, a1 is the first term, and an is the last term. Clearly, we have a1=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 (an) is.
  • 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 1000th term will be
    an = 2 ·1000     =     2000
    Now that we know all the necessary values for the arithmetic series formula, we can just plug in:
    n

    2
    ·(a1 + an)     =     1000

    2
    ·(2 + 2000)     =     500 ·2002     =     1 001 000
1 001 000
Calculate the value of the below sum.
53

i=4 
(8i + 2)
  • (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
    n

    2
    ·(a1 + an),
    where n is the total number of terms in the series, a1 is the first term, and an is the last term. To find the first term, just plug in the lowest value our sigma index can give: i=4.
    a1     ⇒     i = 4     ⇒     8 ·(4) + 2     =     34
    To find the last term, just plug in the upper value our sigma index can give: i=53.
    an     ⇒     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
    ·(a1 + an)     =     50

    2
    ·(34+426)     =     25·460     =     11  500
11 500
Calculate the value of the below.
40

i=1 
(2+5i)  −    22

k=4 
(100−4k)
  • (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:
    n

    2
    ·(a1 + an),
    where n is the total number of terms in the series, a1 is the first term, and an is the last term.
  • Begin by finding the value of the first series. The first term will occur at the lowest index value of i=1:
    a1     ⇒     i = 1     ⇒     2+5(1)     =     7
    The last term occurs at the highest index value of i=40:
    an     ⇒     i = 40     ⇒     2+5(40)     =     202
    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:
    40

    i=1 
    (2+5i)     =     n

    2
    ·(a1 + an)     =     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:
    a1     ⇒     k = 4     ⇒     100−4(4)     =     84
    The last term occurs at the highest index value of k=22:
    an     ⇒     k=22     ⇒     100−4(22)     =     12
    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:
    22

    k=4 
    (100−4k)     =     n

    2
    ·(a1 + an)     =     19

    2
    ·(84+12)     =     19

    2
    ·96     =    912
  • From the above work, we now know
    40

    i=1 
    (2+5i) = 4180              
                  22

    k=4 
    (100−4k) = 912
    Our goal was to find the value of the below
    40

    i=1 
    (2+5i)  −    22

    k=4 
    (100−4k),
    so we just plug in the values that we found:
    40

    i=1 
    (2+5i)  −    22

    k=4 
    (100−4k)     =     4180   −   912     =     3268
3268
An arithmetic series has a first term of a1 = 3 and a common difference of d=4 between each pair of consecutive terms. If the total value of the series is 4095, what is the number of terms in the series, n? What is the value of the final term in the series, an?
  • From the problem, we're adding up some number of terms from an arithmetic sequence that begins at a1=3 with a difference of d=4:
    3,   7,   11,   15,   19,   …
    We know that if we add up some number of terms, we get the below:
    3  + 7  + 11  + … +  ?    =   4095
    Our goal is to figure out how many terms we have to add up and what the value of that last term will be.
  • From the video lesson, we learned that the sum of any arithmetic sequence is
    n

    2
    ·(a1 + an),
    where n is the total number of terms in the series, a1 is the first term, and an is the last term. Thus we can plug in to that and we know that the value it should output is 4095:
    n

    2
    ·(3 + an) = 4095
    The only issue is that the above equation has two unknowns: both n and an. 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 an in terms of the other.
  • Notice that each subsequent term is based off the previous terms. Consider the below:
    a2 = 7 = 4+3,            a3 = 11 = 2·4 + 3,            a4 = 15 = 3·4 + 3
    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 nth 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 an, we have
    an = (n−1)·4 + 3,
    which we can simplify to an = 4n1.
  • Now we can plug in to our arithmetic series formula to get an equation with a single unknown:
    n

    2
    ·(3 + an) = 4095     ⇒     n

    2
    ·(3 + [4n−1]) = 4095
    From here, work towards solving for n:
    n

    2
    ·(3 + [4n−1]) = 4095     ⇒     n

    2
    ·(4n+2) = 4095     ⇒     2n2+n = 4095
    We can solve using the quadratic formula, so set it up:
    2n2 + n − 4095 = 0    ⇒     n =
    −1 ±


    12 − 4·2·(−4095)

    2·2
    Working with a calculator, we get
    n = −1 ±181

    4
    ,
    which provides us with two values for n:
    n = −1+181

    4
    = 180

    4
    = 45              
                  n = −1−181

    4
    = −182

    4
    = − 91

    2
    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.
  • Finally, we can use this value to find the last term. Since
    an = (n−1)·4 + 3,
    we can plug in n=45 to find the value of an:
    an = (45−1) ·4 + 3     =     44 ·4 + 3     =     179
    [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 a1, n, and an:
    n

    2
    ·(a1 + an)     =     45

    2
    ·(3 + 179)     =     45

    2
    ·182     =     4095    
    Great! Since the formula gives the same sum as the problem told us it should have, we know our results for n and an are correct.]
n=45,    an = 179
A national spelling bee competition gives scholarships to those who finish in the top 20 spots. First place gets $5000, second place gets $4800, third gets $4600, and so on. How much scholarship money is given out in total?
  • 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
    ·(a1 + an)
  • From the problem, it's quite clear there are a total of 20 prize-winners, so n=20, and that first place gets $5000, so a1 = 5000. The only difficult part is figuring out the value of the final prize, an. To figure it out, remember it is the 20th 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:
    an = a20 = 19·(−200) + 5000     =     1200
    Therefore the last prize winner gets an=1200.
  • Now that we know all the pertinent values, we can apply the formula:
    n

    2
    ·(a1 + an)     =     20

    2
    ·(5000+1200)     =     10 ·6200     =     62 000
$62 000
An ancient coliseum has a central arena that is ringed by an audience seated in circular rows. The row closest to the arena has a total of 60 seats in it. The next one up has 65 seats, and the next up has 70 seats. If this pattern continues and the coliseum has a total of 23 rows, how many seats does the coliseum have?
  • 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
    ·(a1 + an)
  • From the problem, we know there are a total of 23 rows, so n=23, and that the first row has 60 seats, so a1=60. The only difficult part is figuring out the number of seats in the final row, the 23rd row, a23. To figure it out, remember it is the 23rd 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:
    an = a23 = 22·(5) + 60     =     170
    Therefore the final row has an=170 seats.
  • Now that we know all the pertinent values, we can apply the formula:
    n

    2
    ·(a1 + an)     =     23

    2
    ·(60 + 170)     =     23

    2
    ·230     =     2645
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 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

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 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

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 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