WEBVTT mathematics/algebra-2/eaton
00:00:00.000 --> 00:00:02.200
Welcome to Educator.com.
00:00:02.200 --> 00:00:12.100
In a previous lesson, we talked about arithmetic sequences; and in this lesson, we will continue on with the discussion, to discuss arithmetic s0ries.
00:00:12.100 --> 00:00:18.700
First of all, what are arithmetic series? An **arithmetic series** is the sum of the terms of an arithmetic sequence.
00:00:18.700 --> 00:00:23.900
So, I will briefly review arithmetic sequences; but if you need a complete review of that,
00:00:23.900 --> 00:00:28.500
go and check out the previous lesson, and then move on to arithmetic series.
00:00:28.500 --> 00:00:43.600
Recall that a series is a list of numbers in a particular order; and each term in the series is related to the previous one by a constant called the **common difference**.
00:00:43.600 --> 00:00:51.600
A typical arithmetic sequence would be something like 20, 40, 60, 80, 100.
00:00:51.600 --> 00:01:03.300
Taking 40 - 20 or 60 - 40, I get a common difference of 20, which means that, to go from one term to the next, I add 20.
00:01:03.300 --> 00:01:11.300
So, this is the sequence: it is a list of numbers; the series is actually a sum of numbers,
00:01:11.300 --> 00:01:21.800
so I am going to add a + between each number: 20 + 40 + 60 + 80 + 100.
00:01:21.800 --> 00:01:37.000
So, an arithmetic series is in the general form: first term, a₁ + a₂ + a₃, and on and on until the last term, a < font size="-6" > n < /font > .
00:01:37.000 --> 00:01:46.600
This is a finite arithmetic series, because there is a specific endpoint; there is a finite number of terms.
00:01:46.600 --> 00:01:55.400
As with arithmetic sequences (those could be finite or infinite), you could have finite or infinite arithmetic series.
00:01:55.400 --> 00:02:07.400
So, looking at a different sequence: 100, 200, 300, 400, and then the ellipses to tell me that this is an infinite sequence:
00:02:07.400 --> 00:02:28.000
I could have a corresponding series, 100 + 200 + 300 + 400 + ...on and on; this is an infinite arithmetic series.
00:02:28.000 --> 00:02:34.500
There is a formula (or actually, two formulas) here to allow you to find the sum of an arithmetic series.
00:02:34.500 --> 00:02:42.300
Sometimes you will be asked to find the sum of the first n terms of an arithmetic series--maybe the first 17 terms or the first 10 terms.
00:02:42.300 --> 00:02:45.700
And that would be s < font size="-6" > 17 < /font > or s < font size="-6" > 10 < /font > .
00:02:45.700 --> 00:02:52.800
There are two formulas: the first one involves taking the number of terms, dividing it by 2, and then multiplying it
00:02:52.800 --> 00:02:59.100
by 2 times the first term, plus the quantity (n - 1) times the common difference.
00:02:59.100 --> 00:03:04.800
The second term is the sum of the first n terms: again, it equals the number of terms divided by 2.
00:03:04.800 --> 00:03:10.200
But this time, you are going to multiply that by the sum of the first and last terms.
00:03:10.200 --> 00:03:18.800
So, you can see the difference: you use the first formula if you know the common difference, and the second formula if you know the last term.
00:03:18.800 --> 00:03:22.000
So, you have to choose a formula, depending on what you know.
00:03:22.000 --> 00:03:31.000
For example, let's say that a series has, as its first term, a 5: 5 is the first term.
00:03:31.000 --> 00:03:39.800
And it has a common difference of 4, and you are asked to find the sum of the first 10 terms, s < font size="-6" > 10 < /font > .
00:03:39.800 --> 00:03:48.700
I am looking, and I see that I have the common difference; so I am going to go ahead and use the first formula.
00:03:48.700 --> 00:04:00.800
And let's go ahead and do that: s < font size="-6" > 10 < /font > = 10, because, since I am finding the sum of the first 10 terms,
00:04:00.800 --> 00:04:13.000
n = 10, divided by 2, and that is times 2 times 5 (the first term), plus n (which is 10) minus 1, times the common difference (which is 4).
00:04:13.000 --> 00:04:22.300
So, s < font size="-6" > 10 < /font > = 5(10) +...10 - 1 is, of course, 9, times 4.
00:04:22.300 --> 00:04:31.700
Therefore, s < font size="-6" > 10 < /font > = 5 times...9 times 4 is 36; 36 + 10 is 46.
00:04:31.700 --> 00:04:36.400
And if you multiply that out, or use your calculator, you will find that the sum of the first 10 terms
00:04:36.400 --> 00:04:42.700
in a series with this first term and this common difference is actually 230.
00:04:42.700 --> 00:04:51.900
So, that is one example; in another example, perhaps you are asked the sum of the 16 terms in a series,
00:04:51.900 --> 00:05:04.700
if the first term is 20, and the last term of the ones we are trying to find, a < font size="-6" > 16 < /font > , is equal to -10.
00:05:04.700 --> 00:05:11.600
Well, we are going to use the second formula, because we have the first and last terms, but we don't have the common difference.
00:05:11.600 --> 00:05:23.400
So, s < font size="-6" > 16 < /font > = n; in this case, I am looking for the sum of 16 terms, so n is 16
00:05:23.400 --> 00:05:35.900
(I am going to put that right here--n is 16); that is going to give me 16/2, times the first term, which is 20, plus that last term, which is -10.
00:05:35.900 --> 00:05:47.900
So, this is s < font size="-6" > 16 < /font > = 8, times 20 - 10, which is 10; that gives me 80 as the sum of the first 16 terms.
00:05:47.900 --> 00:05:53.600
You need to learn both formulas and simply know when to use a particular one.
00:05:53.600 --> 00:05:58.700
Sigma notation: a series can be written in a concise form, using what is called sigma notation.
00:05:58.700 --> 00:06:06.200
And what sigma refers to is the Greek letter Σ; and in this case, sigma means "sum."
00:06:06.200 --> 00:06:11.300
So, what this symbol means, in the context in which we are using it, is "sum."
00:06:11.300 --> 00:06:18.200
And you will see it written something like this: you will see a letter here, called the index (that variable is called the index).
00:06:18.200 --> 00:06:25.500
And we have been working with n, but often i is used; it could be n; it could be i; it could be k; it could be something else.
00:06:25.500 --> 00:06:32.700
i = 1; I am just giving an example; and let's say, up here, the upper index is going to be 10.
00:06:32.700 --> 00:06:39.400
And then, there will be the formula for the general term, which we talked about earlier on,
00:06:39.400 --> 00:06:45.700
when we talked about arithmetic sequences and the formula for a < font size="-6" > n < /font > that is particular to a sequence.
00:06:45.700 --> 00:06:49.600
So, if you need to go ahead and review that, it is in the previous lecture.
00:06:49.600 --> 00:06:52.200
This is the formula for the general term of the sequence.
00:06:52.200 --> 00:07:00.400
And this is read as "the sum of a < font size="-6" > n < /font > as i goes from 1 to 10."
00:07:00.400 --> 00:07:09.200
So, it is the sum of the terms that are found, using this particular formula, when you plug in 1, then 2, then 3, and up through 10.
00:07:09.200 --> 00:07:15.000
That is in general; let's talk about a specific example with a specific formula.
00:07:15.000 --> 00:07:28.000
You could have another arithmetic series, written in sigma notation, where n goes from 5 to 12.
00:07:28.000 --> 00:07:33.900
So, n goes from 5 to 12; so I am going to start with 5, and I am going to end with 12.
00:07:33.900 --> 00:07:40.100
That means that there are actually 8 terms; there are 8 terms in the series, because it is 5 through 12, inclusive.
00:07:40.100 --> 00:07:51.800
And the formula, we are going to say, is n + 3: so the formula to find a particular term, a < font size="-6" > n < /font > , is n + 3.
00:07:51.800 --> 00:08:04.800
I could find the terms, then, by saying that for the first term, a₁, I am going to use 5 as my n, so it equals 5 + 3; therefore, it is 8.
00:08:04.800 --> 00:08:17.600
For a₂, I am then going to go to 6: so that is going to be 6 + 3; that is 9.
00:08:17.600 --> 00:08:32.300
a₃ =...then I am going to go to the next value, which is 7: 7 + 3 = 10; and on up.
00:08:32.300 --> 00:08:44.600
So, if I wanted to find the last term, a₈ (because there are 8 terms), then I would put in 12; and 12 + 3 is 15.
00:08:44.600 --> 00:08:46.800
There would be terms in between here, of course.
00:08:46.800 --> 00:08:58.800
This is just a concise way of writing a series; and we have already talked about how to work with these series, and what the different terms mean, and formulas.
00:08:58.800 --> 00:09:00.800
But now, this is just a different way of writing them.
00:09:00.800 --> 00:09:07.300
We need to find the first term of the arithmetic series with a common difference of 3.5 and equal to 20,
00:09:07.300 --> 00:09:13.600
and the sum of the terms, s < font size="-6" > 20 < /font > , equaling 1005.
00:09:13.600 --> 00:09:26.500
Since we know the common difference, we can use this formula: s < font size="-6" > n < /font > = n/2, times 2 times the first term, plus (n - 1)d.
00:09:26.500 --> 00:09:31.400
Except, in this case, I am not looking for the sum: I have the sum; I am looking for the first term.
00:09:31.400 --> 00:09:49.200
Therefore, 1005 = 20/2, times 2 times a₁, plus n (n is given as 20), minus 1, times the common difference of 3.5.
00:09:49.200 --> 00:10:01.400
This gives me 1005 = 10(2a₁) + 19(3.5).
00:10:01.400 --> 00:10:22.200
19 times 3.5 is actually 66.5; therefore, 1005 equals 10 times 2a₁, which is 20a₁, plus 10(66.5), which is 665.
00:10:22.200 --> 00:10:31.100
1005; subtract 665 from both sides to get 340 = 20 times that first term.
00:10:31.100 --> 00:10:36.200
Divide both terms by 20, and you will get that the first term is equal to 17.
00:10:36.200 --> 00:10:45.800
So, we use this formula for the sum of the series; but in this case, we were looking for the first term.
00:10:45.800 --> 00:10:47.500
We had the sum; we were looking for the first term.
00:10:47.500 --> 00:10:53.600
And the solution is that the first term is 17.
00:10:53.600 --> 00:11:06.000
In the second problem, we are asked to find the first three terms of the arithmetic series with n = 17, a < font size="-6" > n < /font > = 103, s < font size="-6" > n < /font > = 1102.
00:11:06.000 --> 00:11:10.200
In order to find the first three terms, I need to find the common difference.
00:11:10.200 --> 00:11:20.500
And I also need to find the first term; I need the first term, and then I need the common difference, to find the second and third terms.
00:11:20.500 --> 00:11:31.300
I can use the formula s < font size="-6" > n < /font > = n/2 times the first term plus the last term, because I don't have the common difference--I am looking for it.
00:11:31.300 --> 00:11:38.100
But what I do have is a < font size="-6" > n < /font > , so my first step is going to be to find the first term.
00:11:38.100 --> 00:11:50.100
The sum is 1102; n is 19; I don't know the first term; and I know that a < font size="-6" > n < /font > is 103.
00:11:50.100 --> 00:12:01.600
I am going to multiply both sides by 2 to get 2204 = 19 times this, a₁ + 103.
00:12:01.600 --> 00:12:09.400
I am then going to divide both sides by 19, and that comes out to 116 = the first term, plus 103.
00:12:09.400 --> 00:12:13.800
And then, I just subtract 103 from both sides; and now I have my first term.
00:12:13.800 --> 00:12:17.400
So, I am asked to find the first three terms: the first term is 13.
00:12:17.400 --> 00:12:20.100
To find the next two terms, I find the common difference.
00:12:20.100 --> 00:12:24.400
What I am going to do is switch to the other formula: that other formula, s < font size="-6" > n < /font > ,
00:12:24.400 --> 00:12:34.000
equals n/2 times 2a₁, the first term, plus n - 1 times the common difference.
00:12:34.000 --> 00:12:41.400
I found the first term; now that I have that, I can find the common difference, because I can put the first term in here.
00:12:41.400 --> 00:12:55.300
So again, s < font size="-6" > n < /font > is 1102; n is 19; and this gives me 2 times 13, plus I have an n of 19 - 1, and I am looking for the common difference.
00:12:55.300 --> 00:13:10.200
I am going to multiply, again, 1102 times 2 to get 2204 = 19...2 times 13 is 26, plus 19 - 1...that gives me 18d.
00:13:10.200 --> 00:13:17.900
I am going to divide both sides by 19 to get 116 = 26 + 18d.
00:13:17.900 --> 00:13:29.800
Subtracting 26 from both sides gives me 90 = 18d; the final step is to divide both sides by 18 to get a common difference of 5.
00:13:29.800 --> 00:13:43.100
I have a₁ is 13; I am going to take 13 + the common difference of 5 to get the second term (that is 18).
00:13:43.100 --> 00:13:55.800
So, a₂ is 18; the third term--I am going to take 18, and I am going to add 5 to that to yield 23.
00:13:55.800 --> 00:14:03.500
So, I was asked to find the first three terms, and I did that by first using this formula to find the first term,
00:14:03.500 --> 00:14:13.900
then going to the other formula and finding the common difference to get 13, 18, and 23 as my solutions.
00:14:13.900 --> 00:14:27.000
The third example: I am asked to find the sum of the series 6 + 11 + 16 + 21, and on and on, with the last term of 126.
00:14:27.000 --> 00:14:33.800
That means that what I have is the first term and the last term.
00:14:33.800 --> 00:14:44.100
I am going to find the sum using this formula, because I can figure out my common difference.
00:14:44.100 --> 00:14:56.500
So, I am going to use the formula n/2, times 2a₁, plus a - 1, times d.
00:14:56.500 --> 00:15:03.800
I could easily find the common difference, because I know I will just take 11 - 6, so I have a common difference of 5.
00:15:03.800 --> 00:15:10.200
Looking at this formula, the only issue is going to be that I don't know n.
00:15:10.200 --> 00:15:16.700
But I can figure out n; and that is because I have another formula.
00:15:16.700 --> 00:15:26.600
Recall the formula for the general term: we discussed this in the lecture on arithmetic sequences.
00:15:26.600 --> 00:15:37.400
a < font size="-6" > n < /font > equals the first term, plus n - 1, times the common difference.
00:15:37.400 --> 00:15:56.400
a < font size="-6" > n < /font > is 126; so that shouldn't be 16--that is 126: a < font size="-6" > n < /font > is 126, and I have the first term equal to 6; and I am solving for n.
00:15:56.400 --> 00:15:59.000
I know that the common difference is 5.
00:15:59.000 --> 00:16:10.300
126...and then I am subtracting 6 from both sides: that gives me 120 = (n - 1)5, so 120 = 5n - 5.
00:16:10.300 --> 00:16:21.800
Adding 5 to both sides gives me 125 = 5n; 125/5 is 25, so I have n = 25.
00:16:21.800 --> 00:16:29.700
Now, again, I am asked to find the sum of the series; and I can use this formula, because I now have n; I have the first term; and I have the common difference.
00:16:29.700 --> 00:16:40.300
So, let's go ahead and use that: it is actually s < font size="-6" > 25 < /font > = n, which is 25, divided by 2, times 2 times that first term
00:16:40.300 --> 00:16:49.000
(which is 6), plus (n - 1) (n is 25, minus 1), times the common difference of 5.
00:16:49.000 --> 00:17:01.900
Now, it is just a matter of simplifying: this is going to give me 25/2 times 12, plus 24 times 5.
00:17:01.900 --> 00:17:13.300
So, the sum equals 25/2, times 12; and then if you multiply out 24 times 5, you will get 120.
00:17:13.300 --> 00:17:30.100
Therefore, the sum equals 25/2; 120 + 12 is going to give you 132.
00:17:30.100 --> 00:17:34.000
Now that I have gotten it down to this point, I can simplify,
00:17:34.000 --> 00:17:45.200
because this is going to give me 132/2, times 25; so that is s < font size="-6" > 25 < /font > = 25 times 66.
00:17:45.200 --> 00:17:53.700
You can multiply it out; or it is a good time to use your calculator to find that the sum equals 1650.
00:17:53.700 --> 00:17:57.300
So again, in order to use this formula, I had my common difference.
00:17:57.300 --> 00:18:01.600
I didn't have n; I solved for n; n equals 25.
00:18:01.600 --> 00:18:13.500
I went back in, substituted those values in, and then came out with s < font size="-6" > 25 < /font > ; the sum of this series is 1650.
00:18:13.500 --> 00:18:18.300
Example 4: we are working with sigma notation, so you need to know how to read this notation.
00:18:18.300 --> 00:18:35.700
And I am going to start with 4 and end with 14; and I can find the first term, because I am also given the formula for a general term in this series.
00:18:35.700 --> 00:18:42.800
So, to find the sum of the series, let's just start out by finding the first term, because we know we are going to need that.
00:18:42.800 --> 00:18:51.100
a₁...we are going to begin with 4, so for the first term, n is going to equal 4.
00:18:51.100 --> 00:19:03.200
It is 2 times 4, minus 3; a₁ = 8 - 3, so the first term is going to be equal to 5.
00:19:03.200 --> 00:19:06.600
Now, recall that we have two formulas that we can use to find the sum.
00:19:06.600 --> 00:19:10.000
We have one formula that involves knowing the common difference.
00:19:10.000 --> 00:19:16.000
We have another formula that requires us to know the first term and the last term.
00:19:16.000 --> 00:19:22.000
I found the first term; since I know that, for the last term, n = 14, I can find that, as well.
00:19:22.000 --> 00:19:32.000
And what that means is that I can use this formula: that the sum of the series is going to be equal to n/2, times the first term, plus the last term.
00:19:32.000 --> 00:19:56.100
Therefore, let me find the last term: a < font size="-6" > n < /font > = 2(14) - 3; this is going to give me a < font size="-6" > n < /font > = 28 - 3, so the last term equals 25.
00:19:56.100 --> 00:20:05.900
Now, what is n? Well, this is telling me that the number of terms...I would have to take each number from 4 through 14, inclusive.
00:20:05.900 --> 00:20:09.600
And if you figure that out, that is actually 11 terms, because you are including 14.
00:20:09.600 --> 00:20:15.100
So, starting with 4 and going up through 14, there are actually 11 terms, so n = 11.
00:20:15.100 --> 00:20:22.000
It is really a < font size="-6" > 11 < /font > = 15 I am asked to find the sum of these 11 terms.
00:20:22.000 --> 00:20:28.600
And I can do that now, because I know that I have n = 11, divided by 2, and then I am going to get the first term
00:20:28.600 --> 00:20:39.200
(that is 5), plus the last term, which is 25; so the sum is 11/2, times 5 + 25 (is 30).
00:20:39.200 --> 00:20:51.000
Therefore, this cancels; I am going to get 11 times 15, and that is simply 165.
00:20:51.000 --> 00:21:00.300
OK, so in sigma notation, this gives me a lot of information, because I saw that I knew the formula to find a particular term.
00:21:00.300 --> 00:21:04.400
And I knew the n for the first term and the n for the last term.
00:21:04.400 --> 00:21:07.900
So, I knew I could use this formula, because I could find the first and last terms.
00:21:07.900 --> 00:21:17.300
So, I made n equal to 4 to find the first term, which is 5; I made n equal to 14, which is to help me find the last term, which is 25.
00:21:17.300 --> 00:21:22.200
And then, I knew that, since it was going from 4 to 14, that n is equal to 11.
00:21:22.200 --> 00:21:30.100
Once I had first term, last term, and n, it was just a matter of calculating the sum, which was 165.
00:21:30.100 --> 00:21:36.000
That finishes up today's lesson on arithmetic series; thanks for visiting Educator.com!