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### Arithmetic Series

- There are two formulas for the sum. Use one if you are given d and the other if you are given a
_{n}. In both cases, you will be given the values of a_{1}and n. - In some problems, you must first use the formula for the sum and then the formula for the nth term.

### Arithmetic Series

d = 4, n = 6, and S

_{6}= 156

- You can use the formula S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - Plug in the values for d, n and S
_{n} - S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - 156 = [6/2]( 2a
_{1}+ (6 − 1)4 ) - 156 = 3( 2a
_{1}+ (5)4 ) - 156 = 3( 2a
_{1}+ 20 ) - 156 = 6a
_{1}+ 60 - 96 = 6a
_{1}

_{1}= 16

d = − 6, n = 9, and S

_{9}= − 288

- You can use the formula S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - Plug in the values for d, n and S
_{n} - S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - − 288 = [9/2]( 2a
_{1}+ (9 − 1)( − 6) ) - − 576 = 9( 2a
_{1}+ (8)( − 6) ) - − 576 = 9( 2a
_{1}− 48 ) - − 576 = 18a
_{1}− 432 - − 144 = 18a
_{1}

_{1}= − 8

d = 5, n = 10, and S

_{9}= 325

- You can use the formula S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - Plug in the values for d, n and S
_{n} - S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - 325 = [10/2]( 2a
_{1}+ (10 − 1)(5) ) - 325 = 5( 2a
_{1}+ (9)(5) ) - 325 = 5( 2a
_{1}+ 45 ) - 325 = 10a
_{1}+ 225 - 100 = 10a
_{1}

_{1}= 10

n = 8, a

_{n}= 46, and S

_{n}= 172

- You can use the formula S
_{n}= [n/2]( a_{1}+ a_{n});S_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - Step 1: Plug in the values for n, a
_{n}and S_{n}to find the first term. - S
_{n}= [n/2]( a_{1}+ a_{n}) - 172 = [8/2]( a
_{1}+ 46 ) - 172 = 4(a
_{1}+ 46) - 172 = 4a
_{1}+ 184 - − 12 = 4a
_{1} - a
_{1}= − 3 - Step 2 - Use the equation S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) to find the common difference - 172 = [8/2]( 2( − 3) + (8 − 1)d )
- 172 = 4( − 6 + 7d )
- 172 = − 24 + 28d
- 196 = 28d
- d = 7
- Step 3 - Find the next three terms using the 1st term and common difference
- a
_{1}= − 3 - a
_{2}= - a
_{3}= - a
_{4}=

_{2}= 4a

_{3}= 11a

_{4}= 18

n = 40, a

_{n}= 175, and S

_{n}= 3880

- You can use the formula S
_{n}= [n/2]( a_{1}+ a_{n}); - Step 1: Plug in the values for n, a
_{n}and S_{n}to find the first term. - S
_{n}= [n/2]( a_{1}+ a_{n}) - 3880 = [40/2]( a
_{1}+ 175 ) - 3880 = 20(a
_{1}+ 175) - 3880 = 20a
_{1}+ 3500 - 380 = 20a
_{1} - a
_{1}= 19 - Step 2 - Use the equation S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) to find the common difference - 3880 = [40/2]( 2(19) + (40 − 1)d )
- 3880 = 20( 38 + 39d )
- 3880 = 760 + 780d
- 3120 = 780d
- d = 4
- Step 3 - Find the next three terms using the 1st term and common difference
- a
_{1}= 19 - a
_{2}= - a
_{3}= - a
_{4}=

_{2}= 23a

_{3}= 27a

_{4}= 31

n = 30, a

_{n}= 74, and S

_{n}= 1350

- You can use the formula S
_{n}= [n/2]( a_{1}+ a_{n});S_{n}= [n/2]( 2a_{1}+ (n − 1)d ) - Step 1: Plug in the values for n, a
_{n}and S_{n}to find the first term. - S
_{n}= [n/2]( a_{1}+ a_{n}) - 1350 = [30/2]( a
_{1}+ 74 ) - 1350 = 15(a
_{1}+ 74) - 1350 = 15a
_{1}+ 1110 - 240 = 15a
_{1} - a
_{1}= 16 - Step 2 - Use the equation S
_{n}= [n/2]( 2a_{1}+ (n − 1)d ) to find the common difference - 1350 = [30/2]( 2(16) + (30 − 1)d )
- 1350 = 15( 32 + 29d )
- 1350 = 480 + 435d
- 870 = 435d
- d = 2
- Step 3 - Find the next three terms using the 1st term and common difference
- a
_{1}= 16 - a
_{2}= - a
_{3}= - a
_{4}=

_{2}= 18a

_{3}= 20a

_{4}= 22

∑

_{x = 1}

^{6}(3x − 11 )

- ∑
_{x = 1}^{6}(3x − 11 ) = (3(1) − 11) + (3(2) − 11) + (3(3) − 11) + (3(4) − 11) + (3(5) − 11) + (3(6) − 11) =

_{x = 1}

^{6}(3x − 11 ) = − 3

∑

_{x = 1}

^{10}(6x − 4 )

- ∑
_{x = 1}^{10}(6x − 4 ) = (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) - ∑
_{x = 1}^{10}(6x − 4 ) = (6(1) − 4) + (6(2) − 4) + (6(3) − 4) + (6(4) − 4) + (6(5) − 4) + (6(6) − 4) + (6(7) − 4) + (6(8) − 4) + (6(9) − 4) + (6(10) − 4)

_{x = 1}

^{10}(6x − 4 ) = 290

∑

_{x = 1}

^{9}(10x − 13 )

- ∑
_{x = 1}^{9}(10x − 13 ) = (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) - ∑
_{x = 1}^{9}(10x − 13 ) = (10(1) − 13) + (10(2) − 13) + (10(3) − 13) + (10(4) − 13) + (10(5) − 13) + (10(6) − 13) + (10(7) − 13) + (10(8) − 13) + (10(9) − 13)

_{x = 1}

^{9}(10x − 13 ) = 333

∑

_{x = 1}

^{6}(10x − 11 )

- ∑
_{x = 1}^{6}(10x − 11 ) = + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11) - ∑
_{x = 1}^{6}(10x − 11 ) = (10(1) − 11) + (10(2) − 11) + (10(3) − 11) + (10(4) − 11) + (10(5) − 11) + (10(6) − 11)

_{x = 1}

^{6}(10x − 11 ) = 144

*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 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
- What are Arithmetic Series? 0:11
- Common Difference
- Example: Arithmetic Sequence
- Example: Arithmetic Series
- Finite/Infinite Series
- Sum of Arithmetic Series 2:27
- Example: Sum
- Sigma Notation 5:53
- Index
- Example: Sigma Notation
- Example 1: First Term 9:00
- Example 2: Three Terms 10:52
- Example 3: Sum of Series 14:14
- Example 4: Sum of Series 18:13

### Algebra 2

### Transcription: Arithmetic Series

*Welcome to Educator.com.*0000

*In a previous lesson, we talked about arithmetic sequences; and in this lesson, we will continue on with the discussion, to discuss arithmetic s0ries.*0002

*First of all, what are arithmetic series? An arithmetic series is the sum of the terms of an arithmetic sequence.*0012

*So, I will briefly review arithmetic sequences; but if you need a complete review of that,*0019

*go and check out the previous lesson, and then move on to arithmetic series.*0024

*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.*0028

*A typical arithmetic sequence would be something like 20, 40, 60, 80, 100.*0043

*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.*0051

*So, this is the sequence: it is a list of numbers; the series is actually a sum of numbers,*0063

*so I am going to add a + between each number: 20 + 40 + 60 + 80 + 100.*0071

*So, an arithmetic series is in the general form: first term, a _{1} + a_{2} + a_{3}, and on and on until the last term, a_{n}.*0082

*This is a finite arithmetic series, because there is a specific endpoint; there is a finite number of terms.*0097

*As with arithmetic sequences (those could be finite or infinite), you could have finite or infinite arithmetic series.*0106

*So, looking at a different sequence: 100, 200, 300, 400, and then the ellipses to tell me that this is an infinite sequence:*0115

*I could have a corresponding series, 100 + 200 + 300 + 400 + ...on and on; this is an infinite arithmetic series.*0127

*There is a formula (or actually, two formulas) here to allow you to find the sum of an arithmetic series.*0148

*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.*0154

*And that would be s _{17} or s_{10}.*0162

*There are two formulas: the first one involves taking the number of terms, dividing it by 2, and then multiplying it*0166

*by 2 times the first term, plus the quantity (n - 1) times the common difference.*0173

*The second term is the sum of the first n terms: again, it equals the number of terms divided by 2.*0179

*But this time, you are going to multiply that by the sum of the first and last terms.*0185

*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.*0190

*So, you have to choose a formula, depending on what you know.*0199

*For example, let's say that a series has, as its first term, a 5: 5 is the first term.*0202

*And it has a common difference of 4, and you are asked to find the sum of the first 10 terms, s _{10}.*0211

*I am looking, and I see that I have the common difference; so I am going to go ahead and use the first formula.*0220

*And let's go ahead and do that: s _{10} = 10, because, since I am finding the sum of the first 10 terms,*0229

*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).*0241

*So, s _{10} = 5(10) +...10 - 1 is, of course, 9, times 4.*0253

*Therefore, s _{10} = 5 times...9 times 4 is 36; 36 + 10 is 46.*0262

*And if you multiply that out, or use your calculator, you will find that the sum of the first 10 terms*0272

*in a series with this first term and this common difference is actually 230.*0276

*So, that is one example; in another example, perhaps you are asked the sum of the 16 terms in a series,*0283

*if the first term is 20, and the last term of the ones we are trying to find, a _{16}, is equal to -10.*0292

*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.*0305

*So, s _{16} = n; in this case, I am looking for the sum of 16 terms, so n is 16*0311

*(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.*0323

*So, this is s _{16} = 8, times 20 - 10, which is 10; that gives me 80 as the sum of the first 16 terms.*0336

*You need to learn both formulas and simply know when to use a particular one.*0348

*Sigma notation: a series can be written in a concise form, using what is called sigma notation.*0353

*And what sigma refers to is the Greek letter Σ; and in this case, sigma means "sum."*0359

*So, what this symbol means, in the context in which we are using it, is "sum."*0366

*And you will see it written something like this: you will see a letter here, called the index (that variable is called the index).*0371

*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.*0378

*i = 1; I am just giving an example; and let's say, up here, the upper index is going to be 10.*0385

*And then, there will be the formula for the general term, which we talked about earlier on,*0393

*when we talked about arithmetic sequences and the formula for a _{n} that is particular to a sequence.*0399

*So, if you need to go ahead and review that, it is in the previous lecture.*0406

*This is the formula for the general term of the sequence.*0409

*And this is read as "the sum of a _{n} as i goes from 1 to 10."*0412

*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.*0420

*That is in general; let's talk about a specific example with a specific formula.*0429

*You could have another arithmetic series, written in sigma notation, where n goes from 5 to 12.*0435

*So, n goes from 5 to 12; so I am going to start with 5, and I am going to end with 12.*0448

*That means that there are actually 8 terms; there are 8 terms in the series, because it is 5 through 12, inclusive.*0454

*And the formula, we are going to say, is n + 3: so the formula to find a particular term, a _{n}, is n + 3.*0460

*I could find the terms, then, by saying that for the first term, a _{1}, I am going to use 5 as my n, so it equals 5 + 3; therefore, it is 8.*0472

*For a _{2}, I am then going to go to 6: so that is going to be 6 + 3; that is 9.*0485

*a _{3} =...then I am going to go to the next value, which is 7: 7 + 3 = 10; and on up.*0497

*So, if I wanted to find the last term, a _{8} (because there are 8 terms), then I would put in 12; and 12 + 3 is 15.*0512

*There would be terms in between here, of course.*0524

*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.*0527

*But now, this is just a different way of writing them.*0539

*We need to find the first term of the arithmetic series with a common difference of 3.5 and equal to 20,*0541

*and the sum of the terms, s _{20}, equaling 1005.*0547

*Since we know the common difference, we can use this formula: s _{n} = n/2, times 2 times the first term, plus (n - 1)d.*0553

*Except, in this case, I am not looking for the sum: I have the sum; I am looking for the first term.*0566

*Therefore, 1005 = 20/2, times 2 times a _{1}, plus n (n is given as 20), minus 1, times the common difference of 3.5.*0571

*This gives me 1005 = 10(2a _{1}) + 19(3.5).*0589

*19 times 3.5 is actually 66.5; therefore, 1005 equals 10 times 2a _{1}, which is 20a_{1}, plus 10(66.5), which is 665.*0601

*1005; subtract 665 from both sides to get 340 = 20 times that first term.*0622

*Divide both terms by 20, and you will get that the first term is equal to 17.*0631

*So, we use this formula for the sum of the series; but in this case, we were looking for the first term.*0636

*We had the sum; we were looking for the first term.*0646

*And the solution is that the first term is 17.*0647

*In the second problem, we are asked to find the first three terms of the arithmetic series with n = 17, a _{n} = 103, s_{n} = 1102.*0653

*In order to find the first three terms, I need to find the common difference.*0666

*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.*0670

*I can use the formula s _{n} = n/2 times the first term plus the last term, because I don't have the common difference--I am looking for it.*0680

*But what I do have is a _{n}, so my first step is going to be to find the first term.*0691

*The sum is 1102; n is 19; I don't know the first term; and I know that a _{n} is 103.*0698

*I am going to multiply both sides by 2 to get 2204 = 19 times this, a _{1} + 103.*0710

*I am then going to divide both sides by 19, and that comes out to 116 = the first term, plus 103.*0721

*And then, I just subtract 103 from both sides; and now I have my first term.*0729

*So, I am asked to find the first three terms: the first term is 13.*0734

*To find the next two terms, I find the common difference.*0737

*What I am going to do is switch to the other formula: that other formula, s _{n},*0740

*equals n/2 times 2a _{1}, the first term, plus n - 1 times the common difference.*0744

*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.*0754

*So again, s _{n} 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.*0761

*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.*0775

*I am going to divide both sides by 19 to get 116 = 26 + 18d.*0790

*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.*0798

*I have a _{1} is 13; I am going to take 13 + the common difference of 5 to get the second term (that is 18).*0810

*So, a _{2} is 18; the third term--I am going to take 18, and I am going to add 5 to that to yield 23.*0823

*So, I was asked to find the first three terms, and I did that by first using this formula to find the first term,*0836

*then going to the other formula and finding the common difference to get 13, 18, and 23 as my solutions.*0843

*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.*0854

*That means that what I have is the first term and the last term.*0867

*I am going to find the sum using this formula, because I can figure out my common difference.*0874

*So, I am going to use the formula n/2, times 2a _{1}, plus a - 1, times d.*0884

*I could easily find the common difference, because I know I will just take 11 - 6, so I have a common difference of 5.*0896

*Looking at this formula, the only issue is going to be that I don't know n.*0904

*But I can figure out n; and that is because I have another formula.*0910

*Recall the formula for the general term: we discussed this in the lecture on arithmetic sequences.*0917

*a _{n} equals the first term, plus n - 1, times the common difference.*0926

*a _{n} is 126; so that shouldn't be 16--that is 126: a_{n} is 126, and I have the first term equal to 6; and I am solving for n.*0937

*I know that the common difference is 5.*0956

*126...and then I am subtracting 6 from both sides: that gives me 120 = (n - 1)5, so 120 = 5n - 5.*0959

*Adding 5 to both sides gives me 125 = 5n; 125/5 is 25, so I have n = 25.*0970

*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.*0982

*So, let's go ahead and use that: it is actually s _{25} = n, which is 25, divided by 2, times 2 times that first term*0990

*(which is 6), plus (n - 1) (n is 25, minus 1), times the common difference of 5.*1000

*Now, it is just a matter of simplifying: this is going to give me 25/2 times 12, plus 24 times 5.*1009

*So, the sum equals 25/2, times 12; and then if you multiply out 24 times 5, you will get 120.*1022

*Therefore, the sum equals 25/2; 120 + 12 is going to give you 132.*1033

*Now that I have gotten it down to this point, I can simplify,*1050

*because this is going to give me 132/2, times 25; so that is s _{25} = 25 times 66.*1054

*You can multiply it out; or it is a good time to use your calculator to find that the sum equals 1650.*1065

*So again, in order to use this formula, I had my common difference.*1074

*I didn't have n; I solved for n; n equals 25.*1077

*I went back in, substituted those values in, and then came out with s _{25}; the sum of this series is 1650.*1081

*Example 4: we are working with sigma notation, so you need to know how to read this notation.*1093

*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.*1098

*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.*1116

*a _{1}...we are going to begin with 4, so for the first term, n is going to equal 4.*1123

*It is 2 times 4, minus 3; a _{1} = 8 - 3, so the first term is going to be equal to 5.*1131

*Now, recall that we have two formulas that we can use to find the sum.*1143

*We have one formula that involves knowing the common difference.*1146

*We have another formula that requires us to know the first term and the last term.*1150

*I found the first term; since I know that, for the last term, n = 14, I can find that, as well.*1156

*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.*1162

*Therefore, let me find the last term: a _{n} = 2(14) - 3; this is going to give me a_{n} = 28 - 3, so the last term equals 25.*1172

*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.*1196

*And if you figure that out, that is actually 11 terms, because you are including 14.*1206

*So, starting with 4 and going up through 14, there are actually 11 terms, so n = 11.*1209

*It is really a _{11} = 15 I am asked to find the sum of these 11 terms.*1215

*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*1222

*(that is 5), plus the last term, which is 25; so the sum is 11/2, times 5 + 25 (is 30).*1228

*Therefore, this cancels; I am going to get 11 times 15, and that is simply 165.*1239

*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.*1251

*And I knew the n for the first term and the n for the last term.*1260

*So, I knew I could use this formula, because I could find the first and last terms.*1264

*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.*1268

*And then, I knew that, since it was going from 4 to 14, that n is equal to 11.*1277

*Once I had first term, last term, and n, it was just a matter of calculating the sum, which was 165.*1282

*That finishes up today's lesson on arithmetic series; thanks for visiting Educator.com!*1290

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