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INSTRUCTORSCarleen EatonGrant Fraser
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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 an. In both cases, you will be given the values of a1 and n.
  • In some problems, you must first use the formula for the sum and then the formula for the nth term.

Arithmetic Series

Find the first term for the arithmetic series with
d = 4, n = 6, and S6 = 156
  • You can use the formula Sn = [n/2]( 2a1 + (n − 1)d )
  • Plug in the values for d, n and Sn
  • Sn = [n/2]( 2a1 + (n − 1)d )
  • 156 = [6/2]( 2a1 + (6 − 1)4 )
  • 156 = 3( 2a1 + (5)4 )
  • 156 = 3( 2a1 + 20 )
  • 156 = 6a1 + 60
  • 96 = 6a1
a1 = 16
Find the first term for the arithmetic series with
d = − 6, n = 9, and S9 = − 288
  • You can use the formula Sn = [n/2]( 2a1 + (n − 1)d )
  • Plug in the values for d, n and Sn
  • Sn = [n/2]( 2a1 + (n − 1)d )
  • − 288 = [9/2]( 2a1 + (9 − 1)( − 6) )
  • − 576 = 9( 2a1 + (8)( − 6) )
  • − 576 = 9( 2a1 − 48 )
  • − 576 = 18a1 − 432
  • − 144 = 18a1
a1 = − 8
Find the first term for the arithmetic series with
d = 5, n = 10, and S9 = 325
  • You can use the formula Sn = [n/2]( 2a1 + (n − 1)d )
  • Plug in the values for d, n and Sn
  • Sn = [n/2]( 2a1 + (n − 1)d )
  • 325 = [10/2]( 2a1 + (10 − 1)(5) )
  • 325 = 5( 2a1 + (9)(5) )
  • 325 = 5( 2a1 + 45 )
  • 325 = 10a1 + 225
  • 100 = 10a1
a1 = 10
Find the first three terms for the arithmetic series with
n = 8, an = 46, and Sn = 172
  • You can use the formula Sn = [n/2]( a1 + an );Sn = [n/2]( 2a1 + (n − 1)d )
  • Step 1: Plug in the values for n, an and Sn to find the first term.
  • Sn = [n/2]( a1 + an )
  • 172 = [8/2]( a1 + 46 )
  • 172 = 4(a1 + 46)
  • 172 = 4a1 + 184
  • − 12 = 4a1
  • a1 = − 3
  • Step 2 - Use the equation Sn = [n/2]( 2a1 + (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
  • a1 = − 3
  • a2 =
  • a3 =
  • a4 =
a2 = 4a3 = 11a4 = 18
Find the first three terms for the arithmetic series with
n = 40, an = 175, and Sn = 3880
  • You can use the formula Sn = [n/2]( a1 + an );
  • Step 1: Plug in the values for n, an and Sn to find the first term.
  • Sn = [n/2]( a1 + an )
  • 3880 = [40/2]( a1 + 175 )
  • 3880 = 20(a1 + 175)
  • 3880 = 20a1 + 3500
  • 380 = 20a1
  • a1 = 19
  • Step 2 - Use the equation Sn = [n/2]( 2a1 + (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
  • a1 = 19
  • a2 =
  • a3 =
  • a4 =
a2 = 23a3 = 27a4 = 31
Find the first three terms for the arithmetic series with
n = 30, an = 74, and Sn = 1350
  • You can use the formula Sn = [n/2]( a1 + an );Sn = [n/2]( 2a1 + (n − 1)d )
  • Step 1: Plug in the values for n, an and Sn to find the first term.
  • Sn = [n/2]( a1 + an )
  • 1350 = [30/2]( a1 + 74 )
  • 1350 = 15(a1 + 74)
  • 1350 = 15a1 + 1110
  • 240 = 15a1
  • a1 = 16
  • Step 2 - Use the equation Sn = [n/2]( 2a1 + (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
  • a1 = 16
  • a2 =
  • a3 =
  • a4 =
a2 = 18a3 = 20a4 = 22
Find the sum of the series
x = 16 (3x − 11 )
  • x = 16 (3x − 11 ) = (3(1) − 11) + (3(2) − 11) + (3(3) − 11) + (3(4) − 11) + (3(5) − 11) + (3(6) − 11) =
x = 16 (3x − 11 ) = − 3
Find the sum of the series
x = 110 (6x − 4 )
  • x = 110 (6x − 4 ) = (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4) + (6( ) − 4)
  • x = 110 (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 = 110 (6x − 4 ) = 290
Find the sum of the series
x = 19 (10x − 13 )
  • x = 19 (10x − 13 ) = (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13) + (10() − 13)
  • x = 19 (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 = 19 (10x − 13 ) = 333
Find the sum of the series
x = 16 (10x − 11 )
  • x = 16 (10x − 11 ) = + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11) + (10() − 11)
  • x = 16 (10x − 11 ) = (10(1) − 11) + (10(2) − 11) + (10(3) − 11) + (10(4) − 11) + (10(5) − 11) + (10(6) − 11)
x = 16 (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

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, a1 + a2 + a3, and on and on until the last term, an.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 s17 or s10.0162

There are two formulas: the first one involves taking the number of terms, dividing it by 2, and then multiplying it0166

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, s10.0211

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: s10 = 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, s10 = 5(10) +...10 - 1 is, of course, 9, times 4.0253

Therefore, s10 = 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 terms0272

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, a16, 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, s16 = n; in this case, I am looking for the sum of 16 terms, so n is 160311

(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 s16 = 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 an 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 an 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, an, is n + 3.0460

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 8.0472

For a2, I am then going to go to 6: so that is going to be 6 + 3; that is 9.0485

a3 =...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, a8 (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, s20, equaling 1005.0547

Since we know the common difference, we can use this formula: sn = 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 a1, plus n (n is given as 20), minus 1, times the common difference of 3.5.0571

This gives me 1005 = 10(2a1) + 19(3.5).0589

19 times 3.5 is actually 66.5; therefore, 1005 equals 10 times 2a1, which is 20a1, 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, an = 103, sn = 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 sn = 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 an, 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 an is 103.0698

I am going to multiply both sides by 2 to get 2204 = 19 times this, a1 + 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, sn,0740

equals n/2 times 2a1, 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, sn 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 a1 is 13; I am going to take 13 + the common difference of 5 to get the second term (that is 18).0810

So, a2 is 18; the third term--I am going to take 18, and I am going to add 5 to that to yield 23.0823

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 2a1, 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

an equals the first term, plus n - 1, times the common difference.0926

an is 126; so that shouldn't be 16--that is 126: an 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 s25 = n, which is 25, divided by 2, times 2 times that first term0990

(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 s25 = 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 s25; 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

a1...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; a1 = 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: an = 2(14) - 3; this is going to give me an = 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 a11 = 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 term1222

(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