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About this Lesson
 Type: Video Tutorial
 Length: 11:12
 Media: Video/mp4
 Use: Watch Online & Download
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 Download: MP4 (iPod compatible)
 Size: 121 MB
 Posted: 06/26/2009
This lesson is part of the following series:
College Algebra: Arithmetic Sequences (2 lessons, $2.97)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Intermediate Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/intermediatealgebra. The full course covers real numbers, equations and inequalities, exponents and polynomials, rational expressions, roots and radicals, relations and functions, the straight line, systems of equations, quadratic equations and quadratic inequalities, conic sections, inverse and exponential and logarithmic functions, and a variety of other AP algebra and advanced algebra.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
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So we're thinking about sequences, and sequences could have any shape they want, they could be anything you want. In fact, you could define them to be real exotic. I want to now share with you a very famous type of sequence, and a very popular type of sequence that has a lot of structure to it, and in fact occurs a lot in nature, and in mathematics. And those are actually called arithmetic sequences. Let me try to inspire it by looking at a very particular example. Let's see if we can figure out what the pattern is, and then try to isolate that pattern as sort of a metatheme for really what makes an arithmetic sequence arithmetic.
So here's an example of a sequence. I'm just going to write the terms out here, and see if we can get the pattern: 4, 7, 10, 13, and now I'll stop, and now it's time for us to guess what the next terms are going to be. Well, what you notice here is how did I get from 4 to 7. It looks like  now I'm just exploring here, there's no right or wrong, I'm just looking at it and saying, well, I added 3. And then if I add 3 again, notice I get to here; notice if I add 3 again, I get to here. That doesn't seem like a coincidence. Maybe, in fact, the next term would be adding 3 again, and getting 16, adding 3 again, getting 19, and so forth. These are all great guesses. An in fact, a sequence that has that property, where all you're doing is adding a fixed particular number to the previous person in order to get the next one, that is actually called an arithmetic sequence, because I'm just using arithmetic, I'm just using addition to add, and I keep upping the ante by a fixed amount.
If you'll notice that we take any two consecutive terms in this sequence and subtract the larger from the smaller, you always get 3. You see that? This minus that, 3; this minus that, 3. There's a constant difference between any two consecutive terms. And that difference in this case is 3. So if you take the next term and subtract the previous term, what I get in this case is 3. And that's a difference which I refer to as d. And if you think about it, how did I get from here all the way out? I take 4, I add 3 once. Then I add 3 again, then I add 3 again, then I add 3 again, then I add 3 again.
So I can actually write out a general formula for this. And how would the general formula for this look? Well, it would look like this: a[n] =  well, I started with 4, and then I'm going to add a whole bunch of threes, and how many threes do I add? Well, some number of them, and how many? Well, at the beginning, I add no threes. Then at the second term, I add 1 three. At the third term, I add 2 threes. At the fourth term, I add 3 threes. So it looks like it's always one less than the index that I'm at.
And let's actually check and see that's really okay. If n = 1, then this is 1  1, which is zero, and it's left with 4. If n = 2, then I have 2  1, which is 1, so I have 4 plus 1 three, which is 7. If I go out to 1, 2, 3, 4  let's go to the fourth term right now. If I go to the fourth term, a[4], then I have a 4  1, which you'll notice is 3; 3 times 3 is 9; and 9 plus 4 is 13. This checks. And, in fact, the general form for an arithmetic sequence is, you have some first person in the sequence, and then plus some common difference d times n  1. Notice it's the exact same thing we found here, but now I'm not telling you what the common difference between two consecutive people is, and I'm not telling you the first term. So this works for any arithmetic sequence. This is the defining trait of an arithmetic sequence. You start somewhere, and then you keep adding multiples of a particular fixed number d.
So let's try an example. Suppose that a[1], the first term, a[1] = 3, and d, the difference, is 5. Let's list the sequence. If it's an arithmetic sequence, it would start with 3, and then what do I do next? Well, then I add a 5. The d is the difference, so that's what I have to add to get the next term. So I add 5 to 3, and I get 8. And now I add another 5, 5 to 8, and get 13. Add another 5, and I get 18, and so on. So, in fact, this is generating this particular arithmetic sequence, starting at 3 and continually adding the 5  5 and 5 and 5 and 5 and so forth.
So there's the general formulation for it. How about this question: Suppose I tell you that a[1] = 2, so I start with the number 2, and that common difference is 11. Now the question is, what is a[21]? Well, one way to do that is to start just as we did before, find the first term is 2, then add 11, then add 11 again, and do that like 21 times and you get the answer. That's fine. However, there's a very easy way of doing that that will avoid doing all that calculation, which is to just write out the general formula using this fact. If I use that fact, what do I see? What I would see is a[n] in general = the first term, which in this case is 2, plus d, that difference, which is known to be 11, times (n 1).
So that's the general term. And if you want the 21st term, all I do is let n be 21. And you see I can compute this now real easily just putting in 21 right there. And what does that equal? That's 2 plus 11 times 20, and that's 2 plus 220, which is 222. That's the answer, that is the 21st term in this arithmetic sequence, starting with 2, with a difference of 11.
So that's how these things go. You can actually figure out any term you want by just knowing the initial person and the difference. In fact, you can even do more. Watch this  you can actually find anything you want as long as you know any two terms. If you know any two terms at all, you'll know the whole story.
Look at this question. How about if I tell you that a[12] in some arithmetic sequence, so here's an arithmetic sequence, the 12th term is 28. So I'm going to tell you that, and I also tell you that the 16th term is 36. That's all I tell you. Notice that the 16th term is not the adjacent term to the 12th term. Like there's there 12th term, the 13th term, the 14th term, the 15th term, and then finally the 16th term. But just knowing that, my question is, question: What is the 18th term? How would you do this? Well, the way that I would do this is to use this general formula, always. So if we go back to this, what I see is, well, a[12] equals the first term, which I don't know what it is, plus the difference  I don't know what that is  times what number? Well, 12 minus 1, which is 11. And I know that the 12th term is 28. I could do the same thing with a[16]; a[16] is equal to the first term, plus the difference, which I don't know, times 15, and I know that number has to work out to be 36. Now what happens if I were to subtract these two things right now? If I subtract, let's see what I get. You see, a[1] minus a[1] is zero, so those drop out, which is great, because I don't know what a[1] equals. And what about here? Here I have d(11) and I'm subtracting d(15) so what is that? Well, that would be negative how many d? Well, 4, 4 d. And what does that equal? Well, it equals 28 minus 36, and so what's 28 minus 36. Oh, goodness, I have no idea  like 8 with a minus sign in front, so 8. And now I can solve for d. I see that therefore d has to equal 2. So I just found the common difference. So what I could do, if I want to get the 18th term is all I have to do is start with 28 and just keeping adding 2 to it from the 12th all the way to the 18th. So let's try that.
So here's a[12], then there would be a[13], a[14]  what am I doing? I want the 18th term  sorry folks. I want the 18th term, that's going to be two more than the 16th, right? Why count all the way up to there? It would work, but lots of work. So etc. Instead, how about starting with a[16], a[17], and then a[18], and then I'm there. So I start with 36, I see the common difference is 2, so I add 2 to this to get the next term. If I add 2 to that, I get 38. If I add 2 to that, I get 40; so, therefore, the 18th term must be 40. So just knowing any two terms, by using this method of subtraction could always figure out what that common difference is. And then, using that common difference, I can figure out any term I want by just continually adding that common difference until I get up to the particular point in the arithmetical sequence that I'm looking for. So arithmetical sequences are really great. If you know any two terms, you know everything about it. It's really rich with structure, rich with a lot of arithmetic, hence the name "arithmetic sequence." Enjoy them for yourselves, make up your own, and I'll see you at the next lecture.
Sequences and Series
Arithmetic Sequences
Finding Termis in an Arithmetic Sequence Page [2 of 2]
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