 Episode 34 of Math 1050 College Algebra, this is a study review for exam number five. This is the last regular exam we have before the final exam, which isn't too far away. Let's see, let's look at the list of objectives for this episode. First of all, for this exam, you should be prepared to solve problems similar to those that have been discussed in class and assigned for homework. You know, we haven't worked an example of every type of homework problem in class, so it's important that you do those homework assignments. You will be needing a calculator on this exam. You remember what was it two exams ago? You could use a calculator for logarithms and exponential functions, but the calculator problems will be given on a separate sheet of paper, and then you'll have to put your calculator away for the rest of the exam. So be sure to bring your calculator with you. Of course, you won't be able to use notes or use your text on this test. Any scratch paper that you're given has to be turned in, even the paper that you don't use. And finally, we encourage you to show your work to justify your answers so you can get partial credit. As a matter of fact, there are some problems where if you don't show any work, you wouldn't get any credit because the work is so important. Now, you know, the material that we've covered for this exam, I think we could divide into three pieces. First of all, there was the conic sections, parabolas, ellipses, and hyperbolas, and those three topics go together, and you might think of that as sort of one block of the exam. Then the next portion of the exam is over sequences and series, and we've been talking about, in particular, arithmetic and geometric sequences and series. And then finally, we talked about annuities, present value of an annuity, and then computing the monthly payments or the regular payments if you're installment buying. There are some formulas there that you'll need to know. So if you think about studying for this exam, there are three major blocks, and we'll be going over some of these ideas today, but I think it helps to sort of distinguish between those three portions. Okay, let's look at sort of a short review of episode 28. Let's bring up the next graphic. This is sort of an abbreviated list of items from episode 28. I wouldn't go strictly by this list, but I think these are some of the key points that you'll need to know for this exam from that very first episode. And that first episode was on parabolas. Now, you remember parabolas, ellipses and hyperbolas are all called conic sections because they can be formed by taking cross sections of a double cone, and if you cut one of the cones parallel to one of the edges, you get a parabola. However, the way we've investigated parabolas was from a different point of view in episode 28, and that was to look at it on the Cartesian, look at parabolas on the Cartesian plane. And so this first item is significant to that definition. You should know that points on a parabola are equidistant from its focus and its directorics. Now, you know, the focus is a point and the directorics is a line, and if you'll come to the green screen here, let me just remind you that on the XY plane, if I pick a point above the origin at coordinates zero, p, and if I pick a line parallel to the x-axis, p units below the x-axis, this would be the line y equals negative p, then when I draw a parabola, that every point on the parabola, let's say we pick this point right here, this point, XY, the distance to the focus and the distance perpendicular to the directorics, those two distances are the same, and that's the beginning of our discussion for parabolas. Now, the second item here under our list of topics from episode 28 is to know two standard equations for a parabola, and these are the two standard equations that involve whether the parabola opens up or down or if the parabola opens left or right. For example, the equation for a parabola that opens up or down is x squared equals four p y. This is the same p that we used to locate the focus in the directorics, and these parabolas, if p is positive, will open up. So this is for p greater than zero, and it'll open down in the case that p is less than zero. On the other hand, the equation y squared equals four p x is the equation of a parabola that opens left or right, and it'll open right if p is positive, and it'll open left if p is negative. So you should be familiar with those two equations and be able to solve problems that involve those equations. For example, let's just take a problem right here while we're on the green screen. Suppose I wanted to graph the equation two x plus y squared equals zero. Well, the first thing I need to do is put it in standard form, that's to isolate the square. So I'll write this as y squared equals negative two p times x, and if you compare that with the equation y squared equals four p x, that tells me that four p is equal to negative two. So if four p is negative two, then p is equal to negative one half. And because I have a y squared in the equation, that tells me this parabola opens to the right or to the left, and p is negative, so it opens to the left. So right below this I'll just say opens to the left. And that means that when I go to graph it, I'm gonna find my focus on the negative x-axis, and so the focus will be at the point negative one half zero. So if this is negative one and this is plus one, then the focus will be at negative one half right here. And what about the directorics? Well, the directorics will be a vertical line on the other side of the y-axis. So the directorics will be the line x equals positive one half. So I'll just draw the directorics coming through right here. And the vertex, of course, is at the origin unless we've shifted the parabola off the origin, and we haven't had any transformations in this equation. So the parabola opens to the left like so. Now, when you draw your parabola, if you don't draw it accurately enough to make it say a little bit narrower, if it should be a little bit wider, as long as it's reasonable, we won't take off for this. So you just wanna make sure that it passes through the vertex at the origin, that you've located the focus and you've located the directorics. And right here I might label this as x equals one half, and I'll just put an F on top of the focus point because there's not enough room to label the coordinates. We have that labeled over here. Okay, let's go back to our list of topics under episode 28. Okay, now, you should also be able to sketch the graph of a parabola when given its standard equation and give its focus and directorics. I guess we've just done that. And then the last item here is to be able to find the equation of a parabola when given information about the parabola, such as its focus or its directorics. Okay, well, let's take an example of that here. Suppose we said that the focus for a parabola was at zero three. And suppose we said that the vertex was at the origin. So given this information, what is the equation of this parabola? Well, I think what I would do is draw an illustration of this. You know, a lot of times drawing a picture, a picture's worth a thousand words. So if we locate this information in a graph, I think we can kind of get our bearings. The focus should be at zero three. So I'll put the focus right here. We'll put a little f beside it. And this is three up here on the y-axis. And the vertex is at the origin. So that tells me that this will be in standard form. There hasn't been a transformation involved because we haven't shifted it off of the origin anyway. And that means the directorics then should be three units below the x-axis. So the directorics should be right across here. This would be the line y equals negative three. Now I ask you, do you think when you draw this parabola, should it be fairly wide or should it be fairly thin? And the answer is that it should be a rather flat, wide parabola because the value of p is relatively large. You see, this is the point zero, this is the point zero p, so p is equal to three. And because the parabola opens up, this is gonna be of the form x squared equals four p y. And if I substitute in p equals three, we have x squared equals 12 y. So this is the answer that I would be expecting to see on your paper for the equation of this parabola. But if we go a little bit further, you know, we could write this as 12 y equals x squared if I turned it around. And if I solve for y, this says 112th x squared. Now, thinking back to what we saw very early in this course, when you put a 112th in front of one of our fundamental functions, what that does is to compress it. And a 112th is a major compression. So this is gonna be a very wide parabola and the parabola is gonna open up, but very, very slowly. And you see points on the parabola, say if I pick a point right here, is the same distance from the focus as it is from the directorics if I've drawn it properly. So this looks like this is a rough sketch of the graph x squared equals 12 y. And again, I'm writing it in the standard form for a parabola as a conic section, whereas down here I've written it in standard form for a transformation that is a compression of a fundamental graph. So, but this is the answer that I'd be looking for on the test. Okay, now there are a few things that aren't on this list. If we go back to that list one more time, there just wasn't enough room to write everything down about parabolas that you need to know, but let me just point out a few other things that you don't see listed here. But if you go to the website, you'll see these things listed, and that is you should be able to graph a transformation of a parabola that's been shifted left or right or up or down. Also, you should know about the reflective property of a parabola. So let's talk about both of those things quickly here on the green screen. Suppose I wanted to graph this parabola. Let's say it's x minus two squared equals negative four times y plus one. Well, you notice this is a variation of the standard equation x squared equals four p, y. Because there's a negative two, there's a two subtracted from the x, that means the graph has been shifted to the right two. And because there's a plus one added to the y, that means it's been shifted down one. So the vertex is no longer at the origin, but we're gonna go over two and down one. So that tells me that the vertex is gonna be at two negative one is the vertex. So let's make a note of that right now. The vertex is at two negative one. Also, let's see, it looks like four p is the same thing as negative four. Four p is negative four. Let's take those arrows out. And that says that p is negative one. So that means I'll be moving one unit away from the vertex to find the focus. And this parabola opens either up or down. That's a negative four, so it opens down. So that says the focus is one unit below. So if I go down one more unit, I'll have the focus right here. So when I go down one from the vertex, I'll be at the point two negative two. So the focus is at two negative two. Now let me ask you, where is the directorics? Well, I'd have to go up one unit from the vertex and it'll be horizontal, so it'll be right on the x-axis. So this is the directorics. So the directorics has the equation y equals zero. And so when I draw my parabola, I'm gonna draw it fairly wide, because for p equals negative one, I think that's a relatively large value of p. So I'm gonna draw my parabola rather wide. But of course there's no accuracy in this without plotting more points, but that's how I would determine how I'd represent it. Okay, so here's an example of a parabola that's been translated to the right and down. Now finally, the last thing you need to remember about a parabola that's not on the list is the reflective property. And let me just draw a quick illustration of that here too. Suppose I have a parabola that opens to the right and let's say its focus is right about here. If you remember, if this were a reflective surface, then if I had light coming in parallel to the central axis, parallel to the x-axis, it would reflect off of the reflective surface and it would come right to the focus. Same thing over here. If I had light coming in there, light coming in here, light coming in here, all of these beams of light would be reflected to the focus. And what's more, if they were all lined up here parallel and if they all cross, if all these rays of light cross this line at the same moment, they would all arrive at the focus at the same moment. This is the reflective property. Now on the other hand, if this were the back of a headlight on your car, if you put the element of the light right here, then the light which went up and bounced off a reflective surface would bounce parallel to the central axis and now it would be going out. And this is why your headlights tend to point straight out along the highway. There is some light, of course, that escapes that goes out here diagonally and misses the end of the headlight. That's why you get sort of an aura of light around the main beam, which is going straight down the highway. Okay, I think we better move on to episode 29 and look at ellipses. Now, you remember, first of all, that we introduced ellipses when we took a cross-section of the double cone and we cut one of the cones so we entered on one side and came out on the other so we got a closed curve and sort of an elongated circle, a stretched circle, and that was the ellipse. But when we introduced ellipses in our textbook, we looked at it from an analytic geometry point of view and that brings us to this first item in episode 29. You should know that the sum of the distances from a point on an ellipse to the two foci is constant. Let me just illustrate that. If I have an ellipse and if I have two focus points, I won't put a coordinate system on here, but if this is focus number one and if this is focus number two, then if you pick a point on the ellipse, then the distance to focus number one added to the distance to focus number two, I'll call this distance D1 plus D2, that distance was constant. And when we derived our equation, we call that total distance 2A. So we said that D1 plus D2 is equal to 2A. Now of course I'm not gonna ask you to derive any of these equations again, but you should know the standard formulas for ellipses and those equations are x squared over a squared plus y squared over b squared equals one. That's one of the standard equations. The other standard equation is x squared over b squared plus y squared over a squared is one. And while I'm drawing my axes, let me be asking you to think about what's the distinction between how we'll draw these ellipses? Well you see in both of these equations we assume that a is greater than b and that each of these ellipses has two foci that are located at plus or minus c on its major axis. Well if the larger value is a, then the larger square of the two is a squared, whichever variable a squared lies under, that's my major axis. So x is the major axis, that means the foci are on that axis. So if I go out of distance c and if I go back a distance negative c, so this is the point c0 and the point negative c0, those are my two foci. Then if I go further out of distance a and if I go back a distance negative a, those are my major axis intercepts and my ellipse passes through those two points. And the minor axis intercepts on the y-axis are b and negative b. Actually that would be zero b and zero negative b. And the relationship between a and b and c is that b squared equals a squared minus c squared. That's the relationship between a and b and the focus value c. Now over in the other equation, if a is the larger value a squared is under y squared, that tells me that the foci will be on the y-axis. So I would go up c and I would go down negative c for the foci and then I would go up even further to a and to negative a and those are my major axis intercepts. And so my ellipse will look like this. And it crosses the minor axis, in this case the y-axis at b and at negative b. Now there are so many possible problems that you could work here with ellipses that we can't work examples of each one of these today but let's take an example of one problem that we might wanna solve. Suppose the question is to graph this ellipse and the equation is x squared over 25 plus y squared over 16 is equal to one. Suppose I wanna graph this ellipse and along the way we wanna decide where are the foci? Where are the major axis intercepts? What are the minor axis intercepts? And what is the eccentricity? What is the eccentricity for this ellipse? Well, let's talk about all these things with this example. First of all, the larger square here is 25 and so that tells me that a squared is 25 and b squared is 16. So in other words that implies that a is equal to five and b is equal to four. And the way I calculate c squared is using my equation b squared equals a squared minus c squared. So we have that 16 equals 25 minus c squared which tells me that c squared is equal to 25 minus 16 if I juggle the terms around a little bit and that's equal to nine. So that tells me that c is equal to three. We'll choose c to be the positive square root like we did for a and b. Okay, so if I erase this, let me write down c is equal to three with this other information. So c squared was nine and c was equal to three. Okay, so now we're ready to draw our graph and the larger square was under the x squared so I know that the foci are on the major axis which is the x axis. So I'm gonna go to the right three units for a focus and I'll go to the left three units and then I'll go further and find the major axis intercepts at five and at negative five. Now if I go up the y axis I'll go up four and I'll go down four and so my ellipse passes through these four points. You know, these intercepts are sort of equivalent to target points that we saw for our fundamental graphs earlier in this course. So if I locate those four intercepts then I'll draw my ellipse through those and every point on the ellipse, let's say if I pick a point right here, every point on the ellipse has the property that the distance to the two foci is constant and the sum of those two distances is two times a and two times a would be 10 so the total distance here would be 10 for the length of that string let's say. Okay, so where are the foci? Well the foci must be at plus or minus three zero that's this point and this point. What are the major axis intercepts? It looks like that would be at plus or minus five zero. What are the minor axis intercepts? It looks like they would be at zero plus or minus four and to find the eccentricity which we abbreviated by the letter E, remember we don't want to get that confused with our natural exponential number. The eccentricity is the ratio of C over A and C was three and A was five so it's three-fifths or as a decimal it'd be 0.6. Okay, so there's one example of a problem using ellipses but I think we should mention the reflective property of an ellipse before we go on to hyperbolas. This isn't on our graphic for episode 29 but let me just remind you that in an ellipse if these are the two foci that if the ellipse represented a reflective surface and if we had a light source at one focus the light that emanates would bounce off and be reflected to the other focus. The same goes in case the light goes the opposite direction it would bounce off the reflective surface and it would head toward the other focus and even if it went past the second focus it would be reflected and come back to the other focus. And this was the basis for a whispering chamber that the sound that emanates from one focus will be collected at the other focus. Okay, now let's go to episode 30. Now this episode was describing hyperbolas and I think hyperbolas probably have a little extra information with those two asymptotes that make the hyperbolas a little bit trickier than parabolas and ellipses. If you remember we first saw hyperbolas when we took a double cone and we cut the double cone parallel to the central axis through the two cones we cut the upper and the lower cone and we got two branches for the hyperbola. But when we investigated hyperbolas from an analytic geometry point of view we said that as it says in the first item here that the difference of the distances from a point on a hyperbola to its two foci is constant. So for the ellipse it was the sum of the distances and now it's the difference of the distances. So let's draw an illustration of this. Suppose I have a hyperbola that looks like this let's say and if I pick a point on the hyperbola say the point x, y and the two foci which lie just inside of each branch focus number one, focus number two if you take the larger distance minus the smaller distance to the foci the difference of those two distances is constant and it's what we called two a. Now what's the equation of a hyperbola such as this? Well the standard equation for a hyperbola that opens to the left and right would be x squared over a squared minus y squared over b squared is one. Now in this case a does not have to be larger than b, a and b can be equal or a can actually be smaller than b. The way I tell that the foci lie on the x-axis is when I place it in standard form one of these two square terms has a plus in front and one of them has a negative in front. The one that's positive is the one on which the foci lie that's referred to as the transverse axis. So the x-axis here is the transverse axis. If the hyperbola had opened up or down up and down I should say then I would be writing this as y squared over a squared minus x squared over b squared is one and I always take the denominator of the positive square term to be a squared and then the term that's negative I'll take that to be b squared. Okay so let's look at an example of hyperbolas how I would graph a hyperbola and look at something more specific here. Okay suppose we have the equation x squared over 16 minus y squared over 25 is equal to one and I wanna graph this hyperbola with its asymptotes. Well because the x squared is the positive term I know that a squared is 16 which means that a is equal to four and b squared is therefore 25 and so b is equal to five. Now the equation that relates a squared, b squared and c squared is slightly different from what it was for ellipses and the equation we wanna use now is that b squared is equal to c squared minus a squared. You notice there's a slight difference I'm placing c squared minus a squared because for hyperbolas c is generally larger than a and therefore 25 is equal to c squared minus 16 and therefore c squared is equal to now if I add these together I get 41 and so c is the square root of 41. Now without using a calculator let me ask you to think about approximately how much is the square root of 41? Like is it more than 10? Well let's see, the square root of 36 is six and the square root of 49 is seven so this must be somewhere between six and seven. I would guess it's probably a little closer to six but it's gonna be, I'll just guess and say it's probably about 6.4 but it may be 6.3 or 6.5 but that sounds like a reasonable guess and you notice that c is bigger than a and it's bigger than b for hyperbola c is always the largest value so when I go to graph the hyperbola I would go out four units to find a and negative a and I would go up five units to find b and negative b and I make a little box right there and the purpose of the box is to help me draw the asymptotes and the asymptotes go through our diagonals through that box and they go right through the origin and my foci by the way are at plus or minus c on the x-axis so I'll go out to about, we'll put, I'll just put a c underneath that and a negative c over here that should be the square root of 41 and my hyperbola goes through its x-intercepted four and it approaches the asymptote the asymptotes on top and bottom on the other side my hyperbola approaches the asymptote as well so this is a rough sketch of the hyperbola x squared over 16 minus y squared over 25 is one now one of the things you should be prepared to see on any of these conic section problems is what if I draw the conic section and I ask you to tell me the equation of the conic so give an information in the graph what is its equation let's try doing that right now for a hyperbola suppose I have a hyperbola that looks like this the numbers here are four and negative four six and negative six and my asymptotes pass through and the hyperbola looks like this now there's enough information in this illustration for you to write the equation of the hyperbola that's represented here obviously it's not a perfect graph but there's enough information for us to figure out its equation let's see now this hyperbola opens up and down so I know that the equation is going to be in the form y squared over a squared minus x squared over b squared is one furthermore I can see that a is equal to four and I can see that b is equal to six and that's enough information to write its equation y squared over sixteen minus x squared over thirty six is equal to one now there are other questions that I could ask you for example where it's foci uh... and let's see to find the foci I'd use the equation b squared equals c squared minus a squared and that would say that uh... thirty six equals c squared minus sixteen and so c squared is equal to fifty two that implies that c is the square root of fifty two which is two times the square root of thirteen now you know the square root of forty nine is seven so the square root of fifty two should be a little bit larger than seven so I'll just say a little larger than seven now what are the coordinates of the foci well the foci should be on the on the uh... transverse axis so the transverse axis is the y-axis so if I go up to zero two square roots of thirteen there's one focus I'll say focus number one and if I go down two zero negative two square roots of thirteen that would be focus number two okay so one example one more example of many types of problems on parabolas or rather hyperbolas and you need to look at your homework and your notes from the uh... television episode to make sure you're aware of how to work uh... the various problems that come up for hyperbolas there was a reflective property for hyperbolas that goes like this if I just draw quickly two branches of hyperbola and put foci inside so here's focus number one and here's focus number two then if you remember light directed toward one focus is reflected toward the other focus if this is a reflective surface along the hyperbola light that approaches one focus that would normally go to one focus would be reflected toward the other toward the other focus in that case so much for the conic sections now um... that that's only uh... the first block of material in this exam that sort of tied together now let's go to the next block of material that has to do with sequences in series so let's look at episode thirty one you should be able to distinguish first of all between a sequence and a series now let's just talk about that for a moment you know a sequence uh... is just a listing of numbers like uh... one two three four five where you separate the numbers by commas whereas the series is a summation of those numbers such as one plus two plus three plus four uh... so you want to be able to distinguish between sequences in series because frequently students get those confused in exams and and apply incorrect formulas at those times uh... the second thing is you should know how to find various terms when given the nth term of a sequence or a series now let's just take examples of both of those sequences and series uh... suppose i told you that the nth term also a sub n is uh... in squared plus one and what if you're supposed to write the first four terms of the sequence now when it's a sequence that means will separate these by commas the sequence may go on indefinitely but we're supposed to find the first four terms well to find the first term would call that a sub one the first term that just means we plug in a one for n and at one squared plus one is two so it goes here to find the second term that would be a sub two what i do is plug in a two for n and i would get four plus one is five so five goes here now let me ask you to think about what are the third and fourth terms the third term is ten and the fourth term is seventeen what if i were to ask you what is the fiftieth term what is the fiftieth term well the fiftieth term would be a sub fifty and that would mean substitute in fifty for n so that would be fifty squared plus one now five squared is twenty five fifty squared is two five zero zero plus one so that would be twenty five oh one so the fiftieth term is twenty five hundred and and one okay and again the sequence may continue beyond that now what if it were a series well you know a series is frequently abbreviated using sigma notation and what if i were to say k goes from one to uh... let's keep it rather short let's say one to four and i'm gonna put that what's called the kth term inside here now the choice of the letter k you may recall this is an index so the choice of the letter k could be another letter like i or j or n those are all common letters used for the index suppose k where suppose the kth term where k plus one over k plus three now this question is comparable to what i asked about sequences but now this is a series and i know it's a series because of the capital sigma or a summation here so what i'll do is write down four terms but the differences i'll be adding these up so when i plug in a one i get two over four which we reduced to be a half and when i plug in two i get three over five and when i plug in three i get four over six which we can reduce and when i plug in four i get five over seven okay so this reduces to be one half plus three-fifths plus two-thirds plus five-sevenths now to finish this problem we need to add these up well the common denominator needs to have a two in it it needs to have a five and a three and a seven you know these are all prime numbers so the common denominator is going to have to have two times five times three times seven it has to have all those in it now the numerator well let's see we already have a two what i need in the numerator is a five a three and a seven to cancel out with the extra five and three and seven so five times three times seven plus the numerator for this fraction there's a three already i'm going to put in a two, a three and a seven that's going to make it two times three squared times seven and in the next numerator let's say i need to put in a two, a five and a seven i've already got a two so that makes it two squared times five times seven and in the last numerator i have a five i need to put in two and a three and a five that'll make it two times three times five squared now how much is that going to be well let's finish this up above so we don't have to erase it looks like the denominator is going to be ten times twenty one that's two hundred and ten and here we have uh... twenty one times five twenty one times five that's a hundred and five and then we have uh... nine times seven is sixty-three sixty-three times two is a hundred and twenty-six and then we have four times five is twenty twenty times seven is a hundred and forty and then we have six times twenty-five is a hundred and fifty so all together this is going to be two hundred and ninety and if i add on a hundred and five, two ninety three ninety-five four ninety-five five twenty-one five twenty-one i don't believe that reduces so i'll leave it at that and say the answer is five twenty-one over two hundred and ten okay that's the summation for this series right here it involves quite a bit more arithmetic than that first example did for this for the sequences okay let's go back to our graphic and look at other things you need to know about sequences in series uh... let's go to the third item you need to be able to determine a reasonable formula for the nth term when given several terms that is if i were to write down the first, second, third and fourth terms i would have some sort of a pattern in mind and i'd want you to recognize the pattern that i have in mind and write down a formula for the nth term you had some problems like that for homework but let's move on to the next item know how to find terms of a recursive sequence let's take an example of that a recursive sequence suppose i told you that the first term of the sequence was equal to two and that for each subsequent term a sub n was three times a sub n minus one minus uh... minus two now this is for n greater than one so you see this first uh... this first term gets it started a one is two and then for subsequent terms in the sequence a sub n is three times its predecessor minus two three times the previous term minus two so let's write down the first uh... five terms of this sequence now we're given the first term we're told that it's two how could i find a sub two well a sub two a sub two would be three times a sub one minus two but i know that a sub one is two so three times a sub one is six six minus two is four so this number will be a four to find a sub three this is three times the previous term a sub two minus two that'll be three times four or twelve minus two is ten okay now let me let ask you to be thinking about what should be the next term of this recursive sequence well we'll be taking three times the previous term that'd be thirty and then subtract two would be twenty eight and to find the next term i would take three times the previous term now let's see three times twenty eight is uh... eighty four i believe minus two is eighty two and so it goes so uh... this sequence continues for all we know and these this looks like these are the first five terms of this recursive sequence okay let's go back to that graphic and look at the next item uh... you should be familiar with the fibonacci sequence primarily because this sequence comes up in a number of other courses in particular in biology the fibonacci sequence uh... it's a recursive sequence in which the first two terms are both one and each term after that is the sum of the two previous terms just prior to that uh... and if you remember a number of uh... plants for example grow according to the fibonacci sequence i mentioned a tree in my front yard i think that's growing according to the fibonacci sequence it seems to branch uh... according to this pattern okay and uh... then finally you should be familiar with the sigma notation for abbreviating series i think we've already looked at one example of uh... using the sigma notation so let's let's move on to episode thirty two okay now in episode thirty two we looked at two special types of sequences in series and those are the arithmetic and the geometric sequences in series so number one you should be able to distinguish between arithmetic and geometric sequences in series let's just talk about this for a minute because it's it's important that you be able to recognize these so that you can apply some formulas that are coming up here that we'll talk about uh... first of all if you have uh... an arithmetic sequence the characteristic of an arithmetic sequence is that you begin with a certain term i'm gonna call the first number just a rather than a one and then what you do is you add on what's called a common difference a plus d so i add on d and then i add on d again so a plus two d and then i add on d again a plus three d so when i get to the nth term i've added on n minus one d is a plus n minus one d and this is the this is the nth term right here now for example an example of an arithmetic sequence would be one three five seven nine etc and you see in this case a is equal to one and the common difference is two because every time i'm adding on two to get to the next term so d is equal to two okay now what about uh... what about a geometric sequence what about a geometric sequence well let's see rather than adding on the same term i'll let you be thinking about what's the difference here between arithmetic and geometric uh... well what distinguishes geometric sequences you multiply by a common factor or a common ratio so if the first term is called a then what i do is i multiply by r which is the common ratio then i multiply by r again and get a squared and then i multiply by r again and get a cube so that when i get out here to the nth term how many r's will i have multiplied by it'll be n minus one it won't be in a lot of people a lot of times students write down uh... a r to the nth power it's actually n minus one and that's because we didn't have an r in the first term we introduced the r's only in the second term so the power here is one behind the position so it's just n minus one now if i were adding up these terms that would make it a geometric series and similarly if i were adding up those geometric terms that would make it an arithmetic series in that case okay let me write down a series and ask you a rather a sequence and ask you whether it's arithmetic or geometric uh... the first term is four uh... the next term is uh... seven the next term is ten and the next term is thirteen so is this an arithmetic or a geometric well this one is arithmetic because the common difference appears to be three in all the terms i've written down so far so this seems to be an arithmetic sequence it looks like a is equal to four and d is equal to three because i'm adding three on every time now on the other hand suppose we had this sequence uh... let's say we had one and then a half and then a fourth and then an eighth i don't think this is arithmetic because the common difference uh... there is no common difference here i added on negative a half here i added on negative a fourth here i added on negative one-eighth instead i believe this one's geometric and it looks like what we're doing is we're multiplying by a half every time if i take half of one i get a half if i take half of a half i get a fourth and if i take half of a fourth i get an eighth it appears to be geometric so it looks like in this case a is equal to one and r is equal to one-half now the reason this is uh... important is because of some formulas that allow us to uh... some arithmetic series and geometric series let's go back to the graphic for episode thirty-two uh... if you look at the third line there it says be able to apply either of two formulas to some in a arithmetic series and then the for the line right after that is be able to use two formulas to some a finite or an infinite uh... geometric series so let's just look at these formulas now you'll have to know these formulas for the uh... for the exam uh... if i'm summing an arithmetic uh... series such as a plus a plus d plus a plus two d and then we go out to the nth term which would be a plus n minus one d now let's say i call the sum s sub n because it's the sum of the first n terms there are two formulas that will compute s sub n the first formula is n over two times two a plus n minus one times d the other formula is s sub n is equal to n over two times now what you do is you add the first in the last terms together so i would call that a one plus a sub n this is an abbreviation for the first term and for the nth term i've added the first and last terms together multiplied by n over two these both give you the sum for an uh... arithmetic series uh... when you know a and d or you know a one and a n the first and the last terms okay let's move on to the formulas for geometric series uh... for a geometric series suppose i were adding up a plus a r plus a r squared and i go out to the nth term which is a r to the n minus one now i would call this the sum of the first n terms or s sub n now if this if this series terminates with n terms then s sub n can be computed by the formula a times one minus r to the n over one minus r and this is where r is between one and negative one oh i'm sorry let's see for a uh... for a finite series there's no restriction on r so this is my formula however if i want to find an infinite summation i'll call it s sub infinity or i think in the textbook that you use an s then in that case this formula is a over one minus r and there is a restriction on r and that is that r is between one and negative one so if you have a finite summation of a geometric for a geometric series you can use this formula regardless of r except that of course r can't be one because you can't divide by zero on the other hand if you have an infinite series you can summit by just taking over uh... a over one minus r now that the proof for these formulas was given in the episodes uh... previous episodes but uh... i want to expect you to derive these and there's not time now to derive these in this episode okay let's go to episode thirty three and this is the last episode before we come to our uh... to our exam and let me just say first of all you should know three formulas for this exam these are the formulas for computing and annuity uh... for computing the present value of an annuity and for computing the payment or what sometimes referred to as the rent on a loan and then you should be able to you should know those formulas niche you should be able to use them to compute an annuity a present value of an annuity or the payment on a loan now here are the three formulas that you need to know and you'll find them in the textbook uh... if you're computing an annuity you remember an annuity is basically a sum of money that's collected in an account that uh... is obtained by regular payments plus interest drawn on those payments and uh... the final amount of the annuity is equal to the rent or the regular payment you're making into the account times one plus i to the nth power minus one all over i okay now um... i represents the interest rate per period which doesn't have to be per year it could be per month or per day in is the number of periods over which the money's being compounded and are is the periodic payment into that account so uh... for example if you're depositing a hundred dollars a month into a savings account are would be a hundred and i would be the annual interest rate over twelve and in would be the number of years in which the these payments are made times twelve because that's the number of payments that are being made now to find present value of an annuity the present value is computed by taking are times one minus one plus i to the negative in power all over okay and uh... the same definitions apply i is the periodic interest rate and is the number of periods over which the money's being compounded and are is the regular payment being made on that now to find the uh... to find the uh... the the rent on an account to solve for our if i solve for our in this formula i get i times a sub p over one minus one plus i to the negative in power okay now you should know all three of these formulas and you should be able to solve problems that use these formulas uh... using your calculator now this portion of the exam the problems would be given uh... separately from uh... uh... from the rest of the test where you can use a calculator okay now i'm just trying to think of uh... of any problem additional problems that might help you in studying for the test and uh... i think there is a problem on infinite geometric series that uh... we we never did one of these in class but they were given in our book we might close with this example suppose i have this decimal number we have this decimal number let's say uh... zero point one eight one eight one eight and the one eight just keeps repeating indefinitely now one of the characteristics about decimal numbers if the decimal either terminates like point five or if it repeats like this one's repeating then it's equal to a fraction for example point five is equal to a half well this decimal is equal to a fraction and suppose would like to find out what fraction is it equal to well i can use the formula for the summation of an infinite geometric series to work this problem uh... you see i could write this number as zero point one eight plus zero point zero zero one eight that takes care of this one eight and this one eight now let's see i need to put in the next one eight that'd be zero point zero zero zero one eight and if i just continue indefinitely that will represent all of the pairs of uh... ones and eight that are expressed in this number if i put these in a column and added them up i would get exactly that answer now it looks like to go from this number to this number what i've done is multiply by one one hundred to get two zeros in here and if i multiply this by one one hundred i get two more zeros so i'm thinking this is a geometric series where a is equal to zero point one eight and r is equal to zero point zero one that's what i'm multiplying by so therefore s is equal to a over one minus r this is an infinite series and you notice the value of r is between one and negative one so i can use this formula so i get zero point one eight over one minus zero point zero one or zero point one eight over zero point ninety nine let me write that a little bit better which is the same thing as eighteen over ninety nine and if i divide by nine this is two over eleven so therefore this decimal is two over eleven okay i hope this helps you to prepare for the exam number five and then i'll see you next time for exam thirty five