 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue discussing certain less trivial, I would say, problems related to sequences and series. I think I have two problems to discuss today. Yes, so let me just start with the first one. Obviously, I do expect everybody who listens to this lecture first try to do these problems just by yourself. Then listen to the lecture and then try to do it yourself again, just to be able to remember it better. All right, so find the quotient of a geometric regression if each of its elements starting from the second equals the difference between its two immediate neighbors. All right, so you have to find a quotient of a geometric progression which has certain property. Now, the property is that if you have geometric progression like this, this is member number m in turn of this sequence. Now, you have to find out what's the quotient q if there is certain property which they satisfy. Namely, every member, every term of this sequence is equal to a difference between the next one and the previous one. But let's think about if we know the formula for any term, any element of a geometric sequence and this is the number nth term, then we can basically have an equation that the nth term of the sequence is equal to n plus first, the next one, which is obviously this one, one greater than this, and minus the previous one. This is one less, power is less than this one. So this is immediately following member after this one and this is immediately preceding. So basically, we have an equation from which we have to find q. Well, obviously from the first glance, we see that we have two different variables, a and q and one equation and also we have some kind of an exponent which we don't really know. But fortunately for us, this is basically canceled out. Now, obviously we can divide everything by a, but we usually assume that the first term of geometric sequence is not equal to 0, otherwise the sequence is trivial, 0, 0, 0, 0. And that allows us to divide both sides of this equation by a. So a is not equal to 0, I can divide, and I have q n minus 1 equals to q to the n minus q to the n minus. Now also, another trivial case when q is equal to 0, we also have to exclude from our geometric progression. This is kind of a typical condition. Whenever you have a problem with geometric sequence, it's reasonable to assume that the first element and the quotient are not equal to 0 to basically exclude all the trivial times. Now this is obviously reducible if you divide both sides of this equation by q to the n minus second degree, you will have, now whenever you divide, you subtract the exponents, right? So you have q n minus 1 minus n minus 2 will be the first degree. n minus n minus 2, it will be the second degree, right? n minus n plus 2. And here it would be just 1. Well, now this is an equation which we can solve for q, and that's exactly what's required. The problem is what's the q if this property is satisfied. So all we have to do right now is to solve this simple quadratic equation, which I will do very simply. q square minus q minus 1 equals to 0, which means the solutions are 1 plus minus square root of 1 plus 4, which is 1 plus minus square root of 5 divided by 2. So this is the quotient. There are two different solutions. So any geometric sequence with any not equal to 0 first term and this or that quotient will have this particular property. That's it. Now, next problem is following. Sequence consists of four elements. Find these elements if it's known that's element 1, 2, and 3 form geometric progression, 2, 3 form arithmetic progression, and some of the first and 4, 14, and the second, 12. Okay, so we have four elements, number one, number two, number three, and number four, which form some kind of a sequence. Now, I'm not telling it's arithmetic sequence but geometric sequence, it's just a sequence, which means it's ordered set of four real numbers. Now, what's known about this? Few things. These first three form geometric progression. Now, next three, I mean starting from 2 to 4, number two, number three, number four, form arithmetic progression. I also know that number one plus number four is equal to 14 and number two plus number three is equal to 12. Okay. So these are conditions which are supposed to allow me to basically determine these four numbers. Well, how can I approach this? Well, I believe that the first thing is to find out what exactly are unknown variables from which these four numbers depend upon using these properties. And then using these two conditions, try to solve certain equations to find these unknown variables. So my first unknown variable would be the first term of this sequence, first element. So x is number one. Now, I know that the first three form geometric progression. Now, geometric progression is defined by the first term, which is x, the first member, and the quotient. So let's just assume that the quotient is just another variable which should be introduced. So number two is equal to x times q, and number three would be x times q squared, right? That's what actually makes the first three elements of my sequence a geometric progression. Okay, fine. Now, so I have expressed one, two and three using two variables, x, which is the number one and q, which is the quotient, the ratio between number two and number one, or number three and number two, which are the same. Now, I have to use the fact that this is a arithmetic progression, number two, number three, and number four. Now, what's the characteristic of arithmetic progression? That every next member is, by certain difference, whatever the letter you can use, difference, as I used before, is greater than or less than the previous one. So you have to add this difference to each element to get to the next. Now, but I already know number two and number three in terms of my two initial variables. So I know what's the difference between them. It's x squared minus x cubed, right? Which means that exactly the same difference between these two. So if I want to find out what's my number four, I have to take number three and add the difference between number three and number four and number two, right? That's basically what it means that minus, number two, number three, and number four numbers make up an arithmetic progression. So to get to this number, I have to add as much as I had to add to get from this to this. So to number three, I add the difference between three and two. Now, in terms which I have already explained before, that would be, let me just write it down, x cubed squared, that's number three, plus x cubed squared minus x cubed. So that's my third, actually fourth, sorry, that's my fourth member. Actually, I can put two x here, right? Two x cubed squared minus x cubed. Okay, these are my, actually, I don't know this. So in terms of x and cubed, the first term of my sequence and the quotient between the first and the second, I have expressed all four numbers. Now, let's use the fact that we have the sum of the first and the fourth equal to 14. So x plus two x cubed squared minus x cubed is equal to 14 and the two middle one is equal to 12. This is a system of two equations with two variables, which I, in theory, can solve. So let's try to solve it and whatever the results will be, will allow us to recreate all four unknown members of our sequence. All right, so how can we do it? Well, obviously, I have to factor out x in the first case and let me write it in this sequence. Two cubed squared minus cubed plus one is equal to 14 and the second one would be also x factor out, all right? Now, again, we'll exclude the case when x is equal to zero because obviously this is not a solution because then you will have first three members equal to zero since it's a geometric progression and then, therefore, the fourth will be equal to zero and obviously the sum will be equal to 14. So x is not equal to zero, it's definitely not a solution. x is not equal to zero, which means I can divide one into another and what do I have? x would cancel out, right? So I will have this ratio of two cubed squared minus cubed plus one over cubed squared plus cubed is equal to 14 over 12. Now, here, I also have to be very careful when I am dividing by cubed squared plus cubed. Now, cubed squared plus cubed should also not be equal to zero. Now, since cubed squared plus cubed cubed times cubed plus one, I have to say that cubed is not equal to zero and cubed is not equal to minus one. Now, again, all these cases when cubed is equal to zero and cubed is equal to plus or minus one are not considered to be usually the candidates for geometric progression. Well, zero is obviously not good because that would give me all zeros and the minus one, well, it's not really a good geometric progression if the quotient is equal to one or minus one because it's trivial. It either gives you the same numbers or gives you numbers which are changing the sign and most likely I will not get 14 and 12. I probably should check it more carefully with minus one. We'll just try to do it very simply. If you have some kind of an A, then you have minus A, then you have plus A and then the difference between these two should be the fourth one which is 2A. Now, the middle will give you zero, definitely not 12. So I can basically check that minus one is not only trivial but really not a solution anyway. Fine. Now, how can I solve this? Well, just make a cross product and let's just think what will happen. It looks like we will probably get some kind of quadratic equation which we've just been able to solve. All right, let's do it. So this times this is equal to this times that. Now, this is one to four Q squared minus 12 Q plus 12 equals 14 Q squared plus 14 Q. All right. Bringing everything to a normal form, quadratic equation, 10 Q squared minus 12 and minus 14 would be minus 26 Q plus 12 equals zero. All right. Now, we're in business. Now, we have just a quadratic equation. First, we reduce by two. So it's 5 Q squared minus 13 Q plus 6 is equal to zero and solutions are double the first quotient. Now, the second plus minus square root of this guy's square is 169 minus 4 times 5 times 6 that's 31 20. These are two solutions. Well, let's simplify it a little bit. 169 minus 20 is 49, square root is 7. So we have 13 plus minus 7 divided by 10 which is either 13 plus 7 is 20 divided by 10 is two or six 10s or three-fifths. So these are two solutions for Q, two and three-fifths. Okay. So we've had the solution. Now, we have to recreate our members. So first, we have to get the X, right? Now, from a simpler case, let's say some of these two is supposed to be equal to 12 and some of these two 14. Well, 12 seems to be easier. So X times Q squared plus Q is equal to 12. Now, for Q equals to 2, I have two square which is 4 plus 2, 6. I have X is equal to 2 and for Q is equal to three-fifths. Three-fifths would be 925 plus three-fifths which is 1525. X equals 12. That's 24. So X 24, 25 is equal to 12. X is equal to 25 divided by 2, am I right? So I have two different pairs of solutions. Now, X equals 2, Q equals 2 or X equals 25-fifths and Q is equal to three-fifths. Now, in theory, we really have to do the checking. And obviously, I don't like to do the checking. For these guys, it's kind of complicated. So let me do the checking for X equals 2. So what's my sequence will be? The first is X which is 2. Then next two members are geometric progression with a Q equals 2 which means 4 and 8. And the next one is an arithmetic progression based on 4 and 8. So the difference is 4. So I have to add another 4 to get 12. So this is my sequence. Now, let's check. Geometric progression of the first three, fine. Arithmetic progression of the next three, fine. Some of the first and the fourth is 14. Some of the middle ones are 12. Everything is fine. How about the second one? Well, yeah, well. Just to demonstrate the time and good spirit, I'll do that too. So I have 25-fifths as the first one, and then I have geometric progression which means the next one will be multiplied by three-fifths, which is... What did I say? No, I'm sorry. 25-fifths? Well, let me just start in the beginning. First times q squared plus q was equal to 12. If q is three-fifths, it's 9-20-fifths plus three-fifths, which is 15-20-fifths. 12. So it's 24-20-fifths equals to 12. Yeah, it's 2. Of course, I see something is wrong. 25 over 2. All right, so let's check this out. The first is 25-second. Next we multiply by 3, so it's 75-fifths-10. Right? Which is, let's just reduce it by 5 would be 15 seconds. Next one, multiply by 3-fifths again, so it's 45-10. Again, reduce by 5, that's 9 seconds. Okay. Next, we will recreate our fourth member using the fact that the difference between this and this should be the same as this and this. So as you see, we are diminishing going down, so the difference is minus 6 seconds. It's minus 3, actually. So I have to subtract 3 from this, so it would be 3 seconds. Is that right? This minus this is 6 seconds, which is 3, right? Now, let's check this out. Now, these three are geometric, these three are arithmetic, we have checked that. Now, the sum of the person, the fourth one, 28 seconds, which is 14, sum of these two is 24 seconds, which is 12. So as you see, we also have satisfied the conditions. Now, in theory, let's be very, very careful about this. In theory, all our transformations were invariant. We have excluded a few cases when x is equal to 0 or q is equal to 0 or minus 1. And in the absence of these cases, everything else, whatever we did, we divided by something, we multiplied by something, but whatever we divided wasn't equal to 0, whenever we multiplied wasn't equal to 0, etc. All our transformations of our original equations were invariant, which means that ultimate solutions which we have for the very last equation are supposed to basically satisfy the beginning as well, since all transformations are invariant. And that's why in theory, we could avoid doing the checking. I mean, I'm not surprised that both solutions fit the problem. However, I do encourage you in your real kind of engagement, whatever you do, what kind of problems you solve, you do the checking. It's always a good thing to do the checking, regardless of how certain you are, the transformations which you made with your equations are invariant. So we've got two solutions for our problem, and that concludes my lecture today. Do again these two problems yourself. And as usually, I encourage you to go to the university.com and register, make sure that you go through enrollment, or your supervisor, your parents should enroll you, and then you go through exams. And it's supposed to be like a process, education is a process. Don't avoid these type of things. It's for your own good. Okay, thanks very much. That's it for today.