 Hi, I'm Zor. Welcome to InDesert Education. This lecture will be about sequences. I have started the previous lecture with introduction to what actually sequences are, and this lecture will be dedicated to a specific type of sequence which is called arithmetic sequence or arithmetic progression. Now, why do we actually start with this one? Well, just because it's a very simple one. I hope I did not scare you during the previous lecture explaining the huge formula for Tibonacci sequence with square root of five to the end of degree, etc., etc. Yeah, the formula was very complex, but the rule for Tibonacci sequence was quite simple. Just add two previous numbers to get the next one, and so is very simple rule for arithmetic sequence. So I'll just explain the rule, and then we will derive the formula which much different from the Tibonacci sequence is really very, very simple. Okay. So what is arithmetic sequence? Very simple. You start at certain point with certain number as the first element of the sequence, and then you basically add the same number again and again and again to get to the second, to the third, to the fourth, etc., elements of this sequence. So number one element is some kind of a number eight, usually it's real number, and then you basically add some number D, which is called a difference of the arithmetic progression. So the next number would be a plus D, the next one would be a plus 2G, etc. So this is number one, this is number two, this is number three. Well, as you probably guess, element number n would be a plus how many times we have added D to our initial element? Well, once to get to the number two, two times to get to number three, and obviously n minus one times to get to element number n. So this is a general formula. Now, in theory, we really have to prove that this formula really does what we have promised in defining the arithmetic progression. Now, what did we say as a property, as a definition actually, a characterized property of arithmetic progression? That the first element is a, and indeed if you will put n equals to one to this formula, you will get, this is zero, so you will get a. And the second property of arithmetic progression, the second characterizing defining property is that the difference between two consecutive elements of this sequence, let's say element n and element n plus one, should be a constant D. Well, let's check it out. According to my formula, the next one would be if instead of n I substitute n plus one would be n plus one minus one, so it would be n times G. And if you will subtract from this one, you will subtract this one, you obviously get G. If you will open all the parentheses, a minus a, nG would be minus nG and plus G, so you will have G, obviously. Well, what I just did, I actually proved the formula by induction. I checked it for n equals one. Then I assumed that the formula is correct for some number n, and then checked that the property is actually retained using this formula for n plus first element. Okay, so we have derived with a formula for a arithmetic progression, which is this one. And this is the formula for arithmetic progression, which starts with number a, and adds G to every consecutive member from the prime, to derive it from the prime number. So this is element number n. Now, first of all, what kind of numbers can be a and g? Usually you can consider that they can be any number which you know, including complex. Well, in reality, people do not really consider complex numbers when talking about arithmetic progression. They usually talk about real numbers, sometimes even integer numbers, while positive and negative. Now, let's just think about how this particular arithmetic progression behaves if the sequence number is increasing to infinity. Well, obviously it depends on the g. If d is positive, which means if we are talking about axis somewhere a, and d is positive, it means we are always stepping forward from a to a plus d, from a plus d to a plus a plus 2d, et cetera, et cetera. And obviously you will go infinitely to the right. Infinitely means that no matter what kind of a number you choose, let's call it z, eventually you will overcome and stay over it. That's what basically means it's infinitely increasing to infinity. On the negative side, if g is negative, it means basically when you are adding a negative number, you are shifting to the left. And that means that eventually you will be smaller in a negative sense than any number z, whatever number you choose. So it will overcome z moving to the left, which means it's smaller. And then all other, all subsequent elements of the same sequence will be smaller. So in this particular case, the value of elements is going to the negative infinity. So let's just evaluate when in the positive case, I will become, my elements will become greater than any number z. And in a negative case, whenever they will become smaller than any z. So let's consider that a is given. And it doesn't really matter whether a is positive or negative. And z is also given. And let's check it to the case when g is greater than 0. All right, so g is greater than 0. I'm looking for element number, sequence number, and such that the n's element would be greater than number z. Can I find n from this inequality? Basically that's my question. Because if I find the first element which overcomes z, then all subsequent elements will also be greater. Because we are always adding something positive. We are talking about only positive d. So let's just find out from this inequality what is the value of n. So when exactly at what number, at what sequence number, my elements, and this is the general formula for an element of a recent progression, will overcome, will become greater than z. Well, very easy. I mean, this is inequality. Inequality can be solved. This is algebra, right? So number 1, from both sides of inequality, we can subtract the same number a. And the sign of inequality remains the same. And I will get a minus a. So I will get only this. And again, let me just emphasize. It doesn't really matter what the signs of z and a. Because this is invariant transformation in inequality. We subtract the same number from both sides. This was greater than this. And with subtract, it will be still greater. Now, how can we solve this particular inequality? Well, obviously, we have assumed that g is greater than 0. Now, again, invariant transformation of the inequality is when you are dividing actually both sides of the inequality by the same positive. That's very important, positive number. Then the inequality actually retains the sign. So both sides I divide by d. And d is positive. I'm explicitly using the d is positive. Because now the sign is exactly the same. And finally, I add 1 to both sides, getting n greater than 1 plus z minus a over b. So this is a condition when my elements would be greater than number z, whatever number z we have chosen. By the way, if by any chance z is negative, already negative, let's say, we have chosen something here. And let's say a is positive and d is positive, which means we start here and move to the right. Well, basically, even the first element would be greater. So basically, it doesn't really matter what kind of z and a are. But for a positive d, we can say that all elements of arithmetic progression with first element a and the positive difference g would be greater than any chosen beforehand number z after this particular sequence number. I mean, all subsequent numbers will also be greater because they're always increasing. OK, so that's for the positive. Now let's consider the negative. What would be the n sequence number when moving to the left? We will have elements smaller than any number z, which we have chosen. All right, so again, let's start from the general formula. a plus m minus 1 b should be less than z. That's what we are looking for number n so that the element would be less than z. And g is negative, all right? So again, we subtract a from both sides, getting n minus 1 d less than z minus a. Now we also divide by d. But considering d is negative now from properties of inequality, if one element is less than, if one number is less than another number, and we both, and we divide both sides by negative number, then they reverse the sum, obviously, right? I mean, if 2 is less than 4 and we divide by minus 2, we will get minus 1 and minus 2 here. And now, obviously, the sign of inequality is reversed. This is less, but this is greater. So this is obvious. These are elementary properties of inequality and invariant transformation. So we divide by g and we get n minus 1. So instead of less, it would be greater than z minus a divided by g. And at 1, we get n greater than 1 plus z minus a divided by g. Now, how interesting. The formula is exactly the same as this one, n is greater. So no matter what the difference d is, positive or negative, we can always say that in case it's positive, it will become greater than any number z after this number. And in case d is negative, it will become smaller than any number z after the same, basically, number calculated by the same formula. Now, and my view, z is basically any number, however big or however small, for positive or negative d, which means that values of elements of arithmetic progression are going always to infinity, positive infinity for positive d difference between elements, and to negative infinity for negative d. So it's not really contained in any particular interval. As number n increasing, we will go to infinity. We will overcome any boundary z which we have with ourselves. And we can find the number from which started from which all the elements would be greater than this particular number z for positive or less than for negative d. All right. So this is basically all, not much, quite frankly, which I wanted to talk about arithmetic sequence. I will definitely talk about some other properties of arithmetic sequence when you sum up the elements of the sequence. But that will be after I will introduce the concept of a series, which is basically the result of the addition of elements of the sequence. So right now I'm not touching this. It will be another lecture which is dedicated to series. So that's it for this particular lecture. Thanks very much. There are some exercises which I present as problems in next lectures. Please do that. I do encourage you to solve all the problems yourself first before you listen to the lectures which are dedicated to the problems. And also what's very important, I believe, is if you register, you will be able to take exams, which is, again, extremely important for your growth, development, and education. Thank you very much. That's it for today.