 Hello and welcome to another session on sequence and series We are going to discuss a new concept today, and that is called arithmetic means now in statistics You would have read about average of two terms and average of many averages of many terms actually now Today also we are now going to understand this concept of arithmetic mean from the perspective of progression sequence and series Okay, and you will see later on then there is some bit of correlation with between whatever we are discussing here and The means and the averages we find in statistics as well So what is arithmetic mean in a common parlance, you know arithmetic mean of any two given number a Nb let's say if you have a Nb so a m Arithmetic mean of a Nb is given by a plus b by 2 Okay arithmetic mean and why is that if you see because a a plus b by 2 and b are always Always in AP you can check Right, let's find out the common difference if at all they are in AP so these are the first two terms and The last two terms right so CD if you try to find out will be a plus b by 2 minus a Which will give you b minus a by 2 right? Let's try to find out the CD in these two terms. So that will be b Minus a plus b by 2 so that will be again b minus a by 2 So if you see the common difference is b minus a by 2 so hence a a plus b by 2 and b will always be in AP and hence We will be calling a plus b by 2 as arithmetic mean some middle mid mid You know in in this three terms AP a plus b by 2 is the middle term So that's the concept of arithmetic mean. So let's say if we have 3 and 7 so arithmetic mean Will be how much 3 plus 7 by 2 so a m is 3 plus 7 by 2 which is 5 now 5 is in Exactly in the middle, you know location of 3 and 7 so 3 and 5 and 7 are in AP Okay, this are always these are always in AP so 5 is equidistant from 3 and 7 right so this is the concept of arithmetic mean Now in sequence and series. We need not have only one term Between a and b which are in AP. Let us say There are between a and b. We have m terms so a 1 a 2 a 3 Dot dot dot let's say a m. There are m terms so in total how many terms are there m a 1 a 2 a 3 2 m 2 a m plus a and b so m plus 2 terms are there totally once again a and b are two terms and in between I am Fitting in m terms these m terms such that such that such that a a 1 a 2 a 3 Dot dot dot a m minus 1 a m and b are In AP let's say if you could figure out these many terms m terms between a and b and Along with a and b. They are in AP then then a 1 a 2 a 3 a m minus 1 and a m are called m arithmetic means between a and b Okay, so there are m arithmetic means between a and b you can insert as many arithmetic means as possible between a and b You have to just ensure that these terms are in AP then we'll say then we'll say a 1 a 2 a 3 Dot dot dot till a m are m m arithmetic means between a and b, okay So if they are in AP cannot can we not say that b is Equal to first term first term is a and then total number of term is m plus 2 Minus one times common difference if all of them are in AP so m plus 2 is a total number of terms, isn't it? So our basic formula is last term term Is equal to first term plus total number of terms minus one times common difference. This is what we have studied in AP Isn't it? So hence if you see this becomes and in this case we had n m plus 2 terms why because there are in if you look at them They are m terms here and along with a and b there are m plus 2 terms, right? So hence we have used m plus 2 here, right? So that means b minus a it will be equal to m plus 1 D therefore the common difference in such kind of an AP will be b minus a by m plus 1 Okay, that means you can find out all the a m's now. So a m's in this case m a m's are there. So first of them is a A plus the common difference that is b minus a by m plus 1, correct? This is the a1 then a2 will be a plus twice the common difference, which is b minus a m plus 1 and Now be very very mindful of this thing that if there is one Here the coefficient is 1 it is 2 here. It is 2. So you now know what is a3 a 3 3rd a m is a plus 3 times the common difference Okay, therefore you proceed in this manner. You'll get a m will be simply a plus now again If 3 was here 3 has to be here. So if it is m here, it will be m times b minus a by m plus 1 right, so you got all the a m's between a And b let's take an example and understand. Let us say we have a is equal to 3 Let's take an example Okay, a is equal to 3 and b is equal to let's say 9 and we have to insert 5 a m's Between A and b Okay, a and b Okay, so how to find out the first So hence m is equal to 5 5 a m's we have to insert so m is equal to 5. So first aim will be a1. You can check the formula here. What is it? first that is 3 a Plus 1 times b minus a so 9 minus 3 divided by 5 a m's right, so 5 plus 1 6 So this one is 3 plus So 9 minus 3 6 and denominator is 6 so 1 so 3 plus 1 that is 4 Right, what will be a2 the formula is a plus 2 times d so 9 minus 3 by 5 plus 1 Right, so this will be simply 5 Then a3 is you can calculate it will be 6 a 4 will be 7 and A5 will be 8 so 5 a m's 5 arithmetic means between 3 and 9 are 4 5 6 7 and 8 why because if you just You know add 3 here and 9 here all of them are in AP Right, so I hope you understood the concept of arithmetic means and hence we can insert as many arithmetic means as possible till m is a Positive integer, right? So that's the concept of arithmetic means we'll solve a few problems on this to understand it better, okay