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let's write an arithmetic sequence in general terms arithmetic sequence so we can start with some number a and then we can keep adding D to it and we that number that we keep adding which could be a positive or negative number we call our common difference so the second term in our sequence will be a plus D the third term in our sequence will be a plus 2 D so we keep adding D all the way to the nth term in our sequence and you already see here that in our first term we added Z we added D zero times our second term we added D once in our third term we added D twice so you see whatever the index of the term is we're adding D one less than that many times so if we go all the way if we were go to the all the way to the nth term we're going to add D 1 less than n time so it's going to be n minus 1 times D fair enough and let me write that this is where this right over here is our nth term now what I want to do is think about what the sum of this arithmetic sequence would be and the sum of an arithmetic sequence we call an arithmetic series so let me write that in yellow the eros that's not yellow so let me color changing is sometimes difficult so Aerith arithmetic sequence I sorry arithmetic series is just the sum of an arithmetic sequence so let's call my arithmetic series let's call it S sub N and let's say well it's just going to be the sum of these terms it's going to be a plus a plus D plus a plus 2d a plus 2d plus all the way to adding the nth term which is a plus n minus 1 times D now I'm going to do the same trick that I did when I added when I did the most basic arithmetic sequence I'm going to add this to itself but I'm going to swap the order in which I write this sum so the S sub n I can write as this but I'm going to write it in reverse order I'm going to write the last term first so the nth term is a plus n minus 1 times D then the second-to-last term is going to be a plus n minus 2 times D the third-to-last is going to be a plus n minus 3 n minus 3 times d and we're going to go all the way down to the first term which is just which is just a now let's add these two equations we are going to get on the left-hand side S sub n plus S sub n you're going to get 2 times S sub n it's going to be equal to well what's the sum what's the sum of these two first terms right over here I'm going to have a plus a plus n minus 1 times D so it's going to be 2 a plus n minus 1 times D now let's add both of these second terms so if I were to add both of these second terms what do I get I'm going to get 2 a plus 2 a and what's d plus n minus 2 times D so you could view it several ways you could view this let me write this over here what is D plus n minus 2 times D well this is just the same thing as 1 D plus n minus 2 times D and so you could just add the coefficients so this is going to be n minus 2 plus 1 times D which is equal to n minus 1 times D so the second term also also becomes also becomes 2 a plus n minus 1 n minus 1 times D now let's add the third term I'll do it in green or the third terms I should say and I think you're going to see a pattern here it's 2a so plus 2a plus if I have 2 plus n minus 3 if something have n minus 3 of something and then I add 2 I'm going to have n minus 1 of that something so plus n minus 1 times D and you're going to keep doing that all the way until your nth pair of terms all the way until you add these two characters over here which is just 2a plus n minus 1 times D so you have this two a plus n minus one D being added over and over and over again and how many times are you're doing that well you had n pairs of terms when you were adding these two equations or in each of them you had n terms this is the first term this is the second term this is the third term all the way to the nth term so I can rewrite this sum as 2 or the 2 times the sum 2 times S sub n is going to be n times this quantity it's going to be n times n times 2 a plus 2 a plus n minus 1 n minus 1 times D and then if we want to solve for S sub n you just divide both sides by 2 and you get S sub n is equal to and we get ourselves a little bit of a drumroll here n times 2 a plus n minus 1 times D all of that over 2 now we've come up with a general formula just a function of what our first term is what our common difference is and how many terms were adding up and so this is the generalized sum of an arithmetic of an arithmetic sequence which we call an arithmetic series but now let's ask ourselves this question this is kind of hard to remember you know n times 2 a plus n minus 1 times D over 2 but in the last video when I did a more concrete example I said well it looks like an arithmetic sequence it looks like the sum of an arithmetic sequence could be written as perhaps could be written as can be written as the average of the first term a 1 plus a n the average of the first term in the last term so that's the average of the first term in the last term times the number of terms that you have so is this actually the case is this actually the case do these two things do these two things gel because this is very easy to remember the average of the first and the last terms multiplied by the number of terms you add and actually makes intuitive sense because you're just increasing by the same amount every time so let's just average the first and or let's just average the first and the last term and then multiply times the number of terms we have well all we have to do is rewrite this a little bit to see that it is indeed the exact same thing is this over here so all we have to do is break out the a so let me rewrite it so this this could be re-written as s of n is equal to n times a plus a plus n minus 1 times D I just broke up this to a into an A plus a all of that all of that over 2 and you see based on how we define to this thing our first term a 1 is a a 1 is a and then our last term a sub n is a plus n minus 1 times D so this whole business all of this business right over here all of this business right over here really is the average of the first and last term's last terms I got my first term adding it to my last term dividing it by 2 and then I'm multiplying by the number of terms we have and that's true for any arithmetic sequence as we've just shown here