 This video will talk about arithmetic sequences if a fixed constant is added to each successive term then we have an arithmetic sequence. So here's the definition. Given a sequence of a's of 1, a's of 2, a's of 3, these are just the different terms all the way up to the nth term where k and n are natural, notice I said natural we talked about that in the last video, numbers and k has to be less than n then if there exists a common difference such that a's of k plus 1 minus a's of k is equal to d for all k so then if that's true we have an arithmetic sequence. So what in the world does this mean? This really just means take the second term and subtract the first one a's of k plus 1 minus a's of k so you're going to take negative 2 minus 1 and that's going to give us negative 3 and we're going to take negative 5 minus a negative 2 which is really plus 2 and get negative 3 and negative 8 minus a negative 5 which is really plus 5 so we have negative 3 or negative 11 minus or plus 8 would give us negative 3, negative 14 plus 11 is going to give us negative 3 so the common difference is negative 3 and this is arithmetic. So we want to write the first five terms in arithmetic sequence given this information where we know we have a's of 1 is equal to and I'm going to rewrite this because it's kind of hard to see it so I'm going to write it as 1 half and d is equal to 0.25 or 1 fourth we know the first one is 1 half and then we have to add 1 fourth to that this is really then 2 over 4 so we have 3 over 4 so 3 fourths and then now we have 3 fourths and we again have to add 1 fourth so that's nice we have 4 over 4 or just 1 and we have to add 0.25 again or 1 fourth more so if we take 4 over 4 and add another 1 over 4 we're going to have 5 over 4 and if we did it in fractions that would be fine it'd get it in decimals so it would really be like 0.5 and then 0.75 and then 1 and this would be 1.25 so you notice in the sequence that we just did the common difference was 0.25 we added 0.25 over and over again and repeatedly adding the same numbers really just multiplying that common difference like this we start with a's of 1 and we want 0 common differences we need it to stay 1 so this would be 1 plus 0 is 1 well we can start with that first term again but this time I would like to have 1 common difference and that would give me 2 times 1 plus 1 or 3 but this is the most important thing that we're looking at right here is what is happening right here so we take our first term and we add to that now 2 common differences and if I have my fourth term I would take my first term and I would add 3 common differences to it so what does that mean in general? well I take my first term whatever that is and I add to that my common difference and then we have to figure out how many of them we need well if you look at this for the a's of 1 term I needed 0 a's of 1 I needed 0 a's of 2 I needed 1 a's of 3 I needed 2 a's of 4 I needed 3 so do you see the pattern happening here? how did these two numbers compare? 2 compared to 1 3 compared to 2 4 compared to 3 a's of n would be I'll put it over here this would be a's of n and we're letting this be if this was 1 and this was 0 1 less 2 minus 1 would be 1 3 minus 1 would be 2 4 minus 1 would be 3 so we're going to say n that's that little number down there minus 1 we need 1 less common difference than the term number and that's what we're going to say is our general formula a's of n let's put it up here again is the first term plus our common difference whatever that is times n minus 1 so we want to find the general term well what do we need? we need an a's of 1 and we need a common difference well a's of 1 is always your first term so we know that's 7 they gave us our first term so we're good and then D is remember we're going to take 4 minus 7 and that's equal to negative 3 and we double check we take 1 minus 4 and that's again negative 3 and if I look at it looks like 1 plus negative 3 we get needed negative 2 negative 2 plus negative 3 we get needed negative 5 so it looks like my common difference is negative 3 then we have here a's of n is going to be equal to our first term 7 plus our common difference of negative 3 times n minus 1 well in your book they're going to want you to take it just a little bit further so they're going to want you to say 7 minus 3n distribute the negative 3 and then negative 3 times negative 1 would be plus 3 so they really want you to say that a's of n is equal to negative 3n plus 10 there's your nth term so the second part of this problem says find the 6th term well that means we want a's of 6 so negative 3 and then we put our 6 in here because that's the n and then plus 10 then that would give us negative 18 plus 10 or a's of 6 is going to be equal to negative 8 you may notice that when you look at these general terms they look very linear and common difference is actually equivalent to the slope of a line so now they're going to ask us to find the number of terms so we need to know what that n is well we know that a's of n is equal to 42 and remember the formula again a's of n is equal to a's of 1 plus your d times n minus 1 well let's plug and chug what we know a's of 1 is 4 and then plus d but we know that to be 2 and we don't know what the rest of that is n minus 1 because we're going to try to solve for n but we do know on this side what a's of n is equal to it's equal to 42 okay so we just plugged in what we knew now we're going to just solve the equation I am going to simplify over here 42 is equal to 4 plus 2n minus 2 so 42 will be equal to 4 minus 2 would be 2 plus 2n subtract the 2 to get it to the other side so I'd have 40 is equal to 2n and if I divide by 2 I find out that n is equal to 20 which means 20 terms and this is find the nth term we know the n is 20 a's of 4 which is a's of 1 plus 2 times 20 minus 1 and we should get 42 if we did this correctly because we already know what a's of n is so 4 plus 2 times 19 and that's going to give us 38 plus 4 which is going to be 42 so we did it right so what else can we do with these arithmetic sequences again a's of n is equal to a's of 1 plus d times n minus 1 keep that in mind find the common difference d and the value of a's of 1 given all of this a's of 5 is equal to negative 17 a's of 11 is equal to negative 2 oh wow we don't really know a whole lot here do we so let's see if we can alter what's going on here let's say that we wanted to start with a's of 11 and I have to plug in as much as I can and then find one variable well I want to first find that common difference so I can say this is equal to a's of 5 plus if I have a's of 5 a's of 11 was bigger and how many more common differences did I need I needed 6 more common differences and that's really because we started with 5 plus 6 will equal 11 this was the 5th term I needed 6 more common differences to actually be at the 11th term now let's plug and check what we know a's of 11 is negative 2 a's of 5 is negative 17 plus 6d and we're trying to solve for d add 17 to this side and you're going to have 15 is equal to 6d dividing by 6 we have 15 divided by 6 but we can reduce that which will make the next part easier so if I reduce that they're both divisible by 3 so d is 5 over 2 now what do we do we got all kinds of things we can do but we kind of thought about this a little bit before I could say that I know let's say a's of 5 because it'll be smaller I don't know what a's of 1 is and I do know what my common difference is I have a's of 1 and I need a's of 5 so how many common differences do I need I need 4 of them or 5 minus 1 would be 4 so let's plug in what we know a's of 5 was negative 17 I could have used negative a's of 11 here as well it really doesn't matter which one I use a's of 1 is what I'm looking for plus when I multiply here I'm going to have the 4 and the 2 cancel each other to have just 5 times 2 or 10 and if I subtract the 10 then I know that a's of 1 is equal to negative 27