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Arithmetic Series - Overview

In mathematics, an arithmetic series is the sum of the terms of an infinite sequence, or the sequence of sums of the terms of such a series. The term arithmetic means that the sequence of numbers is an integer sequence. The sequence can be finite or infinite.


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Video Transcript

Okay, so in this video, we're going to be talking about arithmetic. Siri's. So the first thing that you have to know about arithmetic Siri's is that the change is constant. Change is constant. What that means is that we might have something to 468 10 12 dot dot dot because we know from here it's increasing by two, increasing by two, increasing by two and so on so forth. So in that way, the increase is always to, and it doesn't just apply to an increase we could have 50 40 30 20 dot dot dot, in which case are decrease would be 10 every single time, and it's still constant, so whether you have an increase or decrease doesn't matter. So that's kind of the basis of arithmetic. Siri's. So then let's think about how we can represent that in the equation. The first type of the equation that we have is called the Ex Excuse me, the explicit equation and that goes by the form of a sub en is equal to a sip one plus de times and minus one. So the a sub n is that term the value of the term you're looking for ace of one because that end value is one that's gonna be the first term. D is the difference. So in that 2468 equation, we would be having to as a different since we are increasing by two each time and N would be the number of terms. So then, if we had that 2468 situation and we're looking for the 13th term, you can use this instead of counting. So you could say the first term is to the difference is, too. And the number of terms we're looking for is 13, and then our last or 10th term is what we're solving for. So then, if we do that out 13 minutes, one becomes 12. Now we've got to plus 12 times two, which is equal to 24 plus two, which simplifies down to six. So we know our 13th term is six. That's kind of how you would use the explicit formula. So then let's see what this So then, for the recursive formula, well, the recursive formula is a lot simpler. It's a sub end is equal to a sub end minus one class city, But because of the situation. It's saying, Okay, if we want the 13th term, we have to know the 12th term and add the difference to it. But that also means that you need to know what the previous term is in order to use the recursive formula. So then let's talk about Okay, Well, how would we use this in a situation? So sometimes, instead of giving, being given a pattern, you're given, like a specific term number and the value. So, for example, if we know that ace of 13 is equal to 57 8 of 29 is equal to 101 Well, how would we do this if we're first? I would say you would probably want to put this into a formula. So if you try to put that into a formula, our formula was ace of N is equal to a sev n minus one. It was D times n minus one. At that point, you can plug in what you know. So ace of End would be 57. I'm sorry. This is a sub one. Ace of one is we don't know the ace of one yet because we don't know the first term. We also don't know the difference, But we sure do. You know the number of terms, which is 13. So we have this. So we can't quite simplify for that yet because we have one equation with two unknowns. So let's use our second piece of evidence right here and make another equation. So 101 is going to equal a sub one which we don't know, plus de which we also don't know times 29 minus one. So at this point, you can see that now we have two equations with the two same unknowns, the ace of ones and these. So then you can use a systems of equations to solve for one of the variables and then plug that variable back in. It's pretty much the same thing on saying something like 57 is equal to X plus y times 12 and 101 is equal to X plus y times 28. Well, you know how to solve this because this is from algebra. You just have to use assistance of equations then. So that's how you would simplify that on Let's talk about the sudden equation before arithmetic sequence. They could not only ask you about the sequence, the the components of the sequence, which was the ace of em with a someone with a D or the end. But they could also ask you to add all those numbers in that term. So, for example, if it was 2468 10, 12, 14, they can ask you to add Okay, well, what's the some of the first 20 terms? So the equation for that is going to be s sub n s indicating some is gonna be n over to times a C one plus ace of N. So what that means is, if we're looking for the 1st 20 terms of this some of the 1st 20 terms of this we would plug in. Okay, so s some 20 would become 20 divided by two times the ace of one, which is to plus the 20th term, which we confined by plugging this situation into here, let's first do that. So two plus two times 20 minutes, one then that gives us 19 times to which is 38 plus two, which is 40. I'm gonna put that right into here. So then, from there, that's when you would be able Thio. Just put that into your calculator. So we've got 10 times 42. That becomes 420. So the mechanism behind this let me just show you that really quick. So when you just write out the 1st 22 year terms 2468 12, 14, 16, 18, 20. Let me just do the first time terms, actually. So for the first time terms, they're saying, add the first term in the last term and then if we add this and this so that becomes 22. 4 plus 18 is also 2022. 6 16 is also 22. A plus 14 is also 22. 10 plus 12 is also 22. So essentially what you're doing is you're creating pairs that add up to the same exact thing and then multiplied by the number of pairs, which is going to be half of the numbers that you have. So currently we have 123456789 10 terms, but then we have five pairs, so that's half of 10. So that's kind of the mechanism behind it. If you ever forget, you can kind of use that to try to create this equation again, because that really helps me.