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Arithmetic Series - Example 3

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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Okay, so this is gonna be the third example out of our earth. Nick seats into Siri's. And the question says, find the explicit and recursive formulas given the following information. A sub 10 is equal to 14, and Ace of 37 is equal to 1 22. So now let's go ahead and think about what that means. Um, ace of tennis, equal to 14. All that means is that the 10th term in that sequence is 14 and then the third 37th term in that sequence is 122. So let's go ahead and see what we could do to use that information. So I'm gonna go out and start with the explicit forming last month explicit cursive because explicit is a little bit easier. S. O. R general template for explicit formula is a sub en is equal to a sub one was three times n minus one. So I'm gonna go out and use the first piece of information and plug that into this formula as much as possible so ASAP and is gonna be 14. Lets me know that that's the value of the 10th term. We don't know a sub one and we don't know the devalue difference. But we do know that the term number is 10. We have this. So if I simplify this town I have 14 is equal to a sev one, which is the first term plus D the difference times nine. So I'm gonna use the second equation and make another equation or second and piece of information to use. Make another equation. Then I get 1 22 which is a value is equal to eight of one because we still don't know the ace of one plus de because we still don't know D But we do know the term number is 37. So then again, if we simplify this down, we get 1 22 is equal to a sub one plus D times 36. So at this point, we can use these pieces of information Thio make a systems of equations because we see that we have two equations with two unknowns that are the same. We've got an ace of one here and here. We've also got a d here and here. So then if I combine them, I'm gonna go ahead and isolate for a sub one on both of the equations. So I'm going to subtract 90 from both sides of the equation. That leaves me with 14 minus 90 is equal to a sub one. And then from here, I'm going to subtract 30 60 from both sides of the equation. And that gives me 1. 22 minus 30 60 is equal to a someone at this point, since both of them are equal to ace of one. I'm going to set those two equal to each other eso you could I think of it as substitution or sending equal to each other. So we both got a sub one. So from here and then plug that in right here, and that's going to give me 14. Minus 90 is equal to 1 20 to minus 30 60. I'm here. You just have to try to combine like terms, adding 30 60 to both sides of the equation and then subtracting 14 from both sides of the equation. And that's gonna leave me with 27 d is equal to 100 it at this point you just have to finally divide by 27 on both sides of the equation. so that you could isolate for that devalue on. Then you can get that d is equal to for at that point, you know your D so you can put that back into either of the equations. So I'm gonna use the first equations, and that's a little bit smaller and simpler. So then I have 14 is equal to a sev one plus four times nine. That means it's 14 is equal to a sub one plus 36 and then to isolate for that ace of one variable, I'm going to subtract 36 from both sides of the equation, and that gives me that ace of one is equal to negative 22. Then from here, I could make that explicit formula. So are explicit formula is gonna be, um, a sub em is equal to a sub one, which we just found to be negative 22 plus de which we just want to before times are and minus one. So here is our explicit formula. So now I just have to make our recursive formula and we know are recursive formula happens in the form of a sev n is equal to a sub and minus one plus deep. So that one is easy, because again, we've already found R D. We found our devalue to be four. So when we say, a sub n is equal to a sub end minus one plus four. So once you find explicit, it's easy to find the recursive so you can go from there. Or you could find the recursive first and then go back to the explicit whichever works. So now I found my recursive formula and my explicit formula. So these are are falling formulas. So upon looking at the information provided initially, it looks like there's not that much information given to you. But it's actually enough to be able to piece it together if you are using a systems of equations. So like we did here to speak, sure to fill in each piece of information into its own equation. And then from there, I'm used the systems of equations