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Arithmetic Series - Example 4

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 fourth example out of our earth, Nick Siri's Siri's and it says, given the following information right out the explicit formula recursive formula. Also find that some of the 1st 25 terms and the information that we're given is that Ace of 11 is equal to 1 10, and eight of 37 is equal to 3 70. So what that means is that the 11th term is 110 37. Term is 370. Well, let's go on and try to write out the explicit formulas first before the explicit formula. We know that the template is a sub end is equal to a sub one plus d times n minus one where a sub one is the first term. D is the difference and is the number of terms and a C N is the value of that term number. So, currently, um, if we're looking at this piece of evidence, we don't have the first term and we don't have the difference. But we could plug in the rest of the information because we know that the value is 1 10, so we conclude that information end, and then we also know that that term value is 11, so you can put that in. So then let's make a separate equation for Ace of 37. Well, we know that that value is 370 and we know that we don't know the 81 We don't know the D, But again, it's the 37th term so we can put that in for the end. Well, at this point, you should be able to recognize that we have a systems of equations because you have two equations with two unknowns that air the exact same. We have the ace of one and the D that are missing from both equations. So I'm gonna go ahead and subtract 10 d from both sides of the first equation because I love minds when it's tens of 10 d, and that isolates our ace of one because I got 1 10 minus 10 D is equal to a sub one from the second equation. I'm going to subtract 30 60 from both sides of the equation on. Then that should give me 370. Minus 30 60 is equal to a sub one. So now I've isolated a someone from both sides of the equal for both equations, which allows me to plug it in or substitute. So since we have a sick one there, I'm gonna put that right there. Then that gives me 1 10 minus 10 D is equal to 3. 70 minus 36. The So then, at this point, I can try to combine my life terms. So I'm gonna add 30 60 to both sides, and I'm also going to subtract one time from both sides. And that's gonna leave me with 2060 because least you cancel out is equal to 3. 60 because these to cancel out. And then finally, if I divide both sides of my equation by 26 I should be able to find my D to my d policy. So minus 10 D as this was 3 70 minus 1 10 was to 60. So at that point, if we do to 60 divided by 26 that gives me a D of 10. So now we found our d. But we still have to find our ace of one, which means that I can just take this value of D and plug it into either of the two equations. So I'm gonna plug it into the first equations. Since that looks a little bit easier, I think that that gives me 1. 10 is equal to a sub one plus 10 times 10. So then, if I subtract the 100 since 10 times 10 is 100 from both sides of the equation, I find that ace of one is equal to 10. So now if I want to write out my explicit formula, that's just gonna be Ace of Em is equal to a sub one, which is 10 was de, which we found to be Tena's well and then end minus one. There's are explicit. So then, for our recursive, that one is easy because we just have to write out a sub n is equal to a sub n minus one plus the devalue. When we just found the d over here when we were finding our explicit. So we're gonna go ahead and recycle that and say a seven is equal to a sev n minus one plus 10. So here's a recursive, but then we have to be able to find our some of the 1st 25 terms. Well, if we're looking for the some of the 1st 25 terms, let's go ahead and write out the equation for some, which is s sub n is equal to end over two times Ace of one plus ace of em. So currently we know the end. We know that we're looking for the some of the 1st 25 so it's gonna be s a N is equal to 25/2. We do know our ace of one, because that's just gonna be the very first term which we found a way over here to be 10 and then for our ace of end. That's what we don't know right now. So let's go in and find that because we can use the, um, exclusive formula from earlier to find 25th term. So we're looking for the ace of 25. That's gonna be 10 plus 10 times 25 minus one since we're looking for the 25th term. So if we calculate that out, Ace of 25 is going to be 250. I'm going to put that right into our ace of N So then our equation becomes 25/2 times, 260. If I do that in my calculator, that gives me 3250. So we know that the some of the 1st 25 terms is 3000, 250. And we did need our explicit formula in order to be able to find the sum, because otherwise we wouldn't have known what a 25th term is. So keep in mind that in terms of solving through this, we used the two data points that we had to create an equation for each of them, and we didn't know the first term or the D for either of them. But that's what allowed us to set up a systems of equation on, then soft through it to find the devalue and the ace of one value and then turn that into an explicit and recursive formula. And then, for the some of the 1st 25 terms, we had to use that explicit Formula five 25th term and then plug it in