# Recursive formulas for arithmetic sequences

CCSS Math: HSF.LE.A.2

## Video transcript

- [Voiceover] g is a function that describes an arithmetic sequence. Here are the first few terms of the sequence. So let's say the first term is four, second term is 3 4/5, third term is 3 3/5, fourth term is 3 2/5. Find the values of the missing parameters A and B in the following recursive definition of the sequence. So they say the nth term is going to be equal to A if n is equal to one and it's going to be equal to g of n minus one plus B if n is greater than one. And so I encourage you to pause this video and see if you could figure out what A and B are going to be. Well, the first one to figure out, A is actually pretty straightforward. If n is equal to one, if n is equal to one, the first term when n equals one is four. So A is equal to four. So we could write this as g of n is equal to four if n is equal to one. And now let's think about the second line. The second line is interesting. It's saying it's going to be equal to the previous term, g of n minus one. This means the n minus oneth term, plus B, will give you the nth term. Let's just think about what's happening with this arithmetic sequence. When I go from the first term to the second term, what have I done? Looks like I have subtracted 1/5, so minus 1/5, and then it's an arithmetic sequence so I should subtract or add the same amount to every time, and I am, I'm subtracting 1/5, and so I am subtracting 1/5. And so one way to think about it, if we were to go the other way, we could say, for example, that g of four is equal to g of three minus 1/5, minus 1/5. You see that right over here. g of three is this. You subtract 1/5, you get g of four. You see that right over there and of course I could have written this like g of four is equal to g of four minus one minus 1/5. So if you look at this way, you could see that if I'm trying to find the nth term, it's gonna be the n minus oneth term plus negative 1/5, so B is negative 1/5. Once again, if I'm trying to find the fourth term, if n is equal to four, I'm not gonna use this first case 'cause this has to be for n equals one, so if n equals four, I would use the second case, so then it would be g of four minus one, it would be g of three minus 1/5. And so we could say g of n is equal to g of n minus one, so the term right before that minus 1/5 if n is greater than one. But for the sake of this problem, we see that A is equal to four and B is equal to negative 1/5.