College Algebra: Arithmetic Sequence Problems

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All right. Now I want to tell you that a very special kind of sequence. These are known as Arithmetical Sequences.
Now let me give you an example of an arithmetical sequence and see if you can guess what makes it so arithmetical. How about 3, then 8, then 13, then 18, then 23, and this pattern keeps going. What is the pattern? Well it seems like actually I need to know the previous one and if I know the previous one, how do I get to the next one? Seems like every time I just add 5. So an Arithmetical Sequence is just a sequence where to get the next term you just take the previous term and add some fixed number, in this case we just add 5, 3 + 5 = 8, 8 + 5 = 13, 13 + 5 = 18, and so on. Now, in fact there's a really easy way of expressing this. Think about it. If to get the next term you just take the previous term and add this fixed number, then if that number were let's say, well in fact, let me write this down this way. If I take the n plus first term, that's just the n plus first term in the sequence wherever n is and I subtract off the previous term, what will this have to equal? Well, it would have to equal 5, because to get this term I just take the previous term and add 5 to it. So even though I don't know what n is, no matter which pair I take, if I take their difference, this minus that, it's 5, this minus that, it's 5. This minus that is always 5. That's what makes a sequence an Arithmetic Sequence. The difference of consecutive terms is 5, and notice that's just the generic way of me saying; take a term and the one that comes before it because the subscripts are off by one.
Well, now if you think about it, how can I actually write down a formula that generates these sequences? Well, let's see. Suppose that I have an arithmetical sequence and at each stage I add d. So let's take an arithmetical sequence and each stage I add d, so I have something like this first one, and the next term I'll have a[1] + d. The next one I'll have is a[1] +, and I take this, a[1] + d and add d again, so that's a[1] + 2d. Then I take this and add b to it, that would be a[1] + 3d and then I would have a[1] and I add d and I have a[1 ]+ 4d, and so on. Well, this actually gives us the formula because if this is the first term, and this is the second term, and this is the third term, and this is the fourth term, and so on, we can see that a formula to get the nth term will be just a[1] plus - now what number do I have to multiply d by? Well, in the third I multiply by 2, in the fourth I multiply by 3, so in the nth I multiply by n - 1 and that actually gives me a formula that actually allows me to find the nth term in an Arithmetical Sequence.
For example, let's suppose that we have an arithmetical sequence and a[1] = 3, and that difference between the terms is 2. Then what's a[6]? Well, we can find a[6] real easily now. All we've got to do is say, okay, what is a[6]? Well, a[6], using this formula, would be a[1], which is 3, plus, well this is the 6th term so I take 6 - 1 = 5 x that difference, which is told to be 2. So this is just going to be 3 + 10 = 13. So the 6th term in this particular arithmetical sequence, the sequence that starts, 3, and then 5, and then 7, you go on doing here starting with 3 and I keep adding 2, the 6th term will be 13. So you can find that real easily.
What if I give you just two random terms? I'd tell you I'm thinking of an arithmetical sequence and let's suppose that the 15^th term is 6 and the 17^th term is 2. How could I find out a formula for the sequence in general? I tell you it's arithmetic. Okay, if I tell you it's arithmetic, what would I do? Well, I know the formula, the formula is a[n] = a[1], the first term, x n - 1 x that difference. If I plug this fact in, what do I know? When n = 15, I know this number must be 6, a[15] = 6. So I'd see 6 = a[1] +, and then what's n - 1? Well, if n is 15, this would be 15 - 1 = 14d. So there's a fact, but I don't know what a[1] is or d, those are still unknowns. But I know this fact, I know that a[17] = 2, so that means that if this thing will be 2 when n = 17. So that means that equals a[1], and now I put in a 17 - 1 = 16d.
Well, look what I have; I have 2 equations and 2 unknowns. We know how to solve that now, we can actually figure out exactly what a[1] is and what b is. So what would you do? Well, one thing you can do for example, is just let us subtract these equations. So I'm going to use the elimination method right out of the get go. If I subtract these equations, I see 6 - 2 = 4, a[1] - a[1] = 0, 14 - 16 = -2d. So if I solve for d, I see that d must equal, I divide through by negative 2. And then, when would a[1] be? Well, I would just take -2 and plug it back into one of these things here, I'd plug it back into here, for example, I would see that 6 = a[1] + -2 x 14, which would be -28. So if I take -28 and bring it to that side, I see that a[1] = 34. So, in fact, just knowing two terms in arithmetical sequence, even if they're not the first two, I actually know everything. I start off with 34 and then at every level I deduct 2, the negative signings, I subtract 2. So this sequence looks like this, 34, then 32, because I subtract 2 each, then 30, so this is the sequence that's going to be decreasing, then 28, and then 26, and so on. It turns out that if you keep going, the 15^th term will be 6 and then what's the 16^th term? It's going to be 6 - 2 = 4, and the 17^th term is 4 - 2 = 2 and there's that.
So just knowing two terms you can set up an equation, two equations and two unknowns, and you can actually solve that, and by solving it you can actually find out this sequence exactly. So to understand an arithmetical sequence all you need are two pieces of information. And those pieces of information could be the first term and the d, the thing you add each time, or it could just be two arbitrary terms in the sequence and you could actually then set up using this formula two equations and two unknowns, and figure out the first term and figure out that d, that common difference between consecutive terms.
Give these a try and enjoy basic arithmetic sequences.
Further Topics in Algebra
Sequences
Solving Problems Involving Arithmetic Sequences Page [1 of 2]