Okay, so in this video I want to talk about a little bit more about arithmetic sequences. I want to talk about finding a formula for the nth term. I'm going to put the formula here.
It says to find the nth term of an arithmetic sequence. We use the formula a sub 1 plus n minus 1 times d. a sub 1 is going to represent the first term.
If you want the nth term, you plug in the number n minus 1, and d again is the common difference. So I want to talk about real quick just, you know, kind of a real loose justification of this formula. And then we'll use it and we'll do a couple easy problems and then something maybe a little trickier. So in my first video about arithmetic sequences, we said, yeah, if you've got the sequence 3, 8, 13, 18, 23, 28, and if that continues, we said this is an arithmetic sequence.
They're differing by 5, so that's our d-value. Just a couple things about notation here. The first term we call that, that's our generically a sub 1, this is a sub 2, a sub 3, a sub 4, etc. So let's see if we can kind of justify that formula. Notice, okay, so we said our first term is just equal to 3. Well, our second term, okay, I know it's equal to 8, but I'm going to incorporate the d value into my formula.
So I could say it's 3 plus, well, 5, because that's what they're differing by. a sub 3, well, again, now you've added 5 once, and you've added 5 twice, which I could write as 10, but I'm going to write it as 2 times 5. You've added two 5s to the number 3 to get to the third term. To get to the fourth term, we would have to add yet another 5. So we would have 3 plus, well now we've got 3 5's, and maybe you can kind of see the pattern here, a sub 5. The only thing that's really changing is the number out front of the d value.
So we would get 3 plus 4 times 5. a sub 6 would be 3 plus 5 times 5, maybe one more, a sub 7 would be 3 plus 6 times 5. And just notice the relationship. If we're at the 6th term, we're using one fewer of the d value. If we're at the 7th term, we're using one fewer 6 of that d value.
So generically, it looks like if we have a sub n... It looks like we're using the first term. Again, 3 is a sub 1. Whatever term we're at, we're using 1 fewer, so that would be n minus 1. And again, 5 in this particular sequence represents the d value. Okay, so that's kind of a way that you could justify the formula, or I think produce it real quick if you had forgotten. Alright, so let's use this to do a couple problems here real quick.
So a couple easy ones, or you know, not easy, but I think about as straightforward as they'll come with this. Suppose we have the arithmetic sequence. Let's use what we just had a second ago. 3, 8, 13. 18, 23, 28, 33, etc.
Suppose we wanted to do a couple things. Suppose we wanted to find the 10th term, and suppose we also wanted to find maybe the 202nd term. To find the 10th term, we could almost write it out. So again, this first number, that's our a sub 1, a sub 2, a sub 3, a sub 4, a sub 5, a sub 6. This would be a sub 7. So we could write out a couple more.
Okay, our d value is 5. So we would get 38, 43, 48, and then 53. I think that's how many I need. a sub 8, a sub 9. Whoops, I did one too many. So there's our a sub 10. It should be the value 48. Okay, if we use our formula down here, again, this is what we're plugging in.
It says, hey, it says you need to plug in. The first term, and our a sub 1 in this case is the number 3. So I'm going to plug that in. It says whatever number is downstairs, you subtract 1 away from it. So since we have a 10 here, we'll take 10 minus 1, and then we multiply that by our d value, which we said is 5. So really we're going to get 3 plus 9 times 5. Well, 9 times 5 is 45. 45 plus 3 is going to give us 48. And hey, that's what we said you would get if you actually write them all the way out.
So obviously, a sub 202 here, you wouldn't want to write this one all the way out. Maybe you would. Maybe you're a masochist.
I wouldn't want to. Again, it says just take the first term, which is 3, plus... We subtract 1 away from this number, so we'll take 201. And again, we would just multiply that by our d value, which in this particular example... we set as 5, so that's 3 plus 1,005.
So it looks like our 202nd term would be the number 1,008. Okay, so maybe let's do one just a little bit less straightforward. Suppose we know we have an arithmetic an arithmetic sequence.
And suppose we know that our 8th term is equal to the number 25. And suppose we know that our 14th term is equal to 43. Okay, I want to do a couple things here. Let's find a couple things. Let's find both the first term...
And let's also find our d value for this particular sequence. Okay, all I'm going to do in this problem is just fill in what I know. Okay, so generically a sub 8 is 25, and I'm just going to again fill in our formula, this a sub n equals a1 plus n minus 1 times d.
So this one's going to be a little trickier. There's certainly different ways you could do this. You could find the difference and divide and calculate your d value that way. I'm going to do something a little more algebraic. So it says a sub 8. Well, a sub 8, generically from this formula, it says it would be the first term, a sub 1, plus 8 minus 1, which is 7, times d.
And again, we know that that equals the number 25. Likewise, a sub 14, if I fill in my formula, it would be the first term. You take 1 less than 14, which is 13 times d, and that's going to equal 43. Okay, so what I've done using my information and by filling in what I know in this formula, I've created really a little system of equations. Okay, so this is what I'm going to work with. Now in this problem, so I'm going to write it one more time here, kind of a little bit down here.
So it says we have a1 plus 7d equals 25. We've got a1 plus 13d equals 43. Okay, you've got a system of equations. There's lots of ways to solve these. Substitution, you can put in a matrix, you can do elimination by addition. So I'm not going to discuss this.
...and gory details, but if you're confused about what I do here, look up one of those things. Systems of equations, solving systems of equations using elimination by addition, substitution. I know I have videos on both those. So in my case, I'm going to use substitution.
I'm going to multiply this first equation by negative 1, and it's just going to change the signs. So I would get negative a1 minus 7d equals negative 25. I'm going to leave the bottom one alone. So I'm just going to copy that one down. And now I'm just going to add them up.
The negative a1 and the positive a1 cancel. Negative 7d and 13d is 6d. Negative 25 and 43 is going to give us 18. And if we divide both sides by 6, we get our d value to be 3. Okay, so now we know our d value. That was one thing I wanted to figure out.
And the other thing I think that we were looking for was our a1. And now we can really just use either of our equations. So these were the two equations we just played with. I'm going to take maybe the first one, and I'm going to plug in the d value.
So it says a1 plus 7 times our d value, which we just found over here to be 3. So I'm going to plug that in. equals 25. So a1, 7 times 3 is 21, equals 25. And you could subtract 21 from both sides and get that a1 equals 4. Okay, so a longer video than I anticipated, but a couple different things about using this formula. I think, you know, especially this last one, I think would be kind of a good sort of challenging question. either teaching or if you're a student trying to prepare this to me would be something a little trickier to maybe ask a class in regards to arithmetic sequences. Alright, I hope this video makes some sense.
If you have questions or comments, as always feel free to post them. Hopefully me or somebody else can point you in the right direction.