Everyone, welcome back. So, I hope you'll get a chance to try out this example problem. And if you got a little stuck and didn't know where to go, that's perfectly okay. This one's a little bit of a tricky one. Let's go ahead and get started here. The 4th and 6th terms of a sequence are given to us. We're told that \( a_4 \) is negative 2 and \( a_6 \) is equal to 6. And we're supposed to use this information to find the 18th term of the sequence. Now remember, whenever we're trying to find really high indexes of sequences, you're always going to want to start with a general formula. You don't want to do it recursively. Right? So let's start out there. So we're told that \( a_n \) is going to be \( a_1 + d(n - 1) \). That's the general formula. So do we know \( a_1 \) and \( d \), and can we figure it out from the information that is given to us? Well, not really, because the only thing that we're told here is we're actually just given two numbers in the sequence, and none of them are \( a_1 \). We're told that \( a_4 \) is negative 2 and \( a_6 \) is 6. So in other words, I don't know what the first term in this sequence is, and because I don't even have 2 consecutive terms, \( a_4 \) and \( a_6 \), I don't even know what the common difference is between these terms either. So how do I even get started with this problem? Well, let's just actually use the information that we do know in this example. So from this general formula here, what we're told is that \( a_4 \) is equal to \( a_1 + 3d \). And if we're using \( n = 4 \) here, then this is just going to be 3. \( a_6 \) is just going to be \( a_1 + 5d \). And if we're using \( n - 1 \) and \( n = 6 \), this is going to be 5. Now we actually know what these numbers equal. This is equal to negative 2, and this is equal to 6. So how do I use this information over here to find any one of the variables that I need to know? I need to know \( a_1 \), and I need to know \( d \). So now if you notice here, what I've ended up with is I've ended up with a situation where I have 2 equations. Right? I've got these 2 equations in which these two variables, \( a_1 \) and \( d \), are unknown variables. I've got 2 equations with 2 unknown variables. The whole idea here is that we're going to turn this into a system of equations. So we've got \( a_1 + 3d \) is equal to negative 2, and this is going to be \( a_1 + 5d \) is going to be 6. That's what my two equations say. Right? I just rewrote these 2 equations over here. The whole idea is that this is basically just a system of equations. And so back when we studied systems of equations, we saw different methods to solve these types of problems. We're going to use that here. This sort of system of equations is kind of like so I'm going to put this like when you had a situation where you had something like \( x + 3y \) is equal to negative 2, and \( x + 5y \) is equal to 6. So remember when we had \( x \)'s and \( y \)'s and we used different methods to solve these types of problems? Well, if you already have their coefficients lined up, then we could use here we could use the elimination method, where we basically subtract 2 equations to get rid of 1 of the variables and isolate one of them, and then we can use one equation with 1 variable, and that was easier to solve. That's exactly what we're going to do here. We're going to take these 2 equations here, and we're going to subtract them. So remember, this is the elimination method, so this is elim. Alright? So what I'm going to do here is we're basically going to subtract out the first terms, the \( a_1 \) terms over here, because those coefficients are the same. They're equal but opposite. Alright? And so when you subtract these two equations, what do you get? Well, remember, you're going to sort of subtract everything vertically straight down. Right? So you subtract all these terms. \( 3d - 5d \) gives me \( -2d \), and then \( -2 - 6 \) gives me \( -8 \). So if you divide by \( -2 \) on both sides, what you'll end up with is that \( d \) is equal to 4. So without knowing the general formula, without even knowing the sequence, just knowing 2 of the numbers, I was able to figure out the common difference is equal to 4. And if you think about this, this actually kind of does make sense because if \( a_4 \) is equal to negative 2, then that means that \( a_5 \), the next term in the sequence, would be 2. And now this totally makes sense because from 2 to negative sorry. Negative 2 to positive 2, that's a difference of 4. From 4 to 6, that's another difference of 4. Alright? So the common difference in the sequence is definitely 4. So that's one of our numbers now. We actually have the common difference \( d \), and now we just need to figure out, well, what's \( a_1 \) in this general formula? So we've got \( d \), and we need to figure out \( a_1 \). Well, just as how we use a system of equations where once we found one variable, you could plug it back into either one of these equations to solve for the other. We can do the exact same thing here. I can take this \( d \) and plug it into either one of these two equations, the two equations that I actually do know numbers for, to figure out what the first term of the sequence is. It doesn't matter which one you pick. Feel free to choose whichever one you want. And I'm just going to go ahead and pick the one with the slightly smaller numbers. So I know that \( a_1 + 3d \) over here is going to equal negative 2. Alright? So what I'm going to do here is I'm just basically going to replace \( d \) with 4 so I can figure out \( a_1 \). So what it says here is that \( a_1 + 3 \times 4 \) is going to equal negative 2. So this is going to be \( a_1 + 12 \) equals negative 2, and I can subtract 12 from both sides. What I'll see here is that \( a_1 \) is equal to \( -14 \). So that is the first term in my sequence. And, again, I figured this out by knowing what the common difference is and then plugging it back into one of my two terms that I actually do know. Alright? So that's one of my terms, and that's the other one that I need for the general formula. So now I can actually write out my general formula over here. This is going to be my general formula, and what this says is that the \( n \)th term is going to be the first term, \( -14 \) plus \( d \), in which case this is \( 4 \times (n - 1) \). So this is the general formula over here. I'm going to write this out. This is the general formula. Now, how do we use this to find out the 18th term? Well, actually, this is pretty straightforward. Now, we're just going to use \( a_n \) over here, is going to be \( -14 + 4 \times (17) \). So, if you do \( -14 + 4 \times 17 \), then what you're going to get here is you're going to get an answer of 54. So that means that the 18th term I'm sorry. I actually meant to write 18 over here. The 18th term is going to be positive 54. Alright? So that is the 18th term in the sequence. That's how you take these types of problems where you know 2 terms. You need to write a general formula. It really actually just turns into a system of equations. And then from there, you can solve. Thanks so much for watching. Hopefully, you got this. And if not, that's okay. It's a little bit tricky. Let's go ahead and move on.
Table of contents
- 0. Fundamental Concepts of Algebra3h 29m
- 1. Equations and Inequalities3h 27m
- 2. Graphs1h 43m
- 3. Functions & Graphs2h 17m
- 4. Polynomial Functions1h 54m
- 5. Rational Functions1h 23m
- 6. Exponential and Logarithmic Functions2h 28m
- 7. Measuring Angles39m
- 8. Trigonometric Functions on Right Triangles2h 5m
- 9. Unit Circle1h 19m
- 10. Graphing Trigonometric Functions1h 19m
- 11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
- 12. Trigonometric Identities 2h 34m
- 13. Non-Right Triangles1h 38m
- 14. Vectors2h 25m
- 15. Polar Equations2h 5m
- 16. Parametric Equations1h 6m
- 17. Graphing Complex Numbers1h 7m
- 18. Systems of Equations and Matrices1h 6m
- 19. Conic Sections2h 36m
- 20. Sequences, Series & Induction1h 15m
- 21. Combinatorics and Probability1h 45m
- 22. Limits & Continuity1h 49m
- 23. Intro to Derivatives & Area Under the Curve2h 9m
20. Sequences, Series & Induction
Arithmetic Sequences
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