Video: Pack 4 • Paper 2 • Question 3

Pack 4 • Paper 2 • Question 3

08:15

Video Transcript

The first seven terms of a sequence are shown in the grid. One, one, two, four, seven, 13, and 24. The rule for this sequence is the next term in the sequence is the sum of the previous three terms. Part a) Find the 10th term in the sequence. There is also a part b and a part c that we’ll come on to.

So in order to solve this problem, what we’re gonna start with is actually the eighth term. So as it says, the eighth term is gonna be the previous three terms added together. So the eighth term is going to be seven plus 13 plus 24 which equals 44. And that’s because as we said it’s gonna be the fifth plus the sixth plus the seventh terms. Okay, great, so let’s put this into our grid.

So now, we can move on to the ninth term. And the ninth term is actually gonna be the sixth, seventh, and eighth terms added together. So in this case, it’s gonna be 13 plus 24 plus 44 which is gonna give us a total of 81. Okay, great, so we found the eighth and ninth terms.

So now, we can move on to actually the final term we want to find, which is the 10th term. Well, the 10th term is gonna be the seventh, eighth, and ninth terms added together: so 24 add 44 add 81. This gives us a total of 149. So therefore, we can say that if the first seven terms of a sequence are one, one, two, four, seven, 13, and 24, then the tenth term is gonna be equal to 149. Okay, great, that’s part a solved. Let’s move on to part b.

Now, for part b, we have a bit more information. So we have that the first four terms of another sequence using the same rule, as shown in this grid, we’ve got 𝑎, 𝑏, 𝑎 plus 𝑏, and two 𝑎 plus two 𝑏.

Question b Show that the eighth term in the sequence is 20𝑎 plus 24𝑏.

So for this question, what I’m gonna do is start by finding the fifth term because we’ve got the first four here. And to find the fifth term, I’m gonna add the second, third, and fourth terms. So when I do that, I get 𝑏 plus 𝑎 plus 𝑏 plus two 𝑎 plus two 𝑏. So now, I can actually collect like terms. So I’ve got positive 𝑎 plus two 𝑎 which is gonna give us three 𝑎. And then, I have positive 𝑏 plus 𝑏 plus two 𝑏 which is gonna give us four 𝑏.

So now, we move on to the sixth term. And the sixth term is in fact going to be the third, fourth, and fifth terms added together. So we have 𝑎 plus 𝑏 plus two 𝑎 plus two 𝑏 plus three 𝑎 plus four 𝑏. So again, we’re gonna collect like terms. So we get positive 𝑎 plus two 𝑎 plus three 𝑎 which is gonna give us six 𝑎. And then, we’re gonna have positive 𝑏 plus two 𝑏 plus four 𝑏 which gives us seven 𝑏.

So now, we can move on to our seventh term. And this is gonna be the fourth, fifth, and sixth terms all added together. So we get two 𝑎 plus two 𝑏 plus three 𝑎 plus four 𝑏 plus six 𝑎 plus seven 𝑏. So again, we collect like terms. We’ve got two 𝑎 plus three 𝑎 plus six 𝑎 which is gonna give us 11𝑎. And then, we have positive two 𝑏 plus four 𝑏 plus seven 𝑏 which is actually gonna give us 13𝑏. So then, we’ve actually added our seventh term to the grid, which was 11𝑎 plus 13𝑏.

So now, we can move on to the final term, the one we want to find, which is the eighth term. And this is gonna be the fifth, sixth, and seventh terms added together. So the eighth term is gonna be three 𝑎 plus four 𝑏 plus six 𝑎 plus seven 𝑏 plus 11𝑎 plus 13𝑏. So again, we’re gonna collect like terms. We got three 𝑎 plus six 𝑎 plus 11𝑎, which is gonna give us 20𝑎. And then, we have positive four 𝑏 plus seven 𝑏 plus 13𝑏, which gives us 24𝑏.

So therefore, what we’ve done is we’ve actually solved part b cause we needed to show that the eighth term in the sequence is 20𝑎 plus 24𝑏. And we’ve done that because we’ve got the same answer there. So now, we can move on to part c.

Now for part c, we actually get a bit more information. And it says that the fifth term in this sequence is 23 and the sixth term is 41. Find the value of 𝑎 and 𝑏.

So what I’ve done is add these values to the grid. So we can see that the fifth term is 23 and the sixth term is 41. So now, what we can actually do is actually set up a pair of simultaneous equations to help us find 𝑎 and 𝑏. So what we have are two equations. So the first one is three 𝑎 plus four 𝑏 equals 23 and the second one is that six 𝑎 plus seven 𝑏 is equal to 41. Okay, so the first stage, we’re actually gonna label these one and two.

So now, what we’re gonna do is actually use the elimination method to solve our simultaneous equations. And to do this, we have to have either the same amount of 𝑎s or the same amount of 𝑏s. So what I’m going to do is actually multiply our first equation by two because what this is gonna do is actually give us six 𝑎s in both of our equations. So when I do that, so when I multiply the first equation by two, I get six 𝑎 plus eight 𝑏 equals 46. And I’ve called this equation three.

One thing to note here — this is a common mistake — is students often don’t multiply every single term in the equation by two, but we have to. So we’ve got six 𝑎 and then eight 𝑏 because four 𝑏 multiplied by two is eight 𝑏 and then equals 46 because 23 multiplied by two is 46.

So now, what we’re gonna do is actually equation three minus equation two. We’re gonna do that because what I want to do is actually eliminate one of the terms. And we’re gonna eliminate our 𝑎 terms. And when we do that, we actually get 𝑏 is equal to five. And that’s because if we have six 𝑎 minus six 𝑎, that’s zero. Then eight 𝑏 minus seven 𝑏. That gives us 𝑏. And then, 46 minus 41 is just five. So we get 𝑏 is equal to five.

So now what we’re gonna do is actually substitute 𝑏 equals five into equation one. And it’s actually worth noting you can actually substitute it into any of the equations and it work to help us find 𝑎. I’m just using one cause it’s the simplest for this particular example. So when we substitute it in, we get three 𝑎 plus four multiplied by five equals 23, which is gonna give us three 𝑎 plus 20 equals 23. So then, if we actually subtract 20 from each side of the equation, we’re gonna get three 𝑎 equals three. And then, finally, if we divide both sides of the equation by three, we get 𝑎 is equal to one. So therefore, we actually get the answer 𝑎 is equal to one and 𝑏 is equal to five.

Okay, well, also, what I want to do here is actually quickly show you another way that you could’ve found 𝑎 and 𝑏 and that’s using substitution. So to be able to use the substitution method, we’d actually carry out the first steps exactly the same way. And we get six 𝑎 plus eight 𝑏 equals 46. However, what we’d actually do now is rearrange this to make six 𝑎 the subject. And so we’re gonna do that now. So therefore, when we do that, we actually subtract eight 𝑏 from each side, we get six 𝑎 equals 46 minus eight 𝑏. And this will become our equation three.

Okay, so now, it’s actually where the substitution takes place. Because what we’re gonna do is we’re actually gonna substitute equation three into equation two. So we’re gonna get 46 minus eight 𝑏 plus seven 𝑏 equals 41. And that’s because actually we substituted equation three into equation two. So we knew that six 𝑎 was equal to 46 minus eight 𝑏. So then, if we simplify this, we get 46 minus 𝑏 equals 41 because we had negative eight 𝑏 plus seven 𝑏 which gave us negative 𝑏. Then, we add 𝑏 to each side. We get 46 equals 41 plus 𝑏. And then, finally, we subtract 41 from each side, which gives us five equals 𝑏 or 𝑏 equals five.

So great, this is what we actually used when we found out the value using the elimination method. So therefore, we actually say that “yes, that’s correct.” And then, what we do using this method is carry on to find 𝑎 as we did previously by substituting the value of 𝑏 into the first equation.

So therefore, we can say that part a the 10th term is equal 149, part b we showed that the eighth term was equal to 20𝑎 plus 24𝑏, and part c 𝑎 equals one and 𝑏 equals five.

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