Video: Finding the Terms of a Geometric Sequence given Their Sum and Product

Find the three consecutive numbers of a geometric sequence, given that the sum of the terms is βˆ’14 and the product is 216.

09:30

Video Transcript

Find the three consecutive numbers of a geometric sequence, given that the sum of the terms is negative 14 and the product is 216.

First of all, we need to think about what a geometric sequence, or sometimes called a geometric progression, is. In geometric sequence, each term after the first term is found by multiplying the previous one by a common ratio. If we let our first term be π‘Ž and our common ratio be π‘Ÿ, we can write our consecutive numbers in this way. The first one is π‘Ž. The second one would be π‘Ž times π‘Ÿ, because in a progression, the next term is found by multiplying the previous term by the common ratio. The third term would then be π‘Žπ‘Ÿ, the second term, times π‘Ÿ again. And we can simplify that to π‘Žπ‘Ÿ squared.

We can use this information to write a couple of equations. We know the sum of these values is negative 14. And that means we could say that π‘Ž plus π‘Žπ‘Ÿ plus π‘Žπ‘Ÿ squared is equal to negative 14. Since all three of these terms have an π‘Ž-variable, we can undistribute that to simplify, which would leave us with the statement π‘Ž times one plus π‘Ÿ plus π‘Ÿ squared equals negative 14. And we might wanna rearrange this equation so that the π‘Ÿ squared term comes first and the constant one comes at the end. Because that’s more common when we’re dealing with these kinds of equations.

We haven’t changed any values. We’ve just rearranged the equation. And that’s all we can do with the sum equation for now. We know that the product of these three terms is 216, which means that π‘Ž times π‘Žπ‘Ÿ times π‘Žπ‘Ÿ squared is equal to 216. Because we’re dealing with multiplication, we can multiply π‘Ž times π‘Ž times π‘Ž, which will give us π‘Ž cubed. This is because we’re dealing with π‘Ž to the first power. And π‘Ž to the first power times itself times itself again equals π‘Ž cubed.

We also have π‘Ÿ times π‘Ÿ squared. This would be multiplying π‘Ÿ to the first power times π‘Ÿ squared, which would be π‘Ÿ cubed. π‘Ž cubed times π‘Ÿ cubed equals 216. We can rewrite this as π‘Žπ‘Ÿ cubed equals 216. And then we can take the cube root of both sides of the equation. The cube root of π‘Žπ‘Ÿ cubed equals π‘Žπ‘Ÿ. And the cube root of 216 is six. If π‘Ž times π‘Ÿ is six, that means we found our second number.

But we don’t have enough information yet to find our first or our third number. What we want to do is see if we can find something to plug in to our first equation. We can substitute something in for π‘Ž in terms of π‘Ÿ. Or we can substitute π‘Ÿ in terms of π‘Ž. To do that, we’ll use the statement π‘Ž times π‘Ÿ equals six. If we divide both sides of the equation by π‘Ÿ, we will be able to say that π‘Ž equals six over π‘Ÿ. If we had divided both sides of the equation by π‘Ž, we can also say that π‘Ÿ equals six divided by π‘Ž.

We’re ready to try some substitution. If we plug in six over π‘Ÿ for π‘Ž, we would have the equation six over π‘Ÿ times π‘Ÿ squared plus π‘Ÿ plus one equals negative 14. Or if you plugged in six over π‘Ž in for π‘Ÿ, you would have π‘Ž times six over π‘Ž squared plus six over π‘Ž plus one equals negative 14. Both of these equations would help us solve for our final answer. But the first option will have a little bit simpler calculations. So let’s go with this one.

Right now we have an π‘Ÿ in a denominator. If we multiply both sides of the equation by π‘Ÿ over one, on the left, the π‘Ÿs cancel out. And we’ll have six times π‘Ÿ squared plus π‘Ÿ plus one equals negative 14π‘Ÿ. Then we need to distribute our six. We have six π‘Ÿ squared plus six π‘Ÿ plus six equals negative 14π‘Ÿ.

If we want to solve this equation for π‘Ÿ, we do that by setting it equal to zero. And that means we need to add 14 π‘Ÿ to both sides, which will give us six π‘Ÿ squared plus 20π‘Ÿ plus six equals zero. We notice that all of the coefficients are divisible by two, which means we can divide the entire equation by two or multiply by one-half. And then we have three π‘Ÿ squared plus 10π‘Ÿ plus three equals zero.

We’ll want to try and factor this equation. Usually, when we work with factoring equations, we’re dealing with π‘₯ squared. But it doesn’t change the process just because our variable is π‘Ÿ. Since three is a prime number, we know that we’ll have three π‘Ÿ on one side and π‘Ÿ on the other, as those are the only factors of three. And the same thing is true for our coefficient of three. So we know we’ll be dealing with three and one.

We need to make this middle term be equal to 10π‘Ÿ. If we multiply on the outside, three π‘Ÿ times three, that’s nine π‘Ÿ. And our other two terms multiplied together to equal one are one π‘Ÿ plus nine π‘Ÿ equals 10π‘Ÿ. Everything is positive. And we have three π‘Ÿ plus one times one π‘Ÿ plus three. We set both of these equations equal to zero. On the right, we subtract three from both sides. And we see that when π‘Ÿ equals negative three, the equation equals zero.

On the left side, we need to do two steps. First, we subtract one from both sides. And then we divide both sides by three. So π‘Ÿ is also equal to negative one-third. But what do we do with this information? Well, we know that π‘Ž times π‘Ÿ equals six. And π‘Ÿ will either be negative three or negative one-third.

Let’s consider the first case first, when π‘Ÿ is negative three. To find out what π‘Ž would be equal to if π‘Ÿ was negative three, we divide both sides of the equation by negative three. Six divided by negative three equals negative two. So we want to say when π‘Ÿ is negative three, π‘Ž is negative two. π‘Žπ‘Ÿ is six. That’s our first two terms. And our third term would be equal to Negative two times negative three squared. Negative three squared is nine, times negative two would be negative 18.

And now we’ll consider when π‘Ÿ is negative one-third. To find out what π‘Ž is equal to when π‘Ÿ is negative one-third, we multiply both sides of the equation by negative three, which tells us that π‘Ž is equal to six times negative three. π‘Ž equals negative 18.

Now we’re looking at the case when π‘Ÿ is negative one-third. We already know that the middle term π‘Žπ‘Ÿ must be equal to six. But when π‘Ÿ equals negative one-third, the first term is negative 18. And the third term is π‘Ž times π‘Ÿ squared, which would be negative 18 times negative one-third squared. Negative one-third squared is one over nine. Negative 18 times one over nine is negative two.

This turns out to be really interesting. We have either negative two, six, negative 18 or negative 18, six, negative two. Before we move on, it’s worth checking to make sure that we’ve calculated everything correctly and that the two statements we started with are true with these values. Negative two plus six plus negative 18 does equal negative 14. And because we know that addition is commutative, negative 18 plus six plus negative two is also equal to negative 14. In the same way, negative two times six times negative 18 equals 216. And it doesn’t matter if we change the order. Multiplying these three values still yields 216. Since our question has only asked for the consecutive numbers, either negative two, six, negative 18 or negative 18, six, negative two would be a correct response.

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