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Find the three consecutive numbers of a geometric sequence, given that the sum of the terms is negative 14 and the product is 216.
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First of all, we need to think about what a geometric sequence, or sometimes called a geometric progression, is.
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In geometric sequence, each term after the first term is found by multiplying the previous one by a common ratio.
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If we let our first term be π and our common ratio be π, we can write our consecutive numbers in this way.
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The first one is π.
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The second one would be π times π, because in a progression, the next term is found by multiplying the previous term by the common ratio.
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The third term would then be ππ, the second term, times π again.
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And we can simplify that to ππ squared.
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We can use this information to write a couple of equations.
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We know the sum of these values is negative 14.
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And that means we could say that π plus ππ plus ππ squared is equal to negative 14.
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Since all three of these terms have an π-variable, we can undistribute that to simplify, which would leave us with the statement π times one plus π plus π squared equals negative 14.
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And we might wanna rearrange this equation so that the π squared term comes first and the constant one comes at the end.
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Because thatβs more common when weβre dealing with these kinds of equations.
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We havenβt changed any values.
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Weβve just rearranged the equation.
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And thatβs all we can do with the sum equation for now.
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We know that the product of these three terms is 216, which means that π times ππ times ππ squared is equal to 216.
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Because weβre dealing with multiplication, we can multiply π times π times π, which will give us π cubed.
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This is because weβre dealing with π to the first power.
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And π to the first power times itself times itself again equals π cubed.
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We also have π times π squared.
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This would be multiplying π to the first power times π squared, which would be π cubed.
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π cubed times π cubed equals 216.
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We can rewrite this as ππ cubed equals 216.
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And then we can take the cube root of both sides of the equation.
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The cube root of ππ cubed equals ππ.
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And the cube root of 216 is six.
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If π times π is six, that means we found our second number.
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But we donβt have enough information yet to find our first or our third number.
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What we want to do is see if we can find something to plug in to our first equation.
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We can substitute something in for π in terms of π.
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Or we can substitute π in terms of π.
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To do that, weβll use the statement π times π equals six.
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If we divide both sides of the equation by π, we will be able to say that π equals six over π.
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If we had divided both sides of the equation by π, we can also say that π equals six divided by π.
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Weβre ready to try some substitution.
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If we plug in six over π for π, we would have the equation six over π times π squared plus π plus one equals negative 14.
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Or if you plugged in six over π in for π, you would have π times six over π squared plus six over π plus one equals negative 14.
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Both of these equations would help us solve for our final answer.
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But the first option will have a little bit simpler calculations.
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So letβs go with this one.
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Right now we have an π in a denominator.
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If we multiply both sides of the equation by π over one, on the left, the πs cancel out.
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And weβll have six times π squared plus π plus one equals negative 14π.
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Then we need to distribute our six.
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We have six π squared plus six π plus six equals negative 14π.
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If we want to solve this equation for π, we do that by setting it equal to zero.
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And that means we need to add 14 π to both sides, which will give us six π squared plus 20π plus six equals zero.
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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.
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And then we have three π squared plus 10π plus three equals zero.
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Weβll want to try and factor this equation.
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Usually, when we work with factoring equations, weβre dealing with π₯ squared.
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But it doesnβt change the process just because our variable is π.
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Since three is a prime number, we know that weβll have three π on one side and π on the other, as those are the only factors of three.
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And the same thing is true for our coefficient of three.
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So we know weβll be dealing with three and one.
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We need to make this middle term be equal to 10π.
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If we multiply on the outside, three π times three, thatβs nine π.
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And our other two terms multiplied together to equal one are one π plus nine π equals 10π.
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Everything is positive.
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And we have three π plus one times one π plus three.
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We set both of these equations equal to zero.
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On the right, we subtract three from both sides.
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And we see that when π equals negative three, the equation equals zero.
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On the left side, we need to do two steps.
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First, we subtract one from both sides.
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And then we divide both sides by three.
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So π is also equal to negative one-third.
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But what do we do with this information?
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Well, we know that π times π equals six.
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And π will either be negative three or negative one-third.
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Letβs consider the first case first, when π is negative three.
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To find out what π would be equal to if π was negative three, we divide both sides of the equation by negative three.
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Six divided by negative three equals negative two.
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So we want to say when π is negative three, π is negative two.
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ππ is six.
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Thatβs our first two terms.
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And our third term would be equal to Negative two times negative three squared.
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Negative three squared is nine, times negative two would be negative 18.
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And now weβll consider when π is negative one-third.
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To find out what π is equal to when π is negative one-third, we multiply both sides of the equation by negative three, which tells us that π is equal to six times negative three.
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π equals negative 18.
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Now weβre looking at the case when π is negative one-third.
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We already know that the middle term ππ must be equal to six.
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But when π equals negative one-third, the first term is negative 18.
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And the third term is π times π squared, which would be negative 18 times negative one-third squared.
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Negative one-third squared is one over nine.
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Negative 18 times one over nine is negative two.
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This turns out to be really interesting.
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We have either negative two, six, negative 18 or negative 18, six, negative two.
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Before we move on, itβs worth checking to make sure that weβve calculated everything correctly and that the two statements we started with are true with these values.
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Negative two plus six plus negative 18 does equal negative 14.
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And because we know that addition is commutative, negative 18 plus six plus negative two is also equal to negative 14.
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In the same way, negative two times six times negative 18 equals 216.
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And it doesnβt matter if we change the order.
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Multiplying these three values still yields 216.
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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.