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Find the three consecutive numbers of a geometric sequence, given their sum is seven and the product of their squares is 64.
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First, we need to know what a geometric sequence is.
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Itβs a sequence where each term after the first term is found by multiplying the previous term by a common ratio.
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If we let the first term be π and the common ratio π, then we can write three consecutive terms, such that the first term is π.
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The second term would be the first term multiplied by the common ratio, π times π.
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The third term is the second term multiplied by the common ratio.
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That would be ππ times π.
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And we can simplify that to say ππ squared.
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We can take these three values and use them to write equations for the sum and given product.
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Since we know that the sum of these three values is seven, we can say that π plus ππ plus ππ squared equals seven.
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Each of these terms have a factor of π.
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And if we pull out that factor, we can rewrite this expression to say π times one plus π plus π squared equals seven.
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And because we have a quadratic inside the parentheses, it might be helpful to rearrange it so that the π squared term comes first and the constant one comes at the end.
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This doesnβt change the value of whatβs in the parentheses.
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And itβs a more common format.
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But for now, thatβs all we can do with the equation for sum.
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The product is not just the product of these three values.
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It is the product of their squares and thatβs 64.
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This will mean π squared times ππ squared times ππ squared squared equals 64.
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We need to distribute these squared values, which gives us π squared times π squared π squared times π squared π to the fourth all equal to 64.
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When exponents that have the same base are being multiplied together, we add the exponent values.
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That means here weβll say two plus two plus two.
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And weβll get π to the sixth power.
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We can multiply π squared by π to the fourth by adding two plus four.
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And weβll get π to the sixth power.
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If we know that π to the sixth power times π to the sixth power equal 64, we can say ππ to the sixth power equals 64.
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And then, we can take the sixth root of both sides of the equation.
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We can also note that taking the sixth root is the same thing as taking ππ to the sixth power to the one-sixth power.
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For some calculators, youβll need to enter 64 and then in the exponent value one-sixth.
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The sixth root of ππ to the sixth power is ππ.
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And the sixth root of 64 is two.
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If ππ equals two, then we found our second term.
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What we can do now is take the equation ππ equals two and solve for the variable π or the variable π.
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And once we have that, weβll plug those values in to our first equation.
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But how do we decide if we want to solve for π or for π?
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If we look at this first equation, the variable π is on its own.
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The variable π is squared and then added again.
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Our calculations are bound to be a little bit simpler if we substitute a value in for π instead of substituting a value in for π.
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That doesnβt mean that you couldnβt solve for π and plug it in and still find the answer.
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It just means that the algebra might be a little bit more difficult.
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Because we know that π times π equals two, we want to find an equation that says π equals something.
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If we divide both sides of this equation by π, then weβll see that π equals two over π.
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And so, we take our first equation π times π squared plus π plus one equal seven.
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And we plug in two over π for π.
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Because we havenβt π in the denominator on the left side of the equation, we can get rid of that by multiplying both sides of the equation by π over one.
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On the left, π in the numerator and π in the denominator cancel out.
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And we have two times π squared plus π plus one.
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And on the right, seven times π over one equals seven π.
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Weβll distribute this multiplied by two across the three terms.
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And then, weβll have two π squared plus two π plus two equal seven π.
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To solve for π, weβre going to want to set this entire equation equal to zero.
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And we can do that by subtracting seven π from both sides.
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Where weβll get two π squared minus five π plus two equals zero.
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And then, weβre going to want to factor to find the values of π.
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Two is a prime number.
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So we know that one of the parentheses will be two π and the other one will be π.
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We also have a constant value of two, which means weβll be dealing with either two times one or negative two times negative one.
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Since our middle term is negative five, we know weβll be dealing with negative values.
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Two π times negative two equals negative four π and negative one times π equals negative one π.
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Negative one plus negative four equals negative five.
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And so our factors of this quadratic are two π minus one and π minus two.
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We set them both equal to zero.
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To solve for π, we can add two to both sides of this equation.
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And we find that π equals two.
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So we have one case where π equals two.
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For our second equation, we add one to both sides: two π equals one.
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And then, we divide both sides by two to get π equals one-half.
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We found two cases.
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We found the case where π equals two and a second case where π equals one-half.
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To find the other values, weβll go back to our equation π times π equals two.
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We know that π times π equals two.
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We need to consider what π is if π is two and if π is one-half.
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If π is two, then π is going to be equal to one.
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And if π is one-half, then π is going to be equal to four.
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In the first case, we know that one times two is two and that two times two is four.
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In our second case, we have four times one-half equals two and two times one-half equals one.
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If we look back at our question, itβs only asking for three consecutive numbers in a geometric sequence.
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And either way, the three consecutive numbers will be one, two, and four or four, two, and one.
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Before we leave the problem completely, itβs worth checking to make sure that these three values do meet the requirements we were given.
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The first condition is that the sum of these three values is seven.
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One plus two plus four does equal seven, as does four plus two plus one.
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The second condition is that the product of their square is 64.
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Does one squared times two squared times four squared equals 64?
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That would be one times four times 16 which is 64.
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And because we can multiply in any order, four squared times, two squared times, one squared is also equal to 64.
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And that means we found the set of three consecutive numbers under these two conditions.