Question Video: Finding the Terms of a Geometric Sequence given Their Sum and the Product of Their Squares | Nagwa Question Video: Finding the Terms of a Geometric Sequence given Their Sum and the Product of Their Squares | Nagwa

# Question Video: Finding the Terms of a Geometric Sequence given Their Sum and the Product of Their Squares Mathematics • Second Year of Secondary School

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Find the three consecutive numbers of a geometric sequence, given their sum is 7 and the product of their squares is 64.

08:28

### Video Transcript

Find the three consecutive numbers of a geometric sequence, given their sum is seven and the product of their squares is 64.

First, we need to know what a geometric sequence is. Itβs a sequence where each term after the first term is found by multiplying the previous term by a common ratio. 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. The second term would be the first term multiplied by the common ratio, π times π. The third term is the second term multiplied by the common ratio. That would be ππ times π. And we can simplify that to say ππ squared.

We can take these three values and use them to write equations for the sum and given product. Since we know that the sum of these three values is seven, we can say that π plus ππ plus ππ squared equals seven. Each of these terms have a factor of π. And if we pull out that factor, we can rewrite this expression to say π times one plus π plus π squared equals seven. 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. This doesnβt change the value of whatβs in the parentheses. And itβs a more common format. But for now, thatβs all we can do with the equation for sum.

The product is not just the product of these three values. It is the product of their squares and thatβs 64. This will mean π squared times ππ squared times ππ squared squared equals 64. 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. When exponents that have the same base are being multiplied together, we add the exponent values. That means here weβll say two plus two plus two. And weβll get π to the sixth power. We can multiply π squared by π to the fourth by adding two plus four. And weβll get π to the sixth power.

If we know that π to the sixth power times π to the sixth power equal 64, we can say ππ to the sixth power equals 64. And then, we can take the sixth root of both sides of the equation. We can also note that taking the sixth root is the same thing as taking ππ to the sixth power to the one-sixth power. For some calculators, youβll need to enter 64 and then in the exponent value one-sixth. The sixth root of ππ to the sixth power is ππ. And the sixth root of 64 is two. If ππ equals two, then we found our second term.

What we can do now is take the equation ππ equals two and solve for the variable π or the variable π. And once we have that, weβll plug those values in to our first equation. But how do we decide if we want to solve for π or for π? If we look at this first equation, the variable π is on its own. The variable π is squared and then added again. 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. That doesnβt mean that you couldnβt solve for π and plug it in and still find the answer. It just means that the algebra might be a little bit more difficult.

Because we know that π times π equals two, we want to find an equation that says π equals something. If we divide both sides of this equation by π, then weβll see that π equals two over π. And so, we take our first equation π times π squared plus π plus one equal seven. And we plug in two over π for π. Because we havenβt 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. On the left, π in the numerator and π in the denominator cancel out. And we have two times π squared plus π plus one. And on the right, seven times π over one equals seven π.

Weβll distribute this multiplied by two across the three terms. And then, weβll have two π squared plus two π plus two equal seven π. To solve for π, weβre going to want to set this entire equation equal to zero. And we can do that by subtracting seven π from both sides. Where weβll get two π squared minus five π plus two equals zero. And then, weβre going to want to factor to find the values of π. Two is a prime number. So we know that one of the parentheses will be two π and the other one will be π. We also have a constant value of two, which means weβll be dealing with either two times one or negative two times negative one.

Since our middle term is negative five, we know weβll be dealing with negative values. Two π times negative two equals negative four π and negative one times π equals negative one π. Negative one plus negative four equals negative five. And so our factors of this quadratic are two π minus one and π minus two. We set them both equal to zero. To solve for π, we can add two to both sides of this equation. And we find that π equals two. So we have one case where π equals two. For our second equation, we add one to both sides: two π equals one. And then, we divide both sides by two to get π equals one-half.

We found two cases. We found the case where π equals two and a second case where π equals one-half. To find the other values, weβll go back to our equation π times π equals two. We know that π times π equals two. We need to consider what π is if π is two and if π is one-half. If π is two, then π is going to be equal to one. And if π is one-half, then π is going to be equal to four. In the first case, we know that one times two is two and that two times two is four. In our second case, we have four times one-half equals two and two times one-half equals one.

If we look back at our question, itβs only asking for three consecutive numbers in a geometric sequence. And either way, the three consecutive numbers will be one, two, and four or four, two, and one. Before we leave the problem completely, itβs worth checking to make sure that these three values do meet the requirements we were given.

The first condition is that the sum of these three values is seven. One plus two plus four does equal seven, as does four plus two plus one. The second condition is that the product of their square is 64. Does one squared times two squared times four squared equals 64? That would be one times four times 16 which is 64. And because we can multiply in any order, four squared times, two squared times, one squared is also equal to 64. And that means we found the set of three consecutive numbers under these two conditions.

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