Question Video: Using the Properties of Geometric Sequences to Find an Unknown Term | Nagwa Question Video: Using the Properties of Geometric Sequences to Find an Unknown Term | Nagwa

Question Video: Using the Properties of Geometric Sequences to Find an Unknown Term Mathematics

Find the value of π‘š given the geometric sequence βˆ’4, π‘š, 2π‘š + 3, β‹―.

03:28

Video Transcript

Find the value of π‘š given the geometric sequence negative four, π‘š, two π‘š plus three, continuing.

We’re given the first three terms in a geometric sequence. And the second and third term are given in terms of π‘š. In a geometric sequence, each term is found by multiplying the previous term by a constant. And we call that constant a common ratio. And that means in this sequence, to get from negative four to π‘š, we need to multiply negative four by π‘Ÿ. And also π‘š times π‘Ÿ is going to equal two π‘š plus three.

In order to solve for π‘š, let’s set up a few equations. We could say that π‘š equals negative four times π‘Ÿ and that π‘š times π‘Ÿ equals two π‘š plus three. At this point, we do have two equations. But since we’re solving for π‘š, it would be helpful to be able to substitute an π‘Ÿ-value in terms of π‘š in the second equation. We can do that by rewriting the negative four π‘Ÿ equals π‘š, divide both sides of the equation by negative four, so that we’ve rearranged our first equation to say π‘Ÿ equals negative π‘š over four. We could call this equation three.

If we substitute equation three into equation two, we’ll have π‘š times negative π‘š over four is equal to two π‘š plus three. Multiplying the π‘šβ€™s together, we get negative π‘š squared over four equals two π‘š plus three. From there, we multiply both sides of the equation by four. Negative π‘š squared equals eight π‘š plus 12.

Since we have this π‘š squared, we’re dealing with a quadratic. And we wanna set this equation equal to zero to solve. To do that, we add π‘š squared to both sides of the equation. And we have zero equals π‘š squared plus eight π‘š plus 12.

We can solve this by factoring. We need two factors of 12 that when added together equal eight. That will be two and six. So zero equals π‘š plus two times π‘š plus six. If we set both factors equal to zero, we find that π‘š equals six or π‘š equals negative two. What does that mean in the context of this geometric sequence?

If π‘š equals negative six, then the sequence is negative four, negative six, negative nine, continuing. And if π‘š equals negative two, the sequence is negative four, negative two, negative one. Both of these are true geometric sequences. To get from negative four to negative two, you multiply by one-half. To get from negative two to negative one, you multiply by one-half. On the other hand, to get from negative four to negative six, you multiply by three-halves. And to get from negative six to negative nine, you multiply by three-halves.

Since we aren’t given any other information about this sequence, we have to say that there are two valid solutions for π‘š. Either π‘š equals negative six or π‘š equals negative two.

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