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The first three terms of an infinite geometric sequence are $m - 1$, $6$, $m + 4$, where $m\in\Bbb{Z}$

Write down an expression for the common ratio, $r$.

Do I divide the second term by the first term to get my answer?

Edit: I just realized they’re asking for the expression for r. What does that mean and how do I figure that out?

Jesko Hüttenhain
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Ella
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    Hint : You have to solve $$\frac{m+4}{6}=\frac{6}{m-1}$$ – Peter Aug 07 '18 at 18:49
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    If you do, then you will still have the answer in terms of $m$. You want the answer as an explicit number, not having to do with $m$. – JMoravitz Aug 07 '18 at 18:49
  • @Peter I think I remember seeing this before. Could you tell me why this happens? Is this just what you’re always supposed to do when you encounter an infinite geometric sequence? – Ella Aug 07 '18 at 18:56
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    @Ella Simply because the quotient of consecutive numbers in the sequeunce must be always the same. – Peter Aug 07 '18 at 18:57
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    A geometric sequence with first term $a$ and common ratio $r$ is a sequence of the form $a,ar,ar^2,ar^3,ar^4,ar^5,\dots,ar^n,\dots$. Taking the ratio of two consecutive terms (assuming that $a\neq 0$) yields $\frac{ar^k}{ar^{k-1}}=r$ – JMoravitz Aug 07 '18 at 18:57
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    To elaborate on the earlier hint, only certain values (or only a certain value) of $m$ will be such that $m-1,6,m+4$ are in geometric sequence. That happens precisely when there is some nonzero value $r$ such that simultaneously $m+4=6r=(m-1)r^2$. Through algebraic manipulation, you have $r=\frac{m+4}{6}=\frac{6}{m-1}$. Ignoring $r$ for the moment, we can now manipulate this in order to find what the value of $m$ must be. Then, knowing the value of $m$ we can find the value of $r$. – JMoravitz Aug 07 '18 at 19:03
  • @JMoravitz So, using that fraction, could that be classified as the expression for the common ratio? The second part to the whole question asks me to show how m satisfies the equation $m^2 + 3m - 40 = 0$. And this equation would be after it has been cross multiplied to solve, right? – Ella Aug 07 '18 at 19:14
  • Yes... but it is only marginally better than just giving the answer of $r$ as being simply "$r$." You can, as shown below, say $r=-\frac{2}{3}$ or $r=\frac{3}{2}$ (depending on the value of $m$), which is much more specific than leaving it unsimplified and unspecified. The equation $m^2+3m-40=0$ is precisely the equation that pops out when you cross multiply the equation I mentioned near the end of my previous comment, and it is from this that you can find the possible values of $m$ (as shown explicitly below in the answers). – JMoravitz Aug 07 '18 at 19:21

2 Answers2

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Edit: I just realized you also asked "what does this mean". Frankly, I had to look up the terminology myself:

https://en.wikipedia.org/wiki/Geometric_progression

This informs you that $r(m-1)=6$ and $6r=m+4$. From the second one you get $$ r=\frac{m}6 + \frac23. $$ Now putting that into the first gives you \begin{align*} && 6 &= \left(\frac{m}6+\frac23\right)\cdot(m-1) = \frac{m^2}6-\frac{m}6+\frac{2m}3-\frac23 \\ &\Rightarrow& 36 &=m^2+3m-4 \\ &\Rightarrow& 0 &= m^2+3m-40 \end{align*} That is a quadratic equation in $m$, and the solutions are $m=5$ and $m=-8$.

  • For $m=5$ you get $r=\frac32$ and the sequence $4,6,9,\ldots$
  • For $m=-8$ you get $r=-\frac23$ and the sequence $-9,6,-4,\ldots$.
Jesko Hüttenhain
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  • To find r using the values of m, did you just plug them in or was there some kind of formula? – Ella Aug 07 '18 at 19:36
  • @Ella yes, you just plug in the value of $m$ you found into any of the expressions you found for $r$ (for example $r=\frac{m+4}{6}$) – JMoravitz Aug 07 '18 at 20:49
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The simple answer to your question is "yes"

You have $$r=\frac 6{m-1}$$

But you can also wrote a second expression for $r$ as the ratio of two consecutive terms ie $$r=\frac {m+4}6$$ which would also be correct. Equating these two different expressions for the same thing, will help you to find possible values for $m$ and therefore $r$

For completeness it is also possible to take the ratio of non-consecutive elements to give $$r=\pm \sqrt {\frac {m+4}{m-1}}$$ and this would also be an expression for $r$ in the terms of the question you have been asked.

But it looks to me that this question is the first stage of a longer one, and you will need the simpler expressions without square roots.

Mark Bennet
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