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?
$\endgroup$ 82 Answers
$\begingroup$Edit: I just realized you also asked "what does this mean". Frankly, I had to look up the terminology myself:
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 $-9,6,-4,\ldots$.
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.
$\endgroup$
