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What I'm attempting to do is to rearrange the formula for the sum of a geometric series so as to find the value of its common ratio $r$. I've tried several different methods, all of which have failed; though I can't understand why.

This was my last attempt using logarithms:

$$ S_n=\frac{a(1-r^n)}{1-r}$$ $$S_n(1-r) = a(1-r^n)$$ $$S_n-S_nr = a - ar^n$$ $$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$ $$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$ $$0 = (\log{r})(n-1)$$ $$r = 10^{\frac{0}{n-1}}$$ $$r=1$$

I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ and $a$, then that would be equally fantastic.

Thanks.

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  • $\begingroup$ There is a mistake in the application of log to $(\dfrac{S_n}{a}-1).$ $\endgroup$ – Américo Tavares Oct 31 '10 at 23:04
  • $\begingroup$ Do you want the sum of the (infinite) geometric series, or do you want the $n$th partial sum of the series? Your question implies the former, but the formula you give for $S_n$ implies the latter. $\endgroup$ – Mike Spivey Nov 1 '10 at 2:17
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Added: I assume you don't know both $S_n$ and $S_{n+1}$. Also I thought you had a geometric progression, i.e. not an infinite series. If you mean a series then you have to apply limits to $S_n$.


From

$$\frac{S_{n}}{a}-1=\frac{S_{n}}{a}r-r^{n}$$

you cannot deduce

$$\log \frac{S_{n}}{a}=\log \frac{S_{n}}{a}+\log r-n\log r$$

because

$$\log (x+y)\neq \log x+\log y.$$

The correct property of the $\log $ function is

$$\log (xy)=\log x+\log y.$$

The algebraic identity

$$x^{n}-c^{n}=(x-c)(x^{n-1}+cx^{n-2}+c^{2}x^{n-3}+\cdots +c^{n-2}x+c^{n-1})$$

gives for $c=1$

$$\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x+1.$$

Therefore you have

$$\frac{S_{n}}{a}=\frac{1-r^{n}}{1-r}=r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1.$$

You would have to find $r$ such that

$$r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1-\frac{S_{n}}{a}=0.$$

However this polynomial equation has no algebraic solution, in general, for $n\ge 6$. You need a different method (a numerical one).

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    $\begingroup$ I find it amusing that after all that manipulation, in the end you arrive at what is essentially the most elementary definition of the geometric progression: $a(r^{n-1} + r^{n-2} + \cdots + r + 1) = S_n$. $\endgroup$ – Rahul Nov 1 '10 at 18:27
  • $\begingroup$ The reason is that I have let's say deduced it backwards. So do I, now that you have pointed it out! $\endgroup$ – Américo Tavares Nov 1 '10 at 21:11
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HINT $\ \ $ Consider $\ S_{n+1} - r\ S_n $

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No need to solve the full polynomial as

$$S_{n+1} - rS_n = a$$ (by def. of geometric series)

(I would not leave this as an own entry if I had enough points to comment everywhere, sorry about that)

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