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I know a geometric expression can be written down as:

$$ \sum_{k=1}^{n} ar^{k-1} = \frac{a(1-r^n)}{1-r} $$

Is it possible to solve this for r, knowing the total sum? (how about if you know the last value in the sequence instead of the first? - to me intuitively this would just result in r becoming 1/r)

Wouter
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  • For practical purposes in my case, I am allowed to assume that r is rational and >0. But I suspect @Robert Israel is correct and it would not change the answer to this question. – Wouter Jul 13 '23 at 16:33
  • For others interested, I found a small discussion on this topic from more than 2 decades ago: https://www.math.toronto.edu/mathnet/questionCorner/sumtoratio.html – Wouter Jul 13 '23 at 17:45
  • And for people looking for the pragmatic approach: https://math.stackexchange.com/questions/3473457/how-to-find-r-in-a-finite-geometric-series – Wouter Jul 13 '23 at 19:13
  • So do you know $n$ too? –  Jul 14 '23 at 15:02
  • Just to be clear; are $a$ and $n$ given? – Servaes Jul 14 '23 at 15:13
  • They would be given, but I was interested in the (im)possibility of a general formula/approach where a and n are given. I realize if I actually substitute for a and n , some simplification options could emerge depending on the specific value of a and n. – Wouter Jul 17 '23 at 14:37

2 Answers2

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$\sum_{k=1}^n a r^{k-1} - y$ is a polynomial in $r$ of degree $n-1$, typically irreducible over the rationals. In general it may not be solvable in radicals. For example, try $n=6$, $a=1$, $y=3$: $r^5 + r^4 + r^3 + r^2+r - 2$ has Galois group $S_5$ (according to Maple) and is not solvable in radicals. Of course numerical solution methods can be used: the real root is approximately $0.7090111952$.

EDIT: If you are looking for rational roots, the Rational Root Theorem lets you reduce the problem to finitely many possibilities. It changes things completely. Why didn't you mention that in the first place?

Robert Israel
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It is not possible since the summation is not injective in general, for example

$$\sum_{k=1}^{3} r^{k-1}=7 \implies r=2\;\lor\;r=-3$$

user
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