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Given:
Geometric progression Sum = $S$
First Term = $a$
number of terms = $N + 1$

$$a + aq + aq^2 + ... + aq^N = S$$

So, we need to solve this equation:

$$\sum_{i = 0}^N{q^i} = S/a$$

I need a formula to calculate common ratio $q$. I think this is hard to do, as we need to solve equation of $N$th degree, but this is kind of a special equation, so, maybe there could be a good solution.

Thanks in advance.

Heghine
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2 Answers2

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The $N$-th partial sum is $$ S_N = a \sum_{k=0}^N q^k = \left\{ \begin{align} a\frac{q^{N+1} - 1}{q - 1}, \quad q \ne 1 \\ a (N + 1), \quad q = 1 \end{align}\right. $$

So given a problem instance $(a, S, N)$ the first step is to treat the case $a = 0$ which implies $S=0$. In this case any number will work as value for $q$. For the following we assume $a \ne 0$.

If $a \ne 0$ and $S = 0$ there is no solution $q$. For the following we assume $S \ne 0$ as well.

Next thing would be to check if $$ S = a (N+1) $$ in this case $q=1$ is the solution. For the following we assume $q \ne 1$ too, having $a \ne 0$, $S \ne 0$, $q \ne 1$ as constraints.

If we still got the case $S = a$, then this would require $q = q^{N + 1}$, which would have $q = 0$ as solution and additionally $q = -1$ in case of odd $N$.

Looking for Fixed Points

One method would be to search for a fixed point $q^{*}$ for $$ f(q) = \frac{a}{S} q^{N+1} + 1 - \frac{a}{S} $$ which fullfills $f(q^{*}) = q^{*}$.

This version of the original problem is easier to reason with, because one can treat this as the geometric problem of the graph of $f(q) = u \, q^n + v$ (two cases for even and odd exponents $n$, two cases for positive or negative factor $u$) crossing the diagonal line $\mbox{id}(q) = q$.

For odd $N$ the exponent in $f$ is even, we got some parabolic graph which has zero, one or two crossings with $\mbox{id}$ and thus that many solutions.

For even $N$ the exponent is $f$ is odd and for $N \ge 2$ there might be even a third solution.

mvw
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  • The OP is looking for a formula. I don't think this will do. – user88595 May 21 '14 at 14:09
  • I think there is no general one (thanks to Galois). – mvw May 21 '14 at 14:10
  • Galois proves there is not general solution for polynomials of degree 5 and above but this doesn't mean some don't have formulae. Eg $x^n = b$ has a formula. – user88595 May 21 '14 at 14:12
  • Yes, this is kind of a special case, so, maybe there is a solution to solve this equation. – Heghine May 21 '14 at 14:15
  • I have rewritten it. For $N>4$ I would be surprised if that were a special case. But I am not into Galois theory enough to be able to proof this. Also I don't expect roots for every combination of parameters $(a, N, S)$. – mvw May 21 '14 at 14:19
  • I think it was Niels-Henrik Abel, not Galois... – poolpt May 21 '14 at 14:29
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    OK, according to Wolfram Alpha the problem instance $(a = 1, N = 5, S = 1/2)$ has no real solution $q$. – mvw May 21 '14 at 14:44
  • @mvw : It has solution about $-0.7$. Unless you mean $N+1 = 5$, the number of terms. All odd-degree polynomial will have at least one real solution. – user88595 May 21 '14 at 17:50
  • $N$ should be the highest order, one less than the number of additive terms. Need to check later if i made an error. – mvw May 21 '14 at 18:24
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If the number of terms is $N + 1$ the common ratio satisfies

$S=\frac{a(q^{N+1}-1)}{q-1}$

Your sum has $N+1$ terms.

pertbanking
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poolpt
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