# s = (a(r^n -1))/(r-1) Making 'r' the subject formula..?

Feb 2, 2018

This is not generally possible...

#### Explanation:

Given:

$s = \frac{a \left({r}^{n} - 1\right)}{r - 1}$

Ideally we want to derive a formula like:

$r = \text{some expression in } s , n , a$

This is not going to be possible for all values of $n$. For example, when $n = 1$ we have:

$s = \frac{a \left({r}^{\textcolor{b l u e}{1}} - 1\right)}{r - 1} = a$

Then $r$ can take any value apart from $1$.

Also, note that if $a = 0$ then $s = 0$ and again $r$ can take any value apart from $1$.

Let us see how far we can get in general:

First multiply both sides of the given equation by $\left(r - 1\right)$ to get:

$s \left(r - 1\right) = a \left({r}^{n} - 1\right)$

Multiplying out both sides, this becomes:

$s r - s = a {r}^{n} - a$

Then subtracting the left hand side from both sides, we get:

$0 = a {r}^{n} - s r + \left(s - a\right)$

Assuming $a \ne 0$, we can divide this through by $a$ to get the monic polynomial equation:

${r}^{n} - \frac{s}{a} r + \left(\frac{s}{a} - 1\right) = 0$

Note that for any values of $a , s$ and $n$ one root of this polynomial is $r = 1$, but that is an excluded value.

Let us attempt to factor out $\left(r - 1\right)$...

$0 = {r}^{n} - \frac{s}{a} r + \left(\frac{s}{a} - 1\right)$

$\textcolor{w h i t e}{0} = {r}^{n} - 1 - \frac{s}{a} \left(r - 1\right)$

$\textcolor{w h i t e}{0} = \left(r - 1\right) \left({r}^{n - 1} + {r}^{n - 2} + \ldots + 1 - \frac{s}{a}\right)$

So dividing by $\left(r - 1\right)$ we get:

${r}^{n - 1} + {r}^{n - 2} + \ldots + 1 - \frac{s}{a} = 0$

The solutions of this will take very different forms for different values of $n$. By the time $n \ge 6$, it is not generally solvable by radicals.