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

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Answer 1 · 2018-02-02T22:50:16

This is not generally possible...

Given:

#s = (a(r^n-1))/(r-1)#

Ideally we want to derive a formula like:

#r = "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 = (a(r^color(blue)(1)-1))/(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 #(r-1)# to get:

#s(r-1) = a(r^n-1)#

Multiplying out both sides, this becomes:

#sr-s=ar^n-a#

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

#0 = ar^n-sr+(s-a)#

Assuming #a!=0#, we can divide this through by #a# to get the monic polynomial equation:

#r^n-s/a r+(s/a-1) = 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 #(r-1)#...

#0 = r^n-s/a r+(s/a-1)#

#color(white)(0) = r^n-1-s/a(r-1)#

#color(white)(0) = (r-1)(r^(n-1)+r^(n-2)+...+1-s/a)#

So dividing by #(r-1)# we get:

#r^(n-1)+r^(n-2)+...+1-s/a = 0#

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