Okay cool! Thank you I see now!
Another question haha,
Which term of the series 8, -4 , + 2.... is 1/128?
I'm just particularly struggling with dealing with a negative r value and using log to solve for n
thanks!
Mod Edit [Aaron]: Merged double post. Please edit your previous post if you want to add something.
Hey! So we know that, for a geometric series,

Where a is the first term, r is the geometric difference between two terms, and n is the term number. Thus,
(-\frac{1}{2})^{n-1})
is the definition of each term for this series.
Now, we want to find

So,
(-\frac{1}{2})^{n-1})
^{n-1})
=n-1)
The last line literally means -1/2 to the power of WHAT (ie. n-1) is equal to 1/1024 (at least, that's how I think of logs). Let's keep going
Actually, this is more difficult than it needs to be. Let's take a step back
^{n-1})
^{10}=(-\frac{1}{2})^{n-1})
The negative here doesn't matter; as long as n-1 is even, the term becomes positive.


Let's just check this in the original equation
(-\frac{1}{2})^{11-1}=\frac{1}{128})
As expected