How to find the $n$-th term if the common ratio is negative

Question

Given the following geometric series:

$152-76+38-19+...\frac{-19}{64}$

In order to find the sum of this series, I know I need to find the nth term for $\frac{-19}{64}$. So, to do that, I have this:

$a_n=-304(\frac{-1}{2})^n$

$\frac{-19}{64}=-304(\frac{-1}{2})^n$

$\frac{1}{1024}=(\frac{-1}{2})^n$

Then you take log of both sides, but you can't do log of a negative number, so what is another approach to solve this problem besides just plugging in random $n$ to see which one gives you the solution? I know $ n=10$, but want to know if there is anyway to solve this problem using the log. Thank you.

Answer 1

When you get to $\frac 1{1024}=\left(-\frac 12\right)^n$ you know that $n$ must be even because only an even power of a negative number is positive. An even power does not depend on whether the base is negative or positive, so you can write $$\frac 1{1024}=\left(\frac 12\right)^n$$ and take the base $2$ log of both sides to get $$-10=-n$$

Answer 2

HINT: Since the $n^{\text{th}}$ power of a negative number is positive, $n$ is even. Then, $\left(-1\right)^n$ and $$\left(-\frac{1}{2}\right)^n = \left(-1\right)^n\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{n}.$$ Can you complete it now?

Answer 3

Hint:

Simply factor out the first term $\;152=19\cdot 2^3$: $$152-76+38-19+\dots-\frac{19}{64}=152\biggl(1-\frac12+\frac 1{2^2}-\frac1{2^3}+\dots -\frac1{2^9}\biggr),$$ so you have in the second factor the sum of the terms in a geometric sequence, up to degree $9$.