I know in this particular indeterminate partial sum S = $3^n - 3^{n+1} + 3^{n+2} - 3^{n+3} + \cdots + 3^{3n}$ where $a=3^n$ and $r=-3$. I know if $3^1$ were the first term, there would be $3n$ terms but its not the first term $3^n$ is. So I am missing $3^1,3^2,3^3, \cdots , 3^{n-1}$ = $\,$ $n-1$ terms that are missing to the left of $3^n$. $\,$ Therefore, there are $3n-(n-1) = 2n+1$ terms.
How many terms does S = $3^k + 3^{k-1} + 3^{k-2} + \cdots + 3^{-2k}$ have? In the above I knew that $3^{n-1}$ terms are to the left of $3^n$ because we're adding 1 to the exponent each time. However, here we are subtracting 1 from the exponent each time so what's to the left of $3^k$? How many terms are there missing to the left of $3^k$ I must see the leftward expansion of the missing terms and so therefore there are how many terms are there? Need to follow the same process i did for the example involving n.
You are just counting down from $k$ to $-2k$ by $1$'s. There are $k-(-2k)+1=3k+1$ terms. The $+1$ comes because you count both ends of the interval, like when counting from $5$ to $10$ there are $10-5+1=6$ terms. Try it!