Which term of the series $256 + 128 + 64 + \cdots$ is $\dfrac{1}{2}$?

This is a GP with $a=256$ and ratio $r=\tfrac12$. Setting $256\cdot(\tfrac12)^{n-1}=\tfrac12$ gives $2^{8-(n-1)}=2^{-1}$, so $9-n=-1$ and $n=10$.