An algebra problem by Mohammad Khaza

$128,64,32,...$

The first term in this progression is $a = 128$ and the common ratio is $r = \dfrac{64}{128} = \dfrac 12$ .

The $n^{\text{th}}$ term of a geometric progression is:

$ar^{n-1} =128 \times \left(\frac12 \right)^{n-1} = \frac{2^7}{2^{n-1}} = \frac{1}{2^{n-8}}$

Since the $n^{\text{th}}$ term of this geometric progression is $\dfrac 12$ we can say that,

$\begin{aligned} \dfrac 1{2^{n-8}} & = \dfrac 12 \\2^{-(n-8)} & = 2^{-1} \\-n+8 & = -1 \\-n & = -9 \\n &= \boxed{9}.\\ \end{aligned}$