Q.Find the sum of the first 6 terms of the Geometric Progression $64, 32, 16, 8, \ldots$

Here $r=\frac12<1$, so use $S_n=\frac{a(1-r^n)}{1-r}$.

$S_6=\dfrac{64(1-(1/2)^6)}{1-1/2}=64\left(\dfrac{63}{64}\right)\times2=63\times2=126$.

Here $a=64$, $r=\dfrac{32}{64}=\dfrac12$ (checked: $\dfrac{16}{32}=\dfrac12$, $\dfrac{8}{16}=\dfrac12$ — constant). Since $0<r<1$, use $S_n=\dfrac{a(1-r^n)}{1-r}$.

$S_6=\dfrac{64\left(1-\left(\tfrac12\right)^6\right)}{1-\tfrac12}=\dfrac{64\left(1-\tfrac{1}{64}\right)}{\tfrac12}=\dfrac{64\times\tfrac{63}{64}}{\tfrac12}=\dfrac{63}{\tfrac12}=63\times2=126$.

Check by direct addition: the first 6 terms are $64,32,16,8,4,2$. Adding: $64+32=96$, $+16=112$, $+8=120$, $+4=124$, $+2=126$ — matches exactly.

Using the $r>1$ form $\frac{a(r^n-1)}{r-1}$ unchanged when $r=\frac12$ gives a negative-over-negative expression that is easy to mis-simplify — while mathematically equivalent, the $\frac{a(1-r^n)}{1-r}$ form is far less error-prone when $r<1$, and should be the default choice in that case.