$a)$  

$\frac{a_2}{a_1}=\frac{64}{128}=\frac{1}{2}=q$  

$a_n=a_1\cdot q^{n-1}=128\cdot {\left(\frac{1}{2}\right)}^{n-1}=2^7\cdot {\left(2\right)}^{-n+1}=2^{-n+8}$   

$a_7=2^{-7+8}=2$  

 

$b)$  

$\frac{\frac{1}{4}}{\frac{1}{16}}=4=q$  

$a_n=a_1\cdot q^{n-1}=\frac{1}{16}\cdot 4^{n-1}=4^{-2}\cdot 4^{n-1}=4^{n-3}$  

$a_7=4^4=256$  

 

$c)$  

$\frac{a_2}{a_1}=\frac{-\frac{1}{81}}{\frac{1}{243}}=-3$  

$a_n=a_1\cdot q^{n-1}=\frac{1}{243}\cdot {\left(-3\right)}^{n-1}=\frac{1}{3^5}\cdot {\left(-3\right)}^{n-1}=\frac{{\left(-3\right)}^{n-1}}{3^5}$  

$a_7=3$  

 

$d)$  

$q=\frac{a_2}{a_1}=\frac{\frac{125}{64}}{\frac{625}{256}}=\frac{125}{64}\cdot \frac{256}{625}=1\cdot \frac{4}{5}=\frac{4}{5}$  

$a_n=\frac{625}{256}\cdot {\left(\frac{4}{5}\right)}^{n-1}={\left(\frac{4}{5}\right)}^{-4}\cdot {\left(\frac{4}{5}\right)}^{n-1}={\left(\frac{4}{5}\right)}^{n-5}$  

$a_7={\left(\frac{4}{5}\right)}^2=\frac{16}{25}$  

$e)$  

$q=\frac{a_2}{a_1}=-\frac{2}{\sqrt{2}}$  

$a_n=a_1\cdot q^{n-1}=\sqrt{2}{\left(-\frac{2}{\sqrt{2}}\right)}^{n-1}$  

$a_7=\sqrt{2}\cdot \left(\frac{64}{8}\right)=8\sqrt{2}$  

 

$f)$  

$q=\frac{a_2}{a_1}=\frac{\frac{1}{3}}{\frac{2}{9}}=\frac{1}{3}\cdot \frac{9}{2}=\frac{3}{2}$  

$a_n=\frac{2}{9}\cdot {\left(\frac{3}{2}\right)}^{n-1}$  

$a_7=\frac{2}{9}\cdot \frac{729}{64}=\frac{1458}{9\cdot 64}=\frac{162}{64}=\frac{81}{32}$