given :
[tex]\begin{gathered} a_1=1024\text{ and} \\ a_4=-16 \end{gathered}[/tex]In a GP,
[tex]a_n=ar^{n-1}[/tex]in this problem
[tex]\begin{gathered} n=1\Rightarrow a_1=ar^0 \\ n=4\Rightarrow a_4=ar^3 \end{gathered}[/tex]we divide both equations, we get
[tex]\begin{gathered} \frac{ar^3}{a}=-\frac{16}{1024} \\ \Rightarrow r^3=-\frac{1}{64} \\ \Rightarrow r=-\frac{1}{4} \end{gathered}[/tex]here a=1024 and r =-1/4 and let n=6
[tex]\begin{gathered} a_6=1024(-\frac{1}{4})^5 \\ a_6=1024(-\frac{1}{1024}) \\ a_6=-1 \end{gathered}[/tex]hence the solution is c) -1