Given
[tex]a_6^=8192;a_n=4a_{n-1}[/tex]
Explanation
From the above,
[tex]\begin{gathered} a_6=4a_{6-1}\Rightarrow4a_5 \\ From\text{ the definition of a common ratio} \\ r=\frac{a_6}{a_5}=4 \\ Also; \\ a_6=ar^5 \\ 8192=a4^5 \\ a=\frac{8192}{4^5}=\frac{8192}{1024}=8 \\ \end{gathered}[/tex]
Therefore, we can have the first four terms as
[tex]\begin{gathered} a=8 \\ ar=8\times4=32 \\ ar^2=8\times4^2=128 \\ ar^3=8\times4^3=512 \end{gathered}[/tex]
Answer: Option C
