The sequence is a geometric progression. Therefore,
[tex]\begin{gathered} a_n=ar^{n-1} \\ r=\frac{10}{40}=\frac{1}{4} \\ r=\frac{5}{8}\times\frac{2}{5}=\frac{10}{40}=\frac{1}{4} \\ a_8=40\times(\frac{1}{4})^{8-1} \\ a_8=40\times(\frac{1}{4})^7 \\ a_8=40\times\frac{1}{16384}_{}^{} \\ a_8=\frac{40}{16384}=\frac{20}{8192}=\frac{5}{2048} \\ a_8=\frac{5}{2048} \end{gathered}[/tex]