$\frac{a(1-{{q}^{8}})}{1-q}=\frac{5}{4}\times \frac{a(1-{{q}^{4}})}{1-q}\Rightarrow 1+{{q}^{4}}=\frac{5}{4}\Rightarrow {{q}^{2}}=\frac{1}{2}$ پس $a{{q}^{6}}=a{{(\frac{1}{2})}^{3}}=\frac{1}{8}a$