\[b_{6} = 4;\ \ \ b_{4} = 9\]
\[b_{6} = b_{1} \cdot q^{2};\ \ \ \ b_{4} = b_{1} \cdot q^{3} \Longrightarrow\]
\[\Longrightarrow b_{6}\ :b_{4} = b_{1}q^{5}\ :b_{1}q^{3} = q^{2}\]
\[b_{6}\ :b_{4} = 4\ :9 = \frac{4}{9} \Longrightarrow q^{2} = \frac{4}{9} \Longrightarrow\]
\[\Longrightarrow q_{1} = \frac{2}{3}\ и\ q_{2} = - \frac{2}{3}\]
\[q_{2} < 0 - не\ подходит \Longrightarrow q = \frac{2}{3}\]
\[b_{7} = b_{6} \cdot q = 4 \cdot \frac{2}{3} = \frac{8}{3}\]
\[Ответ:\ \ b_{7} = \frac{8}{3}.\]