\[b_{1} = - 6;\ \ b_{n + 1} = - 2 \cdot \frac{1}{b_{n}}\]
\[b_{2} = - 2 \cdot \frac{1}{b_{1}} = - 2 \cdot \frac{1}{- 6} = \frac{1}{3};\]
\[b_{3} = - 2 \cdot \frac{1}{b_{2}} = - 2 \cdot 3 = - 6;\]
\[b_{4} = - 2 \cdot \frac{1}{b_{3}} = - 2 \cdot \frac{1}{- 6} = \frac{1}{3};\]
\[b_{5} = - 2 \cdot \frac{1}{b_{4}} = - 2 \cdot 3 = - 6.\]
\[Ответ:b_{5} = - 6.\]