Verified answer

[tex]\displaystyle\bf\\b_{3} =25\\\\b_{5} =64\\\\\\:\left \{ {{b_{1}\cdot q^{4} =64} \atop {b_{1}\cdot q^{2}=25 }} \right. \\--------\\q^{2} =\frac{64}{25} \\\\\\q_{1,2} =\pm \ \sqrt{\frac{64}{25} } =\pm \ \frac{8}{5} \\\\\\b_{4} < 0 \ \ \ \Rightarrow \ \ \ q=-\frac{8}{5} \\\\\\b_{4} =b_{3} \cdot q=25\cdot\Big(-\frac{8}{5} \Big)=-5\cdot 8=-40[/tex]