Відповідь: b₁=-10,  q=-2.

Пояснення:

[tex]\displaystyle\\\frac{b_{N+10}}{b_{N+5}}=-32 \ \ \ \ \ \ b_4+b_2=10N\ \ \ \ \ \ N=10\ \ \ \ \ \ b_1=?\ \ \ \ q=?\\\\\\\left \{ {{\frac{b_{10+10}}{b_{10+5}}=-32} } \atop {b_4+b_2=10*10}} \right.\ \ \ \ \left \{ {{\frac{b_{20}}{b_{15}}=-32 } \atop {b_1q^3+b_1q=100}} \right. \ \ \ \ \left \{ {\frac{b_1q^{19}}{b_1q^{14}}=-32 } \atop {b_1*(q^3+q)=100}} \right. \\\\\\[/tex]

[tex]\displaystyle\\\left \{ {{q^5=-32} \atop {b_1=\frac{100}{q^3+q} }} \right. \ \ \ \ \left \{ {{q^5=(-2)^5} \atop {b_1=\frac{100}{q^3+q} }} \right. \ \ \ \ \left \{ {{q=-2} \atop {b_1=\frac{100}{(-2)^3+(-2)} }} \right. \ \ \ \ \left \{ {{q=-2} \atop {b_1=\frac{100}{-8-2} }} \right. \\\\\\\left \{ {{q=-2} \atop {b_1=\frac{100}{-10} }} \right. \ \ \ \ \left \{ {{q=-2} \atop {b_1=-10}} .\right.[/tex]