Question

b1+b3=10
b4+b6=-80
Найдите первый член геометрической прогрессии (b1-?)

Answer 1

$\left \{ {{b_1+b_3=10} \atop {b_4+b_6=-80}} \right. \\ \left \{ {{b_1+b_1 \cdot q^2=10} \atop {b_1 \cdot q^3 + b_1 \cdot q^5=-80}} \right. \\ \left \{ {{b_1(1+q^2)=10} \atop {b_1q^3(1+q^2)=-80}} \right. \\ \frac{b_1q^3(1+q^2)}{b_1(1+q^2)} = \frac{-80}{10} \\ q^3 = -8 \\ q = -2 \\ \\ b_1(1+q^2)=10 \\ b_1(1+4)=10 \\ 5b_1 = 10 \\ b_1 = 2$