Question

геометрическая прогрессия​

Answer 1

$\left \{ {{b_4-b_2=24} \atop {b_2+b_3=6}} \right.\; \; \left \{ {{b_1q^3-b_1q=24} \atop {b_1q+b_1q^2=6}} \right.\; \; \left \{ {{b_1q(q^2-1)=24} \atop {b_1q(1+q)=6\; \; }} \right.\; \; \left \{ {{b_1q(q-1)(q+1)=24} \atop {b_1q(1+q)=6\qquad }} \right. \\\\\frac{b_1q(q-1)(q+1)}{b_1q(1+q)}=\frac{24}{6}\; \; \to \; \; \; (q-1)=4\; ,\; \; q=5\\\\b_1=\frac{6}{q(1+q)}=\frac{6}{5\cdot 6}=\frac{1}{5}\\\\b_4=b_1q^3=\frac{1}{5}\cdot 5^3=25$