Question

Произведение первых трех членов геометрической прогрессии равна 1728, а их сумма чисел равна 63. найдите b1 и q

Answer 1

Ответ:

Пошаговое объяснение:

$\left \{ {{b1*b2*b3=1728} \atop {b1+b2+b3=63}} \right. \\\left \{ {{b1*b1q*b1q^{2}=1728 b} \atop {b+b1q+b1q^{2}=63 }} \right. \\\left \{ {{b_{1}^3*q^{3}=1728 } \atop {b_{1}(1+q+q^{2}) =63}} \right. \\\left \{ {{b1*q=12} \atop {b_{1}(1+q+q^{2}) =63}} \right. \\\left \{ {{b1=\frac{12}{q} } \atop {\frac{12}{q} (1+q+q^{2}) =63}} \right. \\\left \{ {{b1=\frac{12}{q} } \atop {12(1+q+q^{2})=63q}} \right.$

$\left \{ {{b1=\frac{12}{q} } \atop {1+q+q^{2}=5.25q }} \right. \\\left \{ {{b1=\frac{12}{q}} \atop {q^2-4.25q+1=0}} \right. \\\\q^2-4.25q+1=0$

$D=(-4.5)^2-4*1*1=18.0625-4=14.0625\\\sqrt{D} =3.75\\q1=\frac{4.5+3.75}{2} =4\\q2=\frac{4.5-3.75}{2} =0.25\\b1=\frac{12}{q1} =12:4=3\\b2=12:q2=12:0.25=48$