Произведение первых трёх членов прогрессии равен 1728,а сумма=63. найдите первый член и знаменатель.

Ответ:

hiohio

04.10.2020 02:16

$\left \{ {{b_1*b_2*b_3=1728} \atop {b_1+b_2+b_3=63}} \right.;
 \left \{ {{b_1*b_2*qb_2=1728} \atop {b_1+b_2+qb_2=63}} \right.;
 \left \{ {{q*b_1*b_2^2=1728} \atop {b_1+(1+q)b_2=63}} \right.;$

$\left \{ {{q*b_1*(qb_1)^2=1728} \atop {b_1+(1+q)*qb_1=63}} \right.;
 \left \{ {{q*b_1*q^2b_1^2=1728} \atop {b_1*[1+(1+q)*q]=63}} \right.;
 \left \{ {{q^3*b_1^3=1728} \atop {b_1*[1+(1+q)*q]=63}} \right.;$

$\left \{ {{(qb_1)^3-12^3=0} \atop {b_1*[1+q+q^2]=63}} \right.; \left \{ {{[qb_1-12]*[(qb_1)^2+12qb_1+12^2]=0} \atop {b_1*[1+q+q^2]=63}} \right.; \left \{ {{qb_1-12=0} \atop {b_1(1+q+q^2)=63}} \right.;$

$\left \{ {{b_1=\frac{12}{q}} \atop {\frac{12(1+q+q^2)}{q}=63}} \right.; \left \{ {{b_1=\frac{12}{q}} \atop {\frac{12q^2+12q+12-63q}{q}=0}} \right.; \left \{ {{b_1=\frac{12}{q}} \atop {12q^2-51q+12=0}} \right.;$

$\left \{ {{b_1=\frac{12}{q}} \atop {\frac{12(1+q+q^2)}{q}=63}} \right.; \left \{ {{b_1=\frac{12}{q}} \atop {\frac{12q^2+12q+12-63q}{q}=0}} \right.; \left \{ {{b_1=\frac{12}{q}} \atop {12q^2-51q+12=0}} \right.; \left \{ {{b_1=\frac{12}{q}} \atop {4q^2-17q+4=0}} \right.;$

$\left \{ {{b_1=\frac{12}{q}} \atop {4q^2-17q+4=0}} \right.;\left \{ {{b_1=\frac{12}{q}} \atop {4q^2-q-16q+4=0}} \right.;\left \{ {{b_1=\frac{12}{q}} \atop {q(4q-1)-4(4q-1)=0}} \right.;\left \{ {{b_1=\frac{12}{q}} \atop {(q-4)(4q-1)=0}} \right.;$

$\left \{ {{b_1=\frac{12}{q}} \atop {(q-4)(4q-1)=0}} \right.;\left \{ {{b_1=\frac{12}{q}} \atop {q=4,or,q= \frac{1}{4} }} \right.;\left \{ {{b_1= \frac{12}{4} } \atop {q=4}} \right.,or,\left \{ {{b_1=12*4} \atop {q= \frac{1}{4} }} \right.;$

$\left \{ {{b_1=3} \atop {q=4}} \right.,or,\left \{ {{b_1=48} \atop {q= \frac{1}{4} }} \right.$