Question

16.14. В геометрической прогрессии (bn) известно, что:
2) S3=93, b1 b2 b3=3375. Найдите S4.
Помогите пожалуйста ​

Answer 1

Объяснение:

$\displaystyle\\\left \{ {S_3=93} \atop {b_1*b_2*b_3=3375}} \right. \ \ \ \ \ \ \ \left \{ {{b_1*\frac{q^3-1}{q-1}=93 } \atop {b_1*b_1q*b_1q^2=3375}} \right.\ \ \ \ \ \ \left \{ {{b_1*\frac{q^3-1}{q-1}=93 } \atop {b_1^3*q^3=15^3}} \right. \\\\\\\left \{ {{b_1*\frac{(q-1)*(q^2+q+1)}{q-1}=93 } \atop {(b_1*q)^3=15^3}} \right.\ \ \ \ \ \ \left \{ {{b_1*\frac{q^2+q+1}{1}=93 } \atop {b_1*q=15}} \right. \ \ \ \ \ \ \left \{ {{\frac{15*(q^2+q+1)}{q}=93 } \atop {b_1=\frac{15}{q} }} \right.$

$\displaystyle\\\left \{ {{15q^2+15q+15=93q} \atop {b_1=\frac{15}{q} } \right.\ \ \ \ \ \ \left \{ {{15q^2-78q+15=0\ |:3} \atop {b_1=\frac{15}{q}}} \right.\ \ \ \ \ \ \left \{ {{5q^2-26q+5=0} \atop {b_1=\frac{15}{q}}} \right. \\\\\\\left \{ {{5q^2-25-q+15=0} \atop {b_1=\frac{15}{q}}} \right. \ \ \ \ \ \ \left \{ {{5q*(q-5)-(q-5)=0} \atop {b_1=\frac{15}{q}}} \right.$

$\displaystyle\\\\\left \{ {{(q-5)*(5q-1)=0} \atop {b_1=\frac{15}{q}}} \right.\ \ \ \ \ \ \left \{ {{q_1=5\ \ \ \ q_2=\frac{1}{5} } \atop {b_1=3\ \ \ \ b_2=75}} \right. .\ \ \ \ \ \ \Rightarrow\\\\b_4'=b_1*q_1^3=3*5^3=3*125=375\\\\S_4'=S_3+b_4'=93+375=468.\\\\b_4''=b_2*q_2^3=75*(\frac{1}{5})^3 =\frac{75}{125} =\frac{3}{5}=0,6.\\\\S_4''=S_3+b_4''=93+0,6=93,6.$