Ответ: S₅=155.

Объяснение:

[tex]\displaystyle\\\left \{ {{b_4-b_1=35} \atop {b_1+b_2+b_3=35}} \right. \ \ \ \ \ \ \left \{ {{b_1q^3-b_1=35} \atop {b_1+b_1q+b_1q^2=35}} \right. \ \ \ \ \ \ \left \{ {{b_1*(q^3-1)=35} \atop {b_1*(1+q+q^2)=35}} \right. \\\\\\ \left \{ {{b_1*(q-1)*(q^2+q+1)=35\ \ \ \ (1)\atop {b_1*(q^2+q+1)=35\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)}} \right..[/tex]

Разделим уравнение (1) на уравнение (2):

[tex]\displaystyle\\\frac{b_1*(q-1)*(q^2+q+1)}{b_1*(q^2+q+1)}=\frac{35}{35} \\\\q-1=1\\\\q=2.\\\\b_1*(q^3-1)=35\\\\b_1*(2^3-1)=35\\\\b_1*(8-1)=35\\\\b_1*7=35\ |:7\\\\b_1=5.\ \ \ \ \ \ \Rightarrow\\\\S_n=b_1*\frac{q^n-1}{q-1} \\\\S_5=5*\frac{2^5-1}{2-1}=5*\frac{32-1}{1} =5*31=155.[/tex]