Ответ:

[tex]b_n=4*(\frac{2}{5})^{n-2};\\\\b_1=4*(\frac{2}{5})^{1-2}=4*\frac{5}{2}=10;\\\\b_2=4*(\frac{2}{5})^{2-2};4*1=4;\\\\q=\frac{b_2}{b_1}=\frac{4}{10}=\frac{2}{5};|q|<1;\\\\S=\frac{b_1}{1-q}=\frac{10}{1-\frac{2}{5}}=\frac{50}{3}[/tex]