First, we must remember that if a sequence is geometric, that of finding the sum of the first n terms, called Sn, without having to add all the terms.
To do this, we use the following equation:
[tex]S_n=\frac{a_1(1-r^n)}{1-r}[/tex]Where in this case we have:
[tex]\begin{gathered} a_1=125 \\ r=\frac{2}{5} \\ n=4 \end{gathered}[/tex]Now, we replace these values and solve the sum:
[tex]\begin{gathered} S_4=125\cdot\frac{(1-(\frac{2}{5})^4)}{(1-\frac{2}{5})} \\ S_4=125\cdot\frac{(1-\frac{16}{25})}{(1-\frac{2}{5})} \\ S_4=125\cdot\frac{\frac{609}{625}}{\frac{3}{5}} \\ S_4=125\cdot\frac{609\cdot5}{625\cdot3} \\ S_4=125\cdot\frac{203}{125} \\ S_4=203 \end{gathered}[/tex]In conclusion, the sum of the geometric series with a1= 125, r=2/5, n=4 is a total of 203.
