Solution:
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 3/5 gives the next term. In other words,
[tex]a_n\text{ = }a_1r^{n-1}[/tex]where r = 3/5. Now, The sum of a series S_n is calculated using the formula:
[tex]S_n\text{ = }\frac{a(1-r^n)}{1-r}[/tex]For the sum of an infinite geometric series, as n approaches infinity we have that 1-r^n approaches 1. Thus
[tex]\text{ }\frac{a(1-r^n)}{1-r}[/tex]approaches
[tex]\text{ }\frac{a}{1-r}[/tex]then the sum of an infinite geometric series would be:
[tex]S_{\infty}=\frac{a}{1-r}_{}[/tex]The values a = 100 and r = 3/5 can be put in the previous equation:
[tex]S_{\infty}=\frac{100}{1-\frac{3}{5}}_{}=\text{ }250[/tex]then, the correct answer is:
[tex]S_{\infty}=250[/tex]