Answer :
The given geometric series is
[tex]120+20+\frac{10}{3}+\frac{5}{9}+\cdots[/tex][tex]The\text{ common difference r=}\frac{20}{120}=\frac{1}{6}[/tex][tex]\text{The first term is a=120.}[/tex]Recall that the formula for the sum of the infinite geometric series is
[tex]\sum ^{\infty}_{k\mathop=0}ar^k=\frac{a}{1-r}[/tex]Substitute a=120 and r=1/6, we get
[tex]\sum ^{\infty}_{k\mathop=0}120(\frac{1}{6})^k=\frac{120}{1-\frac{1}{6}}[/tex][tex]=\frac{120}{\frac{6-1}{6}}[/tex][tex]=\frac{120}{\frac{5}{6}}[/tex][tex]\text{ Use }\frac{a}{\frac{b}{c}}=a\times\frac{c}{b}\text{.}[/tex][tex]=120\times\frac{6}{5}[/tex][tex]=144[/tex]Hence the sum of the given infinite geometric series is 144.
Hence the third option is correct.