Answer
$=\dfrac {2500}{19}$
Work Step by Step
$\sum ^{\infty }_{k=0}\left( \dfrac {1}{4}\right) ^{k}\times 5^{3-k}=\sum ^{\infty }_{k=0}\left( \dfrac {1}{20}\right) ^{k}\times 5^{3}=\dfrac {a_{1}}{1-r};a_{1}=5^{3};r=\dfrac {1}{20}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {5^{3}}{1-\dfrac {1}{20}}=\dfrac {2500}{19}$
