Answer
$=\dfrac {4}{33}$
Work Step by Step
$0.\overline {12}=0.121212...=\dfrac {12}{100}+\dfrac {12}{100^{2}}+\dfrac {12}{100^{3}}+...=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {12}{100}}{1-\dfrac {1}{100}}=\dfrac {\dfrac {12}{100}}{\dfrac {99}{100}}=\dfrac {12}{99}=\dfrac {4}{33}$
