Answer
$\dfrac{41,333}{33,300}$
Work Step by Step
We have:
$1.24123123123=\dfrac{124}{100}+\dfrac{123}{100000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$
The sum of the geometric series $1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..$ can be found as:
$S=\dfrac{a}{1-r}$
Here, $a=1, r=\dfrac{1}{1000}=0.001$;
Thus,
$S=\dfrac{124}{100}+\dfrac{123}{100000}(\dfrac{1}{1-0.001})=\dfrac{41,333}{33,300}$