Answer
$\dfrac{41,333}{33,300}$
Work Step by Step
Here, $1.24123123123=\dfrac{124}{100}+\dfrac{123}{100000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$
Formula to find the sum of a geometric series$(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ is $S=\dfrac{a}{1-r}$
Since, we have $a=1, r=\dfrac{1}{1000}=0.001$;
Then , we get
$S=\dfrac{124}{100}+\dfrac{123}{100000}(\dfrac{1}{1-0.001})=\dfrac{41,333}{33,300}$