$\color{blue} \boxed{\textbf{SOLUTION}}$ Consider, $S_1,S_2$ be the two sequence. Let, $A={12 112 12 12 112}$ $B={12 112 12 112 12 12 112}$ Then, $S_1={ABAABABABAB...}$ $\color{red} \boxed{\textbf{S1}}$ $A_1,B_1$ denotes number of $1's$ that it contains. $A_1=7, B_1=10$ $A_1B_1A_1A_1B_1A_1B_1A_1B_1A_1B_1A_1=7+10+7+7+10+7+10+7+10+7+10+7=99$ So hundredth $1$ is in the first posion of $B$ after $ABAABABABABA$ $|ABAABABABABA|=12+17+12+12+17+12+17+12+17+12+17+12=169$ So $1$ is in the $169+1=\boxed {\textbf{170}}$ th position. $\color{red} \boxed{\textbf{S2}}$ $|S_1| \leq 1000$ $\implies 12+17+12+[(12+17)k] + m \leq 1000 [m={0,12}]$ $\implies 29k+m \leq 959$ $\implies 29×33=957 \implies m=0,k=33$ So the thousandth digit is the second digit of $A$ that is $\boxed {\textbf{2}}$