Answer :

[tex]n;\ n+2;\ n+4-the\ 3\ consecutive\ odd\ numbers\\\\n+(n+2)+(n+4)=129\\n+n+2+n+4=129\\3n+6=129\ \ \ \ \ |subtract\ 6\ from\ both\ sides\\3n=123\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\boxed{n=41}\\\\Answer:\\n=41\\n+2=43\\n+4=45[/tex]