Question 892292
Let n be an integer (maybe?).

{{{6n+6(n+1)+6(n+2)+6(n+3)=125}}}


{{{24n+6+12+18=125}}}
{{{24n+36=125}}}
{{{24n=125-36}}}
{{{24n=89}}}
{{{n=89/24}}}
n is NOT an integer.


The four multiples of six are 89/4, ....
No.  I am stopping there.
Are you SURE you have the correct problem description?  What do you mean by "consecutive multiples"?  Does each multiple of 6 differ by 1?  If so, then the next or last three numbers are {{{6(89/24+1)}}}, {{{6(89/24+2)}}}, and {{{6(89/24+3)}}}.