let's say our first integer is say 2a, bearing in mind that 2 times any number, gives us an even value, regardless of what that "a" number might be.
let's notice something, 2, 4, 6, 8, 10......, to get a consecutive even number, we simply hop twice either forwards or backwards, namely 6±2 gives us 4 and 8, and both are even, so let's move forwards and use 2a + 2 for the second one.
so if the first integer is 2a, the next consecutive is 2a + 2, and the third integer will be 2a + 4.
we know they all add up to 126, so
[tex]\bf (2a)+(2a+2)+(2a+4)=126\implies 6a+6=126 \\\\\\ 6a=120\implies a=\cfrac{120}{6}\implies a=20 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{largest~~2(20)+4}{44}~\hfill[/tex]
