jmc

Let the common difference of this arithmetic sequence be $d$. The four angles of the trapezoid are $60$, $60+d$, $60+2d$, and $60+3d$. Since the interior angles in a trapezoid add up to equal 360 degrees, we have the equation $60+(60+d)+(60+2d)+(60+3d)=360\Rightarrow d=20$. Thus, the largest angle in the trapezoid is $60+3\cdot 20=\boxed{120}$ degrees.