Answer: D

Small Hint:

A triangle’s angles form an arithmetic progression exactly when its middle angle is $60^\circ$

Big Hint:

Let $\angle DBA = x$ and $\angle ADB = y;$ the four angles of $ABCD$ become $x, y, 180-y, 180-x,$ which must be an arithmetic progression with one triangle angle equal to $60^\circ$

Solution:

The angles of a triangle form an arithmetic progression exactly when the middle one is $60^\circ.$ With $\angle DBA = x$ and $\angle ADB = y,$ the four angles of $ABCD$ are $x, y, 180 - y, 180-x,$ which must itself be an arithmetic progression. In increasing order they are either $x,y,180-y,180-x$ or $x,180-y,y,180-x,$ giving $3y=x+180$ or $3y=360-x.$ One of the triangle angles $x,y,180-x-y$ is $60^\circ.$ Substitution leaves the angle sets $(60,80,100,120)$ and $(45,75,105,135).$ The two largest angles sum to at most $105 + 135 = 240.$ Thus, the correct answer is D.