ABCD is a square. A straight line is drawn through the centre of the square to cut AB at N such that AN/BN = 1/2. M is an arbitrary point inside the square on this line. Show that the distances of M from AB, AD, BC, CD, form an A.P.

consider your square ABCD to have 2 sides  along the x and y axes with side length say a. Then mark the co-ordinates of other points based on the reference axes. Then solve the rest. if the x axis co-ordinate of M is lets say X then the common difference will come out to be (a-2X).[you can verify it]

good luck