Given: The four numbers are in AP such that their sum is 20 and the sum of their squares is 120. Let (a - 3d), (a - d), (a + d) and (a + 3d) be the required four terms. ∵ Their sum is 20. i.e (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20 ⇒ 4a = 20 ⇒ a = 5. ∵ Sum of their squares is 120. i.e (a−3d)2+(a−d)2+(a+d)2+(a+3d)2=120 ⇒a2+5.d2=30 By substituting a=5 in the above equation we get, ⇒25+5.d2=30 ⇒d2=1⇒d=±1 Case 1: When a = 5 and d = 1 then the four numbers are:2, 4, 6, 8 Case 2: When a = 5 and d = - 1 then the four numbers are: 8, 6, 4, 2 Hence, in either case the four numbers are 2, 4, 6, 8.