If the sum of four numbers in AP is 24 and sum of their square is 164, then those numbers are

Consider the four numbers which are in A.P  be a-3d,a-d,a+d,a+3d  Sum of the numbers =(a−3d)+a−d+a+d+a+3d=24⇒4a=24⇒a=6 Sum of the squares of the numbers =(a−3d)2+(a−d)2+(a+d)2+(a+3d)2=164⇒2(a2+9d2)+2a2+d2=164⇒4a2+20d2=164⇒a2+5d2=41⇒36+5d2=41⇒d=±1Required  numbers  are =6-3,6-1,6+1,6+3                                              =  3,5,7,9