Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.

Let the numbers be (a - 3d), (a - d), (a + d), (a + 3d).

Now, sum of the numbers = 28


(a - 3d) + (a - d) + (a + d) + (a + 3d) = 28


4a = 28


a = 7


Now, sum of the squares of the terms = 216


(a - 3d)2 + (a - d)2 + (a + d)2 + (a + 3d)2= 216


a2 + 9d2 - 6ad + a2 + d2 - 2ad + a2 + d2 + 2ad + a2 + 9d2 + 6ad = 216


4a2 + 20d2 = 216


Put the value of a = 7, we get,


4(49) + 20d2 = 216


20d2 = 216 - 196


20d2 = 20


d2 = 1


d = 1


If d = 1, then the numbers are 4, 6, 8, 10.


If d = - 1, then the numbers are 10, 8, 6, 4.


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