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.