Answer: −3, 1, 5, 9
Sum: \(4a + 2d = 12\), so \(2a + d = 6\) ... (I)
\((a + d) + (a + 2d) = 14\), so \(2a + 3d = 14\) ... (II)
(II) − (I): \(2d = 8\), \(d = 4\); then \(a = 1\).
Terms: \(1 - 4, 1, 5, 9\), i.e. −3, 1, 5, 9. Check: sum = 12 and \(5 + 9 = 14\).