Let the four consecutive terms be a – 3d, a – d, a + d and a + 3d
According to the first condition,
a – 3d + a – d + a + d + a + 3d = 12
\(\therefore\) 4a = 12 \(\therefore\) a = 12/4
\(\therefore\) a = 3 ... (i)
According to the second condition,
a + d + a + 3d = 14
\(\therefore\) 2a + 4d = 14
\(\therefore\) 2 x 3 + 4d = 14 ... [From (i)]
\(\therefore\) 4d = 14 – 6
\(\therefore\) 4d = 8
\(\therefore\) d = 2
Thus, a –3d = 3 – 3 x 2 = –3
a – d = 3 – 2 = 1
a + d = 3 + 2 = 5
a + 3d = 3 + 3 x 2 = 9
\(\therefore\) The four consecutive terms of an A.P. are –3, 1, 5 and 9.