Let the first term be a – d

the second term be a

the third term be a + d

the fourth term be a + 2 d

Given sum of consecutive four term is 12

⇒ (a – d) + a + (a + d) + (a + 2d) = 12

⇒ 4 a + 2d = 12

⇒ 2(2 a + d) = 12

⇒ 2a + d = 12/2 = 6

⇒ 2a + d = 6 …. (1)

Also, sum of third and fourth term is 14

⇒ (a + d) + (a + 2d) = 14

⇒ 2a + 3d = 14 …… (2)

Subtracting equation (1) from equation (2) we get,

⇒ (2a + 3d) – (2a + d) = 14 – 6

⇒ 2a + 3d – 2a – d = 8

⇒ 2d = 8

⇒ d = 8/2 = 4

⇒ d = 4

Substituting value of “d” in equation (1) we get,

⇒ 2a + 4 = 6

⇒ 2a = 6 – 4 = 2

⇒ a = 2/2 = 1

⇒ a = 1

Thus, a = 1 and d = 4

Hence, first term a – d = 1 – 4 = – 3

the second term a = 1

the third term a + d = 1 + 4 = 5

the fourth term a + 2 d = 1 + 2×4 = 1 + 8 = 9

Thus, the A.P. is – 3, 1, 5, 9