Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.

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