1. Find four consecutive terms in an A.P. whose sum is 12 and the sum of 3rd and 4th term is 14.
Sol. Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d, a + 3d
As per the first condition,
a – 3d + a – d + a + d + a + 3d = 12
∴ 4a = 12
∴ a = 12/4
∴ a = 3 ...........eq. (i)
As per the second condition,
a + d + a + 3d = 14
∴ 2a + 4d = 14
∴ 2 (3) + 4d = 14 [From eq. (i)]
∴ 6 + 4d = 14
∴ 4d = 14 – 6
∴ 4d = 8
∴ d = 8/4
∴ d = 2
∴ a – 3d = 3 – 3 (2) = 3 – 6 = – 3
a – d = 3 – 2 = 1
a + d = 3 + 2 = 5
a + 3d = 3 + 3 (2) = 9
∴ The four consecutive terms of A.P. are – 3, 1, 5 and 9.
