2. Find four consecutive terms in an A.P. whose sum is –54 and the sum of 1st and 3rd term is – 30.
Sol. Let the four consecutive terms is an A.P. be a – 3d, a – d, a + d and a + 3d
As per first condition,
a – 3d + a – d + a + d + a + 3d = – 54
∴ 4a = – 54
∴ a = –54/4
∴ a = - 13.5 ...........eq. (i)
As per the second condition,
a – 3d + a + d = – 30
∴ 2a – 2d = – 30
∴ 2(a – d) = - 30
∴ a – d = -30/2
∴ a – d = - 15
∴ - 13.5 – d = - 15 [From eq. (i)]
∴ d = 15 – 13.5
∴ d = 1.5
∴ a – 3d = -13.5 – 3 (1.5) = -13.5 – 4.5 = -18
a – d = - 13.5 – 1.5 = - 15
a + d = -13.5 + 1.5 = - 12
a+ 3d = -13.5 + 3(1.5) = -13.5 + 4.5 = - 9
