Find four consecutive terms in an A.P. whose sum is –54 and the sum of 1st and 3rd term is – 30.


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    

 The four consecutive terms of an A.P. are – 18, – 15, – 12 and – 9.