Question

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

Answer 1

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.

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Answer 2

Let the four consecutive terms in 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.1)

As per the second condition,
a+d+a+3d=14
∴ 2a+4d= 14

∴2(3)+4d=14 (from eq.1)
∴ 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+2(3)= 9
∴ The four consecutive terms of A.P. are -3, 1, 5 and 9.

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