Q. 84.0( 32 Votes )

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

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)

Answer :

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 = 6 …..(1)


Also, sum of third and fourth term is 14


(a + d) + (a + 2d) = 14


2a + 3d = 14 ……(2)


Subtracting eq. (1) from eq. (2) we get,


(2a + 3d) – (2a + d) = 14 – 6


2a + 3d – 2a – d = 8


2d = 8



d = 4


Substituting value of “d” in eq. (1) we get,


2a + 4 = 6


2a = 6 – 4 = 2



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


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