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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