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

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Let the four consecutive terms be a – 3d, a – d, a + d and a + 3d

 According to the first condition,

a – 3d + a – d + a + d + a + 3d = 12

\(\therefore\) 4a = 12    \(\therefore\) a = 12/4

\(\therefore\) a = 3 ... (i)

According to the second condition,

a + d + a + 3d = 14

\(\therefore\) 2a + 4d = 14

\(\therefore\) 2 x 3 + 4d = 14 ... [From (i)]

\(\therefore\) 4d = 14 – 6

\(\therefore\) 4d = 8 

\(\therefore\) d = 2

Thus, a –3d = 3 – 3 x 2 = –3

a – d = 3 – 2 = 1

a + d = 3 + 2 = 5

a + 3d = 3 + 3 x 2 = 9

\(\therefore\) The four consecutive terms of an A.P. are –3, 1, 5 and 9.

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