Find four consecutive terms in an A.P. whose sum is 12 and the sum of

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

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

Answer»

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.