Question

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

Answer 1

Let the four consecutive terms in the A.P. be a – 3d, a – d, a + d and a + 3d.
We know that the sum of these four consecutive numbers is 12.
Thus, we have:
a – 3d + a – d + a + d + a + 3d = 12
$\Rightarrow$4a = 12
$\Rightarrow$a = 3 ...(i)
Also,
The sum of the 3^rd and 4^th terms is 14.
Thus, we have:
a + d + a + 3d = 14
$\Rightarrow$2a + 4d = 14
$\Rightarrow$2 × 3 + 4d = 14 [From(i)]
$\Rightarrow$4d = 14 – 6 = 8
$\Rightarrow$d = 2
Now,
a – 3d = 3 – 3 × 2 = 3 – 6 = –3
a – d = 3 – 2 = 1
a + d = 3 + 2 = 5
a + 3d = 3 + 3 × 2 = 3 + 6 = 9

Therefore, the four consecutive numbers in the A.P. are –3, 1, 5 and 9.