Question

7. The sum of four consecutive terms of an
A.P. is 2. The sum of the 3rd and 4th
terms is 11. Find the terms.​

Answer 1

Step-by-step explanation:

Let the four consecutive terms in AP be a - 3d, a - d, a + d, a + 3d

According to the first condition:

Sum of four consecutive terms of an

A.P. is 2.

Therefore,

a - 3d + a - d +a + d + a + 3d = 2

$\therefore \: 4a = 2 \\ \therefore \: a = \frac{2}{4} \\ \therefore \: a =0.5 \\ According \: to \: the \: second \: condition: \\ a + d + a + 3d = 11 \\ \implies \: 2a + 4d = 11 \\ \implies \: 2 \times 0.5 + 4d = 11 \\ \implies \: 1 + 4d = 11 \\ \implies \: 4d = 11 - 1\\ \implies \: 4d = 10 \\ \implies \: d = \frac{10}{4} \\ \implies \: d = 2.5 \\ hence \\ a - 3d = 0.5 - 3 \times 2.5 = 0.5 - 7.5 = - 7 \\ a - d = 0.5 - 2.5 = - 2 \\ a + d = 0.5 + 2.5 = 3 \\ a + 3d = 0.5 + 3 \times 2.5 = 0.5 + 7.5 = 8 \\ thus \: the \: four \: terms \: in \: A.P. \: are : \\ - 7, \: - 2, \: 3, \: 8.$