Find the A.P. whose 4^th term is 9 and the sum of its 6^th term and 13^th term is 40.

Given,

a₄ = 9

⇒ a + (4 - 1)d = 9

⇒ a + 3d = 9

⇒ a = 9 - 3d ……..(1)

Given,

Sum of a₆ + a₁₃ = 40

⇒ a + (6 - 1)d + a + (13 - 1)d = 40

⇒ a + 5d + a + 12d = 40

⇒ 2a + 17d = 40 ……(2)

Substituting value of a from equation (1) in equation (2), we get :

⇒ 2(9 − 3d) + 17d = 40

⇒ 18 − 6d + 17d = 40

⇒ 18 + 11d = 40

⇒ 11d = 22

⇒ d = 2.

Substituting the value of d in equation (1), we get :

⇒ a = 9 - 3(2)

⇒ a = 9 - 6

⇒ a = 3.

So the A.P. is,

3, 5, 7, 9, 11, ….

Hence, A.P is 3, 5, 7, 9, 11, …..