Find an A.P. whose fourth term is 9 and the sum of its sixth term and the thirteenth term is 40. (CBSE 2011)

Let a be the first term and d be the common difference.

Given: a₄ = 9

a₆ + a₁₃ = 40

We first find the common difference.

Now, Consider a₄ = 9

⇒ a + (4 – 1)d = 9

⇒ a + 3d = 9 ……………………….(1)

Consider a₆ + a₁₃ = 40

⇒ a + (6 – 1)d + a + (13 – 1)d = 40

⇒ 2a + 17d = 40 ………………..(2)

Subtracting twice of equation (1) from equation (2), we get,

11d = 22

⇒ d = 2

∴ Common difference = d = 2

Put the value of d in equation (), we get

a = 9 – 3d

= 9 – 6

= 3

∴ a = 3

The AP will be a, a + d, a + 2d, a + 3d,…

Thus the AP will be 3, 5, 7, 9,…