Let a be the first term and d be the common difference of the AP. Then,

a₄ = 9

\(\Rightarrow\) a + (4 - 1) d = 9    [a_n = a + (n - 1)d]

\(\Rightarrow\) a + 3d = 9    ......(1)

Now,

a₆ + a₁₃ = 40    (Given)

\(\Rightarrow\) (a + 5d)+(a + 12d) = 40

\(\Rightarrow\) 2a + 17d = 40    ......(2)

From (1) and (2), we get

2(9 - 3d) + 17d = 40

\(\Rightarrow\) 18 - 6d + 17d = 40

\(\Rightarrow\) 11d = 40 - 18 = 22

\(\Rightarrow\) d = 2

Putting d = 2 in (1), we get

a + 3 x 2 = 9

\(\Rightarrow\) a = 9 - 6 = 3

Hence, the AP is 3, 5, 7, 9, 11,…….

The fourth term is given by 

T₄=9

a+(n-1)d = 9 where a denotes first term and d denotes common difference 

Here for 4th term n=4

Therefore a+3d=9....(1) × 2

2a +6d =18...(2)

Also T₆ + T₁₃ = 40

a+(6-1)d + a+(13-1)d = 40

a + 5d + a +12d = 40

2a + 17d = 40....(3)

Solving equations 2 and 3 we get

(3) - (2)

17d -6d = 40-18

11d = 22

d = 2

Now substituting the value of d in equation (2) we have 

2a+6(2) =18

2a= 18-12 = 6

a=3

Therefore the required AP is

3, 5, 7,9,..