10 points for whoever answer, LoverYourselfFirst took down my question because its not difficult, i would not be putting it here if it was not difficult

Sum of 5th & 8th terms of an AP = 37

11th term = 32

A.P

We know that,

nth term of an AP – an = a + (n - 1)d

Hence,

⟹ a₅ + a₈ = 37

⟹ a + (5 - 1)d + a + (8 - 1)d = 37

⟹ 2a + 4d + 7d = 37

⟹ 2a + 11d = 37 -- equation (1)

Similarly,

⟹ a₁₁ = 32

⟹ a + (11 - 1)d = 32

⟹ a + 10d = 32

⟹ a = 32 - 10d

Substitute the value of a in equation (1).

⟹ 2(32 - 10d) + 11d = 37

⟹ 64 - 20d + 11d = 37

⟹ 64 - 37 = 20d - 11d

⟹ 27 = 9d

⟹ 27/9 = d

⟹ 3 = d

Substitute the value of d in equation (1).

⟹ 2a + 11(3) = 37

⟹ 2a + 33 = 37

⟹ 2a = 37 - 33

⟹ 2a = 4

⟹ a = 4/2

⟹ a = 2

Now,

General form of an ap = a , a + d , a + 2d...

⟶ Required AP = 2 , 2 + 3 , 2 + 2(3)...

⟶ Required AP = 2 , 5 , 2 + 6...

⟶ Required AP = 2 , 5 , 8...

I hope it will help you.

Regards.