Check whether -150 is a term of the A.P. 11, 8, 5, 2, …

Here, a₂ - a₁ = 8 - 11 = -3,     a₃ - a₂ = 5 - 8 = -3,

          a₄ - a₃ = 2 - 5 = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = 11 and d = -3.

Now we want to find whether there exists a natural number n for which a_n = -150

⇒ a + (n - 1)d = -150
⇒ 11 + (n - 1)(-3) = -150
⇒ 11 - 3n + 3 = -150
⇒ 14 - 3n = -150
⇒ 3n = 14 + 150
⇒ 3n = 164
⇒ n = $54\dfrac{2}{3}$, which is not a natural number.

Hence, there is no term in the given list of numbers which is -150.