If $$100$$ times the $$100^{th}$$ term of an $$AP$$ with non zero common difference equals the $$50$$ times its $$50^{th}$$ term, then the $$150^{th}$$ term of this $$AP$$ is

Let $$a$$ be the first term and $$d (\neq 0)$$ be the common difference of the AP.
Then the $$n^{\text{th}}$$ term is $$T_n = a + (n-1)d$$.

Given condition:
$$100 \times T_{100} = 50 \times T_{50}$$

Write the two terms explicitly:
$$T_{100} = a + 99d,$$
$$T_{50} = a + 49d.$$

Substitute in the condition:
$$100\left(a + 99d\right) = 50\left(a + 49d\right).$$

Expand both sides:
$$100a + 9900d = 50a + 2450d.$$

Bring all terms to one side:
$$100a - 50a + 9900d - 2450d = 0,$$
$$50a + 7450d = 0.$$

Solve for $$a$$:
$$a = -\frac{7450d}{50} = -149d.$$

Now find the $$150^{\text{th}}$$ term:
$$T_{150} = a + 149d = \bigl(-149d\bigr) + 149d = 0.$$

Therefore the $$150^{\text{th}}$$ term of the AP is $$0$$.

Option D which is: zero