If the 10$$^{th}$$ term of an A.P. is $$\frac{1}{20}$$, and its 20$$^{th}$$ term is $$\frac{1}{10}$$, then the sum of its first 200 terms is.

Let the first term of the A.P. be $$a$$ and the common difference be $$d$$. The general formula for the $$n^{\text{th}}$$ term of an A.P. is stated first:

$$T_n = a + (n-1)d.$$

We are told that the $$10^{\text{th}}$$ term equals $$\dfrac{1}{20}$$. Substituting $$n = 10$$ in the formula, we have

$$T_{10} = a + (10-1)d = a + 9d = \dfrac{1}{20} \quad \text{(1)}.$$

Similarly, the $$20^{\text{th}}$$ term equals $$\dfrac{1}{10}$$, so with $$n = 20$$ we obtain

$$T_{20} = a + (20-1)d = a + 19d = \dfrac{1}{10} \quad \text{(2)}.$$

Now we subtract equation (1) from equation (2) to eliminate $$a$$:

$$\bigl(a + 19d\bigr) - \bigl(a + 9d\bigr) = \dfrac{1}{10} - \dfrac{1}{20}.$$

On the left, $$a$$ cancels and we are left with $$10d$$. On the right, we find a common denominator:

$$10d = \dfrac{1}{10} - \dfrac{1}{20} = \dfrac{2}{20} - \dfrac{1}{20} = \dfrac{1}{20}.$$

So

$$d = \dfrac{1}{20} \div 10 = \dfrac{1}{200}.$$

We substitute this value of $$d$$ back into equation (1) to find $$a$$:

$$a + 9\left(\dfrac{1}{200}\right) = \dfrac{1}{20}.$$

Thus

$$a = \dfrac{1}{20} - \dfrac{9}{200}.$$

We convert $$\dfrac{1}{20}$$ to a denominator of 200 to combine the fractions:

$$\dfrac{1}{20} = \dfrac{10}{200}, \quad \text{so} \quad a = \dfrac{10}{200} - \dfrac{9}{200} = \dfrac{1}{200}.$$

Now we know both the first term and the common difference:

$$a = \dfrac{1}{200}, \qquad d = \dfrac{1}{200}.$$

To find the sum of the first 200 terms, we use the sum formula for an A.P., stated as

$$S_n = \dfrac{n}{2}\bigl[2a + (n-1)d\bigr].$$

Here $$n = 200$$, so

$$S_{200} = \dfrac{200}{2}\Bigl[2\!\left(\dfrac{1}{200}\right) + (200-1)\!\left(\dfrac{1}{200}\right)\Bigr].$$

The factor $$\dfrac{200}{2}$$ simplifies to 100, hence

$$S_{200} = 100\Bigl[2\!\left(\dfrac{1}{200}\right) + 199\!\left(\dfrac{1}{200}\right)\Bigr].$$

We calculate each term inside the bracket:

$$2\!\left(\dfrac{1}{200}\right) = \dfrac{2}{200} = \dfrac{1}{100},$$

$$199\!\left(\dfrac{1}{200}\right) = \dfrac{199}{200}.$$

Adding them yields

$$\dfrac{1}{100} + \dfrac{199}{200} = \dfrac{2}{200} + \dfrac{199}{200} = \dfrac{201}{200}.$$

Therefore

$$S_{200} = 100 \times \dfrac{201}{200} = \dfrac{100}{200} \times 201 = \dfrac{1}{2} \times 201 = \dfrac{201}{2}.$$

Writing $$\dfrac{201}{2}$$ as a mixed number gives $$100\dfrac{1}{2}$$.

Hence, the correct answer is Option D.