It the `10^(th)` term of an A. P. is `1/20` and its `20^(th)` term is `1/10,` then the sum of its first 200 terms is :

Options

A. 50
B. `100(1)/(2)`
C. 100
D. `50(1)/(4)`

Answer

To solve the problem step by step, we need to find the first term \( A_1 \) and the common difference \( D \) of the arithmetic progression (A.P.) given that the 10th term is \( \frac{1}{20} \) and the 20th term is \( \frac{1}{10} \).

### Step 1: Write the formulas for the 10th and 20th terms of the A.P.
The \( n^{th} \) term of an A.P. can be expressed as:
\[ A_n = A_1 + (n-1)D \]
For the 10th term (\( n = 10 \)):
\[ A_{10} = A_1 + 9D = \frac{1}{20} \]
For the 20th term (\( n = 20 \)):
\[ A_{20} = A_1 + 19D = \frac{1}{10} \]

### Step 2: Set up the equations
From the above, we have two equations:
1. \( A_1 + 9D = \frac{1}{20} \) (Equation 1)
2. \( A_1 + 19D = \frac{1}{10} \) (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2
Subtract Equation 1 from Equation 2 to eliminate \( A_1 \):
\[ (A_1 + 19D) - (A_1 + 9D) = \frac{1}{10} - \frac{1}{20} \]
This simplifies to:
\[ 10D = \frac{1}{10} - \frac{1}{20} \]
To perform the subtraction on the right side, find a common denominator:
\[ \frac{1}{10} = \frac{2}{20} \quad \text{so} \quad \frac{1}{10} - \frac{1}{20} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20} \]
Thus, we have:
\[ 10D = \frac{1}{20} \]

### Step 4: Solve for \( D \)
Dividing both sides by 10 gives:
\[ D = \frac{1}{200} \]

### Step 5: Substitute \( D \) back into Equation 1 to find \( A_1 \)
Now substitute \( D \) back into Equation 1:
\[ A_1 + 9 \left(\frac{1}{200}\right) = \frac{1}{20} \]
This simplifies to:
\[ A_1 + \frac{9}{200} = \frac{1}{20} \]
Convert \( \frac{1}{20} \) to a fraction with a denominator of 200:
\[ \frac{1}{20} = \frac{10}{200} \]
Now we have:
\[ A_1 + \frac{9}{200} = \frac{10}{200} \]
Subtract \( \frac{9}{200} \) from both sides:
\[ A_1 = \frac{10}{200} - \frac{9}{200} = \frac{1}{200} \]

### Step 6: Find the sum of the first 200 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (A_1 + A_n) \]
We need to find \( A_{200} \):
\[ A_{200} = A_1 + 199D = \frac{1}{200} + 199 \left(\frac{1}{200}\right) = \frac{1}{200} + \frac{199}{200} = \frac{200}{200} = 1 \]
Now, substituting \( n = 200 \), \( A_1 = \frac{1}{200} \), and \( A_{200} = 1 \):
\[ S_{200} = \frac{200}{2} \times \left(\frac{1}{200} + 1\right) = 100 \times \left(\frac{1}{200} + \frac{200}{200}\right) = 100 \times \frac{201}{200} \]
Calculating this gives:
\[ S_{200} = 100 \times \frac{201}{200} = \frac{20100}{200} = 100.5 \]

### Final Answer
The sum of the first 200 terms is:
\[ \boxed{100.5} \]