For an A.P. `T_(10) = 1/20 , T_20 = 1/10`. Find sum of first 200 term.

Options

A. `201 1/2`
B. `101 1/2`
C. `301 1/2`
D. `100 1/2`

Explanation

To solve the problem, we need to find the sum of the first 200 terms of an arithmetic progression (A.P.) given the 10th and 20th terms.

### Step 1: Define the general term of the A.P.
The general term \( T_n \) of an A.P. can be expressed as:
\[ T_n = A + (n-1)D \]
where \( A \) is the first term and \( D \) is the common difference.

### Step 2: Write the equations for the given terms.
From the problem, we know:
- \( T_{10} = A + 9D = \frac{1}{20} \) (Equation 1)
- \( T_{20} = A + 19D = \frac{1}{10} \) (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2.
Subtracting Equation 1 from Equation 2 gives:
\[ (A + 19D) - (A + 9D) = \frac{1}{10} - \frac{1}{20} \]
This simplifies to:
\[ 10D = \frac{1}{10} - \frac{1}{20} \]

### Step 4: Simplify the right-hand side.
To simplify \( \frac{1}{10} - \frac{1}{20} \), we find a common denominator:
\[ \frac{1}{10} = \frac{2}{20} \]
Thus,
\[ 10D = \frac{2}{20} - \frac{1}{20} = \frac{1}{20} \]

### Step 5: Solve for \( D \).
Now, we can solve for \( D \):
\[ D = \frac{1}{20} \div 10 = \frac{1}{200} \]

### Step 6: Substitute \( D \) back to find \( A \).
Now, substitute \( D \) back into Equation 1:
\[ A + 9D = \frac{1}{20} \]
Substituting \( D \):
\[ A + 9 \left(\frac{1}{200}\right) = \frac{1}{20} \]
This simplifies to:
\[ A + \frac{9}{200} = \frac{10}{200} \]
Thus,
\[ A = \frac{10}{200} - \frac{9}{200} = \frac{1}{200} \]

### Step 7: Find the sum of the first 200 terms.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
For \( n = 200 \):
\[ S_{200} = \frac{200}{2} \left(2 \cdot \frac{1}{200} + (200-1) \cdot \frac{1}{200}\right) \]
This simplifies to:
\[ S_{200} = 100 \left(\frac{2}{200} + \frac{199}{200}\right) = 100 \left(\frac{201}{200}\right) \]
Thus,
\[ S_{200} = \frac{20100}{200} = 100.5 \]

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