For an A.P. `T_(10) = 1/20 , T_20 = 1/10`. Find sum of first 200 term. (A) 201`1/2` (B) 101`1/2` (C) 301`1/2` (D) 100`1/2`

Options

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

Explanation

To solve the problem step by step, we will follow the given information about the arithmetic progression (A.P.) and use the formulas for the terms and the sum of the series.

### Step 1: Write the general term of an A.P.
The general term \( T_n \) of an arithmetic progression is given by:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations using the given terms
We are given:
- \( T_{10} = \frac{1}{20} \)
- \( T_{20} = \frac{1}{10} \)

Using the general term formula, we can write:
1. For \( T_{10} \):
\[ a + 9d = \frac{1}{20} \quad \text{(1)} \]
2. For \( T_{20} \):
\[ a + 19d = \frac{1}{10} \quad \text{(2)} \]

### Step 3: Subtract the equations to eliminate \( a \)
Subtract equation (1) from equation (2):
\[ (a + 19d) - (a + 9d) = \frac{1}{10} - \frac{1}{20} \]
This simplifies to:
\[ 10d = \frac{1}{10} - \frac{1}{20} \]
Finding a common denominator (20):
\[ 10d = \frac{2}{20} - \frac{1}{20} = \frac{1}{20} \]
Thus, we have:
\[ 10d = \frac{1}{20} \]
Dividing both sides by 10 gives:
\[ d = \frac{1}{200} \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d \) back into equation (1):
\[ a + 9\left(\frac{1}{200}\right) = \frac{1}{20} \]
This simplifies to:
\[ a + \frac{9}{200} = \frac{10}{200} \]
Subtracting \( \frac{9}{200} \) from both sides gives:
\[ a = \frac{10}{200} - \frac{9}{200} = \frac{1}{200} \]

### Step 5: Find the sum of the first 200 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
For \( n = 200 \):
\[ S_{200} = \frac{200}{2} \left(2\left(\frac{1}{200}\right) + (200-1)\left(\frac{1}{200}\right)\right) \]
This simplifies to:
\[ S_{200} = 100 \left(\frac{2}{200} + \frac{199}{200}\right) = 100 \left(\frac{201}{200}\right) \]
Calculating this gives:
\[ S_{200} = \frac{20100}{200} = 100.5 \]

### Final Answer
The sum of the first 200 terms is:
\[ \boxed{100 \frac{1}{2}} \]