Satellite TV manufacturing businesses tend to have what economists call "economies of scale." When economies of scale exist, bigness can be its own reward. The more TV's you manufacture in a single run, lower the costs per unit, which in turn increases your bottom-line margins.
Keeping that in mind, a T.V. manufacturing company increases its production uniformly by fixed number every year. The company produces 8000, sets in the `6^(th)` year and 11,300 sets in the `9^(th)` year.
The company's production of the first year is:

Options

A. 2000
B. 2500
C. 3000
D. 5000

Answer

To solve the problem, we need to determine the production of the first year based on the information given about the production in the 6th and 9th years.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the production in the first year be \( x \) and the fixed increase in production each year be \( d \).

2. **Establish the Production Equations**:
- In the 6th year, the production can be expressed as:
\[ \text{Production in 6th year} = x + 5d = 8000 \]
- In the 9th year, the production can be expressed as:
\[ \text{Production in 9th year} = x + 8d = 11300 \]

3. **Set Up the System of Equations**:
We now have two equations:
\[ (1) \quad x + 5d = 8000 \]
\[ (2) \quad x + 8d = 11300 \]

4. **Subtract the First Equation from the Second**:
To eliminate \( x \), we subtract equation (1) from equation (2):
\[ (x + 8d) - (x + 5d) = 11300 - 8000 \]
This simplifies to:
\[ 3d = 3300 \]

5. **Solve for \( d \)**:
Divide both sides by 3:
\[ d = \frac{3300}{3} = 1100 \]

6. **Substitute \( d \) Back into One of the Equations**:
Now substitute \( d = 1100 \) back into equation (1):
\[ x + 5(1100) = 8000 \]
Simplifying this gives:
\[ x + 5500 = 8000 \]

7. **Solve for \( x \)**:
Subtract 5500 from both sides:
\[ x = 8000 - 5500 = 2500 \]

### Conclusion:
The production of the first year is \( \boxed{2500} \).