The production of TV in a factory increases uniformly by a fixed number every year. It produced 8000 sets in `6^(th)` year and 11300 in `9^(th)` year. Find the total production in 6 years.

Options

A. 40500
B. 20000
C. 20500
D. 31500

Answer

To solve the problem, we need to find the total production of TVs in the factory over the first 6 years given the production in the 6th and 9th years.

### Step-by-step Solution:

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

2. **Set Up the Equations**:
- The production in the 6th year can be expressed as:
\[ P + 5d = 8000 \]
- The production in the 9th year can be expressed as:
\[ P + 8d = 11300 \]

3. **Subtract the First Equation from the Second**:
To eliminate \( P \), we subtract the first equation from the second:
\[ (P + 8d) - (P + 5d) = 11300 - 8000 \]
Simplifying this gives:
\[ 3d = 3300 \]
Therefore, we can solve for \( d \):
\[ d = \frac{3300}{3} = 1100 \]

4. **Substitute \( d \) Back to Find \( P \)**:
Now, substitute \( d \) back into the first equation to find \( P \):
\[ P + 5(1100) = 8000 \]
Simplifying this gives:
\[ P + 5500 = 8000 \]
Therefore,
\[ P = 8000 - 5500 = 2500 \]

5. **Calculate Total Production Over 6 Years**:
The total production over the first 6 years can be calculated using the formula for the sum of an arithmetic series:
\[ \text{Total Production} = 6P + 15d \]
Substituting the values of \( P \) and \( d \):
\[ \text{Total Production} = 6(2500) + 15(1100) \]
Calculating this gives:
\[ = 15000 + 16500 = 31500 \]

### Final Answer:
The total production in 6 years is **31,500 sets**.