The production of TV in a factory increases uniformly by a fixed number every year .It produced 8000 TV's in `6^(th)` year & 11300 in `9^(th)` year , find the production in `8^(th)` year .

Options

A. 10500
B. 9800
C. 9700
D. 10200

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
The production of TVs in a factory increases uniformly by a fixed number every year, forming an arithmetic progression (AP). We know the production in the 6th year and the 9th year.

### Step 2: Define the Variables
Let:
- \( A \) be the first term (the production in the first year).
- \( d \) be the common difference (the fixed increase in production each year).
- \( A_n \) be the production in the nth year.

From the problem:
- \( A_6 = 8000 \) (production in the 6th year)
- \( A_9 = 11300 \) (production in the 9th year)

### Step 3: Write the Equations
Using the formula for the nth term of an AP:
\[ A_n = A + (n - 1)d \]
We can write:
1. For the 6th year:
\[ A_6 = A + (6 - 1)d = A + 5d = 8000 \quad \text{(Equation 1)} \]
2. For the 9th year:
\[ A_9 = A + (9 - 1)d = A + 8d = 11300 \quad \text{(Equation 2)} \]

### Step 4: Subtract the Equations
Now, we can subtract Equation 1 from Equation 2 to eliminate \( A \):
\[ (A + 8d) - (A + 5d) = 11300 - 8000 \]
This simplifies to:
\[ 3d = 3300 \]

### Step 5: Solve for \( d \)
Now, divide both sides by 3:
\[ d = \frac{3300}{3} = 1100 \]

### Step 6: Substitute \( d \) Back to Find \( A \)
Now, substitute \( d \) back into Equation 1 to find \( A \):
\[ A + 5(1100) = 8000 \]
This simplifies to:
\[ A + 5500 = 8000 \]
Subtract 5500 from both sides:
\[ A = 8000 - 5500 = 2500 \]

### Step 7: Find the Production in the 8th Year
Now we can find the production in the 8th year using the formula:
\[ A_8 = A + (8 - 1)d = A + 7d \]
Substituting the values of \( A \) and \( d \):
\[ A_8 = 2500 + 7(1100) = 2500 + 7700 = 10200 \]

### Final Answer
The production in the 8th year is **10,200 TVs**.

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