A manufacture of TV set produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:
(i) the production in the first year
(ii) the production in the 10th year.
(iii) the total production in first 7 years.

Answer

To solve the problem step by step, we will use the concept of arithmetic progression (AP) since the production of TV sets increases uniformly by a fixed number each year.

### Step 1: Define the variables
Let:
- \( A \) = production in the first year
- \( D \) = fixed increase in production each year

The production in the nth year can be expressed as:
\[ P_n = A + (n - 1)D \]

### Step 2: Set up the equations based on given information
From the problem, we know:
- In the 3rd year, \( P_3 = 600 \):
\[ P_3 = A + (3 - 1)D = A + 2D = 600 \tag{1} \]
- In the 7th year, \( P_7 = 700 \):
\[ P_7 = A + (7 - 1)D = A + 6D = 700 \tag{2} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( A + 2D = 600 \)
2. \( A + 6D = 700 \)

We can subtract equation (1) from equation (2):
\[ (A + 6D) - (A + 2D) = 700 - 600 \]
This simplifies to:
\[ 4D = 100 \]
Now, divide both sides by 4:
\[ D = 25 \]

### Step 4: Substitute \( D \) back to find \( A \)
Now substitute \( D = 25 \) back into equation (1):
\[ A + 2(25) = 600 \]
\[ A + 50 = 600 \]
Subtract 50 from both sides:
\[ A = 550 \]

### Step 5: Find the production in the first year
Thus, the production in the first year is:
\[ \text{Production in 1st year} = A = 550 \]

### Step 6: Find the production in the 10th year
Now, we can find the production in the 10th year using the formula:
\[ P_{10} = A + (10 - 1)D = A + 9D \]
Substituting the values of \( A \) and \( D \):
\[ P_{10} = 550 + 9(25) = 550 + 225 = 775 \]

### Step 7: Find the total production in the first 7 years
The total production in the first 7 years can be calculated using the formula for the sum of the first n terms of an AP:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
Here, \( n = 7 \):
\[ S_7 = \frac{7}{2} \times (2(550) + (7 - 1)(25)) \]
Calculating inside the parentheses:
\[ = \frac{7}{2} \times (1100 + 150) = \frac{7}{2} \times 1250 \]
\[ = \frac{7 \times 1250}{2} = \frac{8750}{2} = 4375 \]

### Final Answers
(i) The production in the first year is **550 sets**.
(ii) The production in the 10th year is **775 sets**.
(iii) The total production in the first 7 years is **4375 sets**.