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The problem describes a situation where the manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. The production increases uniformly, meaning it follows an arithmetic progression (AP).

### Let's break down the problem:

Let:
- \( a \) be the production in the first year (the first term of the AP),
- \( d \) be the common difference (the yearly increase in production).

From the information given:
- Production in the third year: \( a + 2d = 600 \),
- Production in the seventh year: \( a + 6d = 700 \).

### Now, solving for \( a \) and \( d \):
1. From \( a + 6d = 700 \) and \( a + 2d = 600 \), subtract the second equation from the first:
   \[
   (a + 6d) - (a + 2d) = 700 - 600
   \]
   \[
   4d = 100 \quad \Rightarrow \quad d = 25.
   \]

2. Substitute \( d = 25 \) into \( a + 2d = 600 \):
   \[
   a + 2(25) = 600 \quad \Rightarrow \quad a + 50 = 600 \quad \Rightarrow \quad a = 550.
   \]

Now that we have \( a = 550 \) and \( d = 25 \), we can solve the three parts of the question:

### (i) Production in the 1st year:
This is simply \( a = 550 \).

### (ii) Production in the 10th year:
The production in the 10th year is given by:
\[
a + 9d = 550 + 9(25) = 550 + 225 = 775.
\]

### (iii) Total production in the first 7 years:
The total production over the first 7 years is the sum of the first 7 terms of the AP:
\[
S_7 = \frac{7}{2} \times [2a + (7 - 1)d]
\]
\[
S_7 = \frac{7}{2} \times [2(550) + 6(25)] = \frac{7}{2} \times [1100 + 150] = \frac{7}{2} \times 1250 = 7 \times 625 = 4375.
\]

### Summary of answers:
- (i) The production in the 1st year: 550 sets,
- (ii) The production in the 10th year: 775 sets,
- (iii) The total production in the first 7 years: 4375 sets.

Would you like further details or clarifications?

Here are some related questions:
1. What is the formula for the \(n\)-th term of an arithmetic progression?
2. How would you calculate the sum of the first \(n\) terms in a general arithmetic progression?
3. If the production increased by 30 sets each year instead of 25, what would the production be in the 10th year?
4. What would be the total production for the first 10 years in this problem?
5. How does the uniform increase in production affect the long-term manufacturing output?

**Tip**: When solving arithmetic progression problems, always try to express unknown terms using known formulas for the \(n\)-th term and the sum of terms.

A manufacturer of TV sets 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 1st year, (ii) the production in the 10th year, (iii) the total production in the first 7 years.

Solve an arithmetic progression problem involving TV production over several years. Learn how to calculate the production in the 1st year, 10th year, and the total production in the first 7 years. Suitable for students in Grades 9-11.

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