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 first 7 years
Step-by-Step Solution
Key Concept: The production each year forms an arithmetic progression (AP). Use the AP term formula $a_n = a + (n-1)d$ to find the common difference $d$ and the first term $a$. Then use the sum formula $S_n = \frac{n}{2}[2a+(n-1)d]$ (or $S_n = \frac{n}{2}(a_1 + a_n)$) to obtain the total production.
1. Let $a$ be the production in the 1st year (first term) and $d$ be the common increase each year.\
2. Using the term formula for the 3rd and 7th years:
$$a_3 = a + 2d = 600 \quad\text{(i)}$$
$$a_7 = a + 6d = 700 \quad\text{(ii)}$$
3. Subtract (i) from (ii):
$$ (a+6d) - (a+2d) = 700 - 600 \Rightarrow 4d = 100 \Rightarrow d = 25.$$\
4. Substitute $d = 25$ into (i) to find $a$:
$$a + 2(25) = 600 \Rightarrow a = 600 - 50 = 550.$$\
5. (i) Production in the 1st year: $a = \boxed{550}$ sets.\
6. (ii) Production in the 10th year:
$$a_{10} = a + 9d = 550 + 9\times25 = 550 + 225 = \boxed{775}\text{ sets}.$$\
7. (iii) Total production in the first 7 years:
Using $S_7 = \frac{7}{2}(a_1 + a_7)$:
$$S_7 = \frac{7}{2}(550 + 700) = \frac{7}{2}\times1250 = 7\times625 = \boxed{4375}\text{ sets}.$$\
8. All required quantities are obtained.
Correct Answer: (i) 550 sets, (ii) 775 sets, (iii) 4375 sets