Question

What is the company’s production in the first year ?

Answer 1

The production forms an arithmetic progression (AP) where:
\[a = \text{Production in the first year}, \quad d = \text{Increase in production every year}.\]
The production in the $n$-th year is given by:
\[a_n = a + (n-1)d.\]
For the 6th year:
\[800 = a + (6 - 1)d \implies 800 = a + 5d. \tag{1}\]
For the 9th year:
\[1130 = a + (9 - 1)d \implies 1130 = a + 8d. \tag{2}\]
Subtract (1) from (2):
\[1130 - 800 = a + 8d - (a + 5d) \implies 330 = 3d \implies d = 110.\]
Substitute $d = 110$  into (1):
\[800 = a + 5(110) \implies 800 = a + 550 \implies a = 250.\]
Correct Answer: Production in the first year = 250 rollers.

Answer 2

From the above calculations, the increase in production every year ($d$) is:
\[Correct Answer: \text{Increase in production every year } = 110 \text{ rollers.}\]

Answer 3

(iii) (a) What was the company's production in the 8th year?
Solution:
For the 8th year:
\[a_8 = a + (8 - 1)d = 250 + 7(110) = 250 + 770 = 1020.\]
Correct Answer: Production in the 8th year = 1020 rollers.

---

(iii) (b) What was the company's total production in the first 6 years?
Solution:
The total production in the first $n$ years is given by:
\[S_n = \frac{n}{2}[2a + (n-1)d].\]
For the first 6 years:
\[S_6 = \frac{6}{2}[2(250) + (6 - 1)(110)] = 3[500 + 5(110)] = 3[500 + 550] = 3(1050) = 3150.\]
Correct Answer: Total production in the first 6 years = 3150 rollers.