Question

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

Answer 1

Assuming production follows an arithmetic progression (AP), where:
\[a = \text{Initial year's production}, \quad d = \text{Annual production increment}.\]
The formula for production in the $n$-th year is:
\[a_n = a + (n-1)d.\]
For the 6th year's production:
\[800 = a + (6 - 1)d \implies 800 = a + 5d. \tag{1}\]
For the 9th year's production:
\[1130 = a + (9 - 1)d \implies 1130 = a + 8d. \tag{2}\]
Subtracting equation (1) from equation (2) yields:
\[1130 - 800 = (a + 8d) - (a + 5d) \implies 330 = 3d \implies d = 110.\]
Substituting $d = 110$ into equation (1):
\[800 = a + 5(110) \implies 800 = a + 550 \implies a = 250.\]
The initial year's production was 250 rollers.

Answer 2

Step 1: Define variables.
Let \(P_n\) be the production of road rollers in year \(n\). Since production increases uniformly, it follows an arithmetic progression: \( P_n = P_1 + (n-1) \times d \), where \(P_1\) is the production in the first year and \(d\) is the constant annual increase.
Step 2: Utilize given data.
We are provided with:- Production in year 6: \(P_6 = 800\). Thus, \( P_1 + (6-1)d = P_1 + 5d = 800 \).
- Production in year 9: \(P_9 = 1130\). Thus, \( P_1 + (9-1)d = P_1 + 8d = 1130 \).
Step 3: Solve the system of equations.
We have the system:1. \( P_1 + 5d = 800 \)2. \( P_1 + 8d = 1130 \)Subtracting equation 1 from equation 2 yields:$$ (P_1 + 8d) - (P_1 + 5d) = 1130 - 800 $$$$ 3d = 330 $$$$ d = 110 $$
Step 4: Conclusion.
The annual increase in production is \(d = 110\) rollers.

Answer 3

(a) Objective: Determine the road roller production in the 8th year.
Step 1: Identify the production formula. The annual production, \(P_n\), in the \(n\)-th year is given by an arithmetic sequence: \( P_n = P_1 + (n-1) \times d \), where \(P_1\) is the first-year production and \(d\) is the annual increase. We are given \(d = 110\) and \(P_6 = 800\).
Step 2: Calculate the first-year production (\(P_1\)). Using the formula for the 6th year: \( P_6 = P_1 + 5d = 800 \). Substituting \(d = 110\): \( P_1 + 5 \times 110 = 800 \), which simplifies to \( P_1 + 550 = 800 \). Solving for \(P_1\) yields \( P_1 = 800 - 550 = 250 \) rollers.
Step 3: Calculate the 8th-year production (\(P_8\)). With \(P_1 = 250\) and \(d = 110\), we use the formula: \( P_8 = P_1 + (8-1) \times d \). Substituting the values: \( P_8 = 250 + 7 \times 110 = 250 + 770 = 1020 \).
Step 4: Result. The production in the 8th year was 1020 rollers.


(b) Objective: Calculate the total road roller production over the first 6 years.
Step 1: Recall the production formula. The production in year \(n\) is \( P_n = P_1 + (n-1) \times d \), with \(P_1 = 250\) and \(d = 110\).
Step 2: Determine production for the first 6 years.
1. \(P_1 = 250\)
2. \(P_2 = 250 + (2-1) \times 110 = 360\)
3. \(P_3 = 250 + (3-1) \times 110 = 470\)
4. \(P_4 = 250 + (4-1) \times 110 = 580\)
5. \(P_5 = 250 + (5-1) \times 110 = 690\)
6. \(P_6 = 250 + (6-1) \times 110 = 800\)
Step 3: Sum the annual productions. The total production is the sum of the first 6 years' outputs: \( \text{Total Production} = P_1 + P_2 + P_3 + P_4 + P_5 + P_6 = 250 + 360 + 470 + 580 + 690 + 800 = 3150 \).
Step 4: Result. The total production over the first 6 years was 3150 rollers.