Question

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

Answer 1

Assuming production follows an arithmetic progression (AP), let \(a\) represent the production in the first year and \(d\) represent the annual increase in production. The production in the \(n\)-th year is calculated as \(a_n = a + (n-1)d\).
For the 6th year, production was 800 units, so: \[800 = a + (6 - 1)d \implies 800 = a + 5d. \tag{1}\]
For the 9th year, production was 1130 units, so: \[1130 = a + (9 - 1)d \implies 1130 = a + 8d. \tag{2}\]
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 production in the first year was 250 rollers.

Answer 2

Step 1: Problem Definition:
The company's annual production increases by a constant amount. The following data is provided: Production in year 6 was 800 rollers. Production in year 9 was 1130 rollers. The objective is to determine the yearly increase in production, which corresponds to the common difference of an arithmetic progression (A.P.).

Step 2: Variable Declaration:
Let \(a\) represent the production in the first year, and let \(d\) represent the constant annual increase (the common difference of the A.P.). The production in the \(n\)-th year is calculated using the formula:
\[ a_n = a + (n-1) \cdot d \]

Step 3: Information Application:
Given: \(a_6 = 800\) and \(a_9 = 1130\). Applying the formula yields two equations:
1. For year 6: \(a + (6-1)d = a + 5d = 800 \quad \text{(Equation 1)}\)
2. For year 9: \(a + (9-1)d = a + 8d = 1130 \quad \text{(Equation 2)}\)

Step 4: Equation System Solution:
The system of equations is:
1. \( a + 5d = 800 \)
2. \( a + 8d = 1130 \)
Subtracting Equation 1 from Equation 2 to eliminate \(a\):
\[ (a + 8d) - (a + 5d) = 1130 - 800 \]
\[ 3d = 330 \]
\[ d = \frac{330}{3} = 110 \]

Step 5: Final Result:
The annual increase in the company's production is 110 rollers.

Answer 3

Step 1: Problem Definition: Determine the company's production in the 8th year and the total production over the initial 6 years.

Step 2: Given Data: Annual production increases uniformly by 110 rollers (\( d = 110 \)). Production in the 6th year is 800 rollers. This defines an arithmetic progression (A.P.) with:
- \( a_6 = a + 5d = 800 \)
- \( d = 110 \)

Step 3: Calculate 8th Year Production: The nth term of an A.P. is given by \(a_n = a + (n-1) \cdot d\). For the 8th year (\( n = 8 \)), \(a_8 = a + 7d\).
From \( a_6 = a + 5d = 800 \), substitute \( d = 110 \):
\( a + 5 \times 110 = 800 \)
\( a + 550 = 800 \)
\( a = 800 - 550 = 250 \)
Now, calculate \( a_8 \) using \( a = 250 \) and \( d = 110 \):
\( a_8 = 250 + 7 \times 110 = 250 + 770 = 1020 \)
The production in the 8th year is 1020 rollers.

Step 4: Calculate Total Production (First 6 Years): The sum of the first \(n\) terms of an A.P. is \(S_n = \frac{n}{2} \times (2a + (n-1) \cdot d)\). For the first 6 years (\(n = 6\)):
\(S_6 = \frac{6}{2} \times (2 \times 250 + (6-1) \cdot 110)\)
\(S_6 = 3 \times (500 + 5 \times 110)\)
\(S_6 = 3 \times (500 + 550)\)
\(S_6 = 3 \times 1050 = 3150 \)
The total production in the first 6 years is 3150 rollers.

Step 5: Summary of Results:
1. The company's production in the 8th year is 1020 rollers.
2. The company's total production in the first 6 years is 3150 rollers.