If the lines `x=k, k=1,2:…., n` meet the line `y=3x+4` at the points `A_(k)(x_(k), y_(k)), k=1,2,….,n` then the ordinate of the centre of Mean position of the points `A_(k), k=1,2,…,n` is

Options

A. `(n+1)/(2)`
B. `(3n+11)/(2)`
C. `(3(n+1))/(2)`
D. none of these

Answer

To solve the problem, we need to find the ordinate (y-coordinate) of the center of the mean position of the points \( A_k \) where the lines \( x = k \) intersect the line \( y = 3x + 4 \).

### Step-by-Step Solution:

1. **Identify the Points of Intersection**:
The points \( A_k \) are given by the intersection of the vertical lines \( x = k \) (where \( k = 1, 2, \ldots, n \)) with the line \( y = 3x + 4 \).
- For each \( k \), the coordinates of the point \( A_k \) can be expressed as:
\[ A_k = (k, y_k) \quad \text{where } y_k = 3k + 4 \]

2. **Calculate the Ordinates**:
The ordinates \( y_k \) for \( k = 1, 2, \ldots, n \) are:
- \( y_1 = 3(1) + 4 = 7 \)
- \( y_2 = 3(2) + 4 = 10 \)
- \( y_3 = 3(3) + 4 = 13 \)
- ...
- \( y_n = 3(n) + 4 = 3n + 4 \)

3. **Mean of the Ordinates**:
To find the ordinate of the center of the mean position, we need to calculate the average (mean) of the ordinates \( y_1, y_2, \ldots, y_n \):
\[ \text{Mean} = \frac{y_1 + y_2 + \ldots + y_n}{n} \]

4. **Sum of the Ordinates**:
The sum of the ordinates can be calculated as follows:
\[ y_1 + y_2 + \ldots + y_n = 7 + 10 + 13 + \ldots + (3n + 4) \]
This is an arithmetic series where:
- The first term \( a = 7 \)
- The last term \( l = 3n + 4 \)
- The number of terms \( n \)

The sum of an arithmetic series is given by:
\[ S_n = \frac{n}{2} (a + l) \]
Substituting the values:
\[ S_n = \frac{n}{2} (7 + (3n + 4)) = \frac{n}{2} (3n + 11) \]

5. **Calculate the Mean**:
Now, substituting \( S_n \) into the mean formula:
\[ \text{Mean} = \frac{S_n}{n} = \frac{\frac{n}{2} (3n + 11)}{n} = \frac{3n + 11}{2} \]

Thus, the ordinate of the center of the mean position of the points \( A_k \) is:
\[ \frac{3n + 11}{2} \]

### Final Answer:
The ordinate of the center of the mean position of the points \( A_k \) is \( \frac{3n + 11}{2} \).