If `p, x_(1),x_(2)…x_(i),….` and `q, y_(1),y_(2)…,y_(i),….` are in A.P.m with common difference a and b respectively, then the centre of mean position of the points `A_(i)(x_(i),y_(i)), i=1,2….n` lies on the line
Note : Centre of Mean Position `((Sigma xi)/(n),(Sigma yi)/(n))`

Options

A. `ax-by=aq-bp`
B. `bx-ay=ap-bq`
C. `bx-ay=bp-aq`
D. `ax-by=bq-ap`

Answer

To solve the problem, we need to find the center of mean position of the points \( A_i(x_i, y_i) \) where \( x_i \) and \( y_i \) are in arithmetic progression (A.P.) with common differences \( a \) and \( b \) respectively. Let's break down the steps:

### Step 1: Define the sequences
Given that \( p, x_1, x_2, \ldots, x_n \) are in A.P. with common difference \( a \), we can express the terms as:
- \( x_1 = p + a \)
- \( x_2 = p + 2a \)
- \( x_3 = p + 3a \)
- ...
- \( x_n = p + na \)

Similarly, for \( q, y_1, y_2, \ldots, y_n \) in A.P. with common difference \( b \):
- \( y_1 = q + b \)
- \( y_2 = q + 2b \)
- \( y_3 = q + 3b \)
- ...
- \( y_n = q + nb \)

### Step 2: Calculate the summation of \( x_i \)
The sum of \( x_i \) from \( i = 1 \) to \( n \) can be calculated as follows:
\[ \sum_{i=1}^{n} x_i = \sum_{i=1}^{n} (p + ia) = np + a(1 + 2 + \ldots + n) \]
Using the formula for the sum of the first \( n \) natural numbers:
\[ 1 + 2 + \ldots + n = \frac{n(n+1)}{2} \]
Thus,
\[ \sum_{i=1}^{n} x_i = np + a \cdot \frac{n(n+1)}{2} \]

### Step 3: Calculate the summation of \( y_i \)
Similarly, for \( y_i \):
\[ \sum_{i=1}^{n} y_i = \sum_{i=1}^{n} (q + ib) = nq + b(1 + 2 + \ldots + n) = nq + b \cdot \frac{n(n+1)}{2} \]

### Step 4: Find the center of mean position
The center of mean position \( (X, Y) \) is given by:
\[ X = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{np + a \cdot \frac{n(n+1)}{2}}{n} = p + \frac{a(n+1)}{2} \]
\[ Y = \frac{\sum_{i=1}^{n} y_i}{n} = \frac{nq + b \cdot \frac{n(n+1)}{2}}{n} = q + \frac{b(n+1)}{2} \]

### Step 5: Determine the line equation
To find the line on which the center of mean position lies, we can express it in the form \( ax - by + c = 0 \):
- Substitute \( X \) and \( Y \):
\[ Y - q = \frac{b}{a} \left( X - p \right) \]
Rearranging gives:
\[ aY - bX + (bp - aq) = 0 \]

### Conclusion
The center of mean position of the points \( A_i(x_i, y_i) \) lies on the line given by the equation \( aY - bX + (bp - aq) = 0 \).

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