`p,x_(1), x_(2) ,. . . x_(n) and q, y_(1),y_(2), . . . ,y_(n)` are two arithmetic progressions with common differences a and b respectively. If `alpha and beta` are the arithmetic means of `x_(1), x_(2), . . . . x_(n), and y_(1), y_(2), . . . . y_(n)` respectivley . then the locus of `p(alpha, beta)` is

Answer

To solve the problem, we need to find the locus of the point \( P(\alpha, \beta) \) where \( \alpha \) and \( \beta \) are the arithmetic means of two arithmetic progressions (APs). Let's go through the steps systematically.

### Step 1: Define the Arithmetic Progressions
Let the first arithmetic progression be:
- \( x_1 = p \)
- \( x_2 = p + a \)
- \( x_3 = p + 2a \)
- ...
- \( x_n = p + (n-1)a \)

The \( n \)-th term can be expressed as:
\[ x_n = p + (n-1)a \]

For the second arithmetic progression:
- \( y_1 = q \)
- \( y_2 = q + b \)
- \( y_3 = q + 2b \)
- ...
- \( y_n = q + (n-1)b \)

The \( n \)-th term can be expressed as:
\[ y_n = q + (n-1)b \]

### Step 2: Calculate the Arithmetic Means
The arithmetic mean \( \alpha \) of the first AP is given by:
\[ \alpha = \frac{x_1 + x_2 + \ldots + x_n}{n} = \frac{p + (p + a) + (p + 2a) + \ldots + (p + (n-1)a)}{n} \]

The sum of the first AP can be simplified:
\[ \text{Sum} = np + a(0 + 1 + 2 + \ldots + (n-1)) = np + a \cdot \frac{(n-1)n}{2} \]
Thus,
\[ \alpha = \frac{np + a \cdot \frac{(n-1)n}{2}}{n} = p + \frac{a(n-1)}{2} \]

Similarly, the arithmetic mean \( \beta \) of the second AP is:
\[ \beta = \frac{y_1 + y_2 + \ldots + y_n}{n} = \frac{q + (q + b) + (q + 2b) + \ldots + (q + (n-1)b)}{n} \]
The sum of the second AP can be simplified:
\[ \text{Sum} = nq + b(0 + 1 + 2 + \ldots + (n-1)) = nq + b \cdot \frac{(n-1)n}{2} \]
Thus,
\[ \beta = \frac{nq + b \cdot \frac{(n-1)n}{2}}{n} = q + \frac{b(n-1)}{2} \]

### Step 3: Eliminate the Parameter \( n \)
From the expressions for \( \alpha \) and \( \beta \):
\[ \alpha - p = \frac{a(n-1)}{2} \quad \text{(1)} \]
\[ \beta - q = \frac{b(n-1)}{2} \quad \text{(2)} \]

Now, we can divide equation (1) by equation (2):
\[ \frac{\alpha - p}{\beta - q} = \frac{a}{b} \]

### Step 4: Rearranging to Find the Locus
Cross-multiplying gives:
\[ b(\alpha - p) = a(\beta - q) \]
Expanding this:
\[ b\alpha - bp = a\beta - aq \]
Rearranging leads to:
\[ b\alpha - a\beta - bp + aq = 0 \]

### Final Step: Replace \( \alpha \) and \( \beta \)
Replacing \( \alpha \) with \( x \) and \( \beta \) with \( y \) gives us the locus:
\[ bx - ay - bp + aq = 0 \]

### Conclusion
The locus of the point \( P(\alpha, \beta) \) is given by:
\[ bx - ay + (aq - bp) = 0 \]