$p, x_1, x_2 \ldots, x_n$ and $q, y_1, y_2, \ldots, y_{\mathrm{n}}$ are two arithmetic progressions with…
$p, x_1, x_2 \ldots, x_n$ and $q, y_1, y_2, \ldots, y_{\mathrm{n}}$ are two arithmetic progressions with common differences a and $\mathrm{b}$ respectively. If $\alpha$ and $\beta$ are the arithmetic means of $x_1, x_2, \ldots x_n$, and $y_1, y_2, \ldots, y_n$ respectively. Then the locus of $P(\alpha, \beta)$ is
$\mathrm{a}(x-p)=\mathrm{b}(y-q)$
$\mathrm{b}(x-p)=\mathrm{a}(y-q)$
$\alpha(x-p)=\beta(y-q)$
$p(x-\alpha)=q(y-\beta)$
Solution
It is given that $p, x_1, x_2, x_3 \ldots x_n$ and $q, y_1, y_2, y_3 \ldots y_n$ are in A.P. whose common difference are $a$ and $b$ respectively.
$
\begin{aligned}
\therefore \quad x_1 & =p+a, x_n=p+n a \\
y_1 & =q+b, y_n=q+n b
\end{aligned}
$
Also, given $\alpha$ is A.M. of $x_1, x_2, x_3 \ldots x_n$
$
\begin{array}{rlrl}
\therefore & \alpha & =\frac{x_1+x_2+x_3+\ldots .+x_n}{n} \\
\Rightarrow & \alpha=\frac{n}{2} \frac{\left(x_1+x_n\right)}{n} \\
\Rightarrow & \alpha=\frac{x_1+x_n}{2}
\end{array}
$
Similarly, $\beta$ is A.M. of $y_1, y_2, y_3 \ldots y_n$. So,
$
\beta=\frac{y_1+y_n}{2}
$
Thus, substituting the value of $x, x_n, y$ and $y_n$, we get
From Eqs. $(i)$ and (ii) eliminate $(n+1)$, we get
$
\begin{aligned}
& \frac{2 \alpha-2 p}{a}=\frac{2 \beta-2 q}{b} \\
\Rightarrow \quad & b(\alpha-p)=a(\beta-q)
\end{aligned}
$
Hence, locus of $p(\alpha, \beta)$ is $b(x-p)=a(y-q)$
