$p, x_{1}, x_{2}, \dots \dots x_{n}$ and $q, y_{1}, y_{2} \dots \dots, y_{n}$ at r two arithmetic progression with common difference $α$ and $β$ respectively. If $α$ and $β$ are the arithmetic means of $x_{1}, x_{2}, \dots \dots \dots x_{n}$ and $y_{1}, y_{2}, \dots \dots y_{n}$ repsectively, then the locus of $P (α, β)$ is

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$α (x - p) = β (y - q)$

$p (x - α) = q (y - β)$

$a (x - p) = b (y - q)$

$b (x - p) = a (y - q)$

answer is D.

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Detailed Solution

Given $α = \frac{x_{1} + x_{2} + x_{3} + . . . . . . . . . x_{n}}{n} - (1)$

$T_{n} = a + (n - 1) d$

Now $x_{n} = p + (n + 1 - 1) a = p + na$

$\Rightarrow x_{1} = p + a$

and $β = \frac{y_{1} + y_{2} + y_{3} + . . . . + y_{n}}{n}$

Now $y_{n} = q + nb$

$\Rightarrow y_{1} = q + b$

$from (1) α = \frac{1}{n} (\frac{n}{2} (x_{1} + x_{n})) = \frac{1}{2} (p + a + p + na) = \frac{1}{2} (2 p + a (n + 1)) α = p + \frac{a (n + 1)}{2} \Rightarrow α - p = \frac{a (n + 1)}{2} similarly β = q + \frac{b (n + 1)}{2} β - q = \frac{b (n + 1)}{2} N o w \frac{α - p}{β - q} = \frac{\frac{a (n + 1)}{2}}{\frac{b (n + 1)}{2}} = \frac{a}{b} \Rightarrow b (α - p) = a (β - q) ∴ locus of (α, β) is b (x - p) = a (y - q)$