If $p, x_{1}, x_{2} . . . x_{i} . . .$ and $q, y_{1}, y_{2} . . . y_{i} . . .$ are in A.P. with common difference $a$ and $b$ respectively, then locus of the centre of mean position of the points $A_{i} (x_{i}, y_{i}), i = 1, 2... n$ is

a x - b y = a q - b p

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b x - a y = a p - b q

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b x - a y = b p - a q

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a x - b y = b q - a p

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Solution

The correct option is B $b x - a y = b p - a q$
[Note. Centre of Mean Position is $(\frac{\sum x_{i}}{n}, \frac{\sum y_{i}}{n})$ ]
Let the coordinates of the centre of mean position of the points
$A_{i} = 1, 2, . . . n$ be $(x, y)$ , then
$x = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}$ , $y = \frac{y_{1} + y_{2} + . . . + y_{n}}{n}$
$\Rightarrow x = \frac{n p + a (1 + 2 + . . . + n)}{n}$ , $y = \frac{n q + b (1 + 2 + . . . + n)}{n}$
$\Rightarrow x = p + \frac{n (n + 1)}{2 n} a$ , $y = q + \frac{n (n + 1)}{2 n} b$
$\Rightarrow 2 \frac{(x - p)}{a} = 2 \frac{(y - q)}{b} \Rightarrow b x - a y = b p - a q .$

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