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Question

$A_{1} (x_{1}, y_{1}), A_{2} (x_{2}, y_{2}), A_{3} (x_{3}, y_{3})$ _____ are $n$ points a plane such that
$A_{1} A_{2}$ is bisected at $G_{1}, G_{2}, A_{2}$ is divided in the ration $1 : 2$ at $G_{2}, G_{2}, A_{4}$ is divided in the ratio $1 : 3$ at $G_{3}$ . The process is continued until all $n$ points are exhausted, then find the co-ordinates of the final point $G_{n}$ .

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Solution

Ans. $(\frac{Σ x_{i}}{n}, \frac{Σ y_{i}}{n})$
$G - 1$ bisects $A - 1 A_{2}$ $∴ G_{1} = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$G_{2}$ divides $G_{1} A_{3}$ in the ratio $1 : 2$
$∴ G_{2} = \frac{1}{3} [2 \frac{x_{1} + x_{3}}{2} + 1 . x_{3}], \frac{1}{3} [2. \frac{y_{1} + y_{3}}{2} + 1. y_{3}]$
$G_{2} = (\frac{Σ x_{i}}{3}, \frac{Σ y_{i}}{3})$
Continuing like this, $G_{n} = (\frac{Σ x_{i}}{n}, \frac{Σ y_{i}}{n})$

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