$A_1,A_2,A_3\dots A_n$ are n points whose coordinates are $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)\dots \left(x_n,y_n\right)\cdot A_1{A}_2$ is bisected at  $G_1.G_1{A}_3$ is divided in the ratio 1:2 at  $G_2.G_2{A}_4$is divided in the ratio 1:3 at $G_3$ and so on until all the points are exhausted. Then the coordinates of the final point so obtained is

$$\begin{gathered} \text{ The coordinates of }{G}_1\text{ are }\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\text{ Now, }{G}_2 \text{ divides }{G}_1{A}_3\text{ in the ratio }1:2 \\ \text{ Therefore, the coordinates of }{G}_2\text{ are }\left(\frac{1}{3}\left(\frac{2\left(x_1+x_2\right)}{2}+x_3\right)\right),\frac{1}{3}\left(\frac{2\left(y_1+y_2\right)}{2}+y_3\right) \\ \Rightarrow \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \\ \text{ Again, }{G}_3\text{ divided }{G}_2{A}_4\text{ in the ratio }1:3\text{. Therefore, the coordinates of }{G}_3 are \\ \left(\frac{1}{4}\left(\frac{3\left(x_1+x_2+x_3\right)}{3}+x_4\right)\right),\frac{1}{4}\left(\frac{3\left(y_1+y_2+y_3\right)}{3}+y_4\right) \\ \Rightarrow \left(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4}\right) \\ \text{ Proceeding in this manner, we can show that the coordinates of the final point }G\text{. So, obtained will be } \\ \left(\frac{1}{n}\left(x_1+x_2+\dots +x_n,\frac{1}{n}\left(y_1+y_2+\dots +y_n\right)\right)\right. \end{gathered}$$

$\Rightarrow G_n=\left(\frac{\sum x_1}{n},\frac{\sum y_1}{n}\right)$