Problem 3. On the ray $\left(O B\right.$, points $A_{1}, A_{2}, \ldots, A_{k}, \ldots$ are considered such that $O B=2 \text{ cm}, O A_{1}=1 \text{ cm}$ and $A_{n} A_{n+1}=\frac{1}{2^{n}} \text{ cm}$, for any non-zero natural number $n$.

a) What is the greatest possible length of the segment $A_{1} A_{10}$? What is the smallest?

b) Show that, regardless of their arrangement (under the conditions of the problem), all points $A_{n}\left(n \in \mathbb{N}^{*}\right)$ belong to the segment $(O B)$.
## Solution and Scoring

a) $\left(A_{1} A_{10}\right)_{\max }=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{9}}=1-\frac{1}{2^{9}}=\frac{511}{512} \text{ cm}$ $2 \text{p}$

$\left(A_{1} A_{10}\right)_{\min }=\frac{1}{2}-\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{9}}\right)=\frac{1}{2^{9}}=\frac{1}{512} \text{ cm}$ $2 \text{p}$

b) Point $A_{1}$ is the midpoint of segment $(O B)$. It needs to be shown that, regardless of the arrangement of the points (under the conditions of the problem), the maximum length of segment $A_{1} A_{n}$ is less than $1 \text{ cm}$ (where $n$ is arbitrary) $2 \text{p}$

$A_{1} A_{n}=\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{n-1}}=1-\frac{1}{2^{n-1}}<1 \text{ cm}$ $1 \text{p}$