# 2021 JMPSC Accuracy Problems/Problem 9

## Problem

If $x_1,x_2,\ldots,x_{10}$ is a strictly increasing sequence of positive integers that satisfies $$\frac{1}{2}<\frac{2}{x_1}<\frac{3}{x_2}< \cdots < \frac{11}{x_{10}},$$ find $x_1+x_2+\cdots+x_{10}$.

## Solution

Say we take $x_1,x_1,x_3,...,x_{10}$ as $4,5,6,...,13$ as an example. The first few terms of the inequality would then be: $$\frac{1}{2}<\frac{2}{4}<\frac{3}{5}<\frac{4}{6}$$ But $\frac{3}{5}<\frac{4}{6}$, reaching a contradiction.

A contradiction will also be reached at some point when $x_1\geq 4$ or when $x_1\leq 2$, so that must mean $x_1=3$.

$\implies 3+4+5+...+12=\frac{10\cdot 15}{2}=\boxed{75}$ $\linebreak$ ~Apple321

## Solution 2

We recall the identity that $\frac{x}{x+1}$ is monotically increasing. Here, we have the same case, $x_1=3$, $x_2=4$, and so on. The answer is $\frac{12(13)}{2}-3=\boxed{75}$

~Geometry285