Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 9"
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| + | Say we take <math>x_1,x_1,x_3,...,x_{10}</math> as <math>4,5,6,...,13</math> as an example. The first few terms of the inequality would then be: | ||
| + | <cmath>\frac{1}{2}<\frac{2}{4}<\frac{3}{5}<\frac{4}{6}</cmath> | ||
| + | But <math>\frac{3}{5}<\frac{4}{6}</math>, reaching a contradiction. | ||
| + | |||
| + | A contradiction will also be reached at some point when <math>x_1\geq 4</math> or when <math>x_1\leq 2</math>, so that must mean <math>x_1=3</math>. | ||
| + | |||
| + | <math>\implies 3+4+5+...+12=\frac{10\cdot 15}{2}=\boxed{75}</math> | ||
| + | <math>\linebreak</math> | ||
| + | ~Apple321 | ||
Revision as of 10:12, 11 July 2021
Problem
If 
is a strictly increasing sequence of positive integers that satisfies ![\[\frac{1}{2}<\frac{2}{x_1}<\frac{3}{x_2}< \cdots < \frac{11}{x_{10}},\]](http://latex.artofproblemsolving.com/a/d/1/ad108b31c7b45c0ccab98e7ff977b0ef95f8c89f.png)
find 
.
Solution
Say we take 
as 
as an example. The first few terms of the inequality would then be:
![\[\frac{1}{2}<\frac{2}{4}<\frac{3}{5}<\frac{4}{6}\]](http://latex.artofproblemsolving.com/3/8/8/38840b3754b4082d8437320d392037b6a34dae42.png)
But 
, reaching a contradiction.
A contradiction will also be reached at some point when 
or when 
, so that must mean 
.
~Apple321
