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# ISI MStat 2015 PSA Problem 17 | Basic Inequality

This is a problem from ISI MStat 2015 PSA Problem 17. First, try the problem yourself, then go through the sequential hints we provide.

## Basic Inequality - ISI MStat Year 2015 PSA Question 17

Let $$X=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}$$. Then,

• x<1
• $$x>\frac{3}{2}$$
• $$1<x< \frac{3}{2}$$
• None of these

### Key Concepts

Basic Inequality

Answer: is $$1<x< \frac{3}{2}$$

ISI MStat 2015 PSA Problem 17

Precollege Mathematics

## Try with Hints

Take it easy. Group things up. Use $$\frac{1}{n+k} < \frac{1}{n}$$ for all natural (k)

$$\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000}$$
$$\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}$$

$$\Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2}$$

Again see that we can write , x=$$(\frac{1}{1001}+\frac{1}{3001}) + (\frac{1}{1002}+\frac{1}{3000})+ \dots + (\frac{1}{2000}+\frac{1}{2002}) + \frac{1}{2001}$$ > $$\frac{2}{2001} +\frac{2}{2001} + \dots + \frac{2}{2001} +\frac{1}{2001} > \frac{2001}{2001}=1$$

Hence , $$1<x< \frac{3}{2}$$.