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

Check the Answer

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

Similar Problems and Solutions

ISI MStat 2015 PSA Problem 17 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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

Check the Answer

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

Similar Problems and Solutions

ISI MStat 2015 PSA Problem 17 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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