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
But try the problem first…
Answer: is \( 1<x< \frac{3}{2} \)
ISI MStat 2015 PSA Problem 17
Precollege Mathematics
Try with Hints
First hint
Take it easy. Group things up. Use \(\frac{1}{n+k} < \frac{1}{n}\) for all natural (k)
Second Hint
\( \frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000} \)
\( \frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}\)
\( \Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2} \)
Final Step
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 , \( 1<x< \frac{3}{2} \).
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