21This is based on the fact that the harmonic mean of n integers lies between the smallest and the largest numbers in the list.
\hspace{-16}\mathbb{I}$f $\bf{n=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+........+\frac{1}{2012}}}$. Then $\bf{\lfloor n \rfloor}$ is \\\\\\ Where $\bf{\lfloor x \rfloor = }$Floor Sum.
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2 Answers
21Let H be the HM of 1980,1981,...,2012.
Then the expression is \lfloor H/33\rfloor
As, 2012>H>1980.
We get answer 60.
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