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# The H.M. of the reciprocal of first n natural numbers isA) $\dfrac{{n + 1}}{2}$B) $\dfrac{n}{{\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + .... + \dfrac{1}{n}} \right)}}$C) $\dfrac{2}{{n + 1}}$D) None of these

Last updated date: 10th Aug 2024
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Hint: The reciprocal of first n natural numbers is given by
$\Rightarrow$Reciprocal of first n natural numbers$= 1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},........,\dfrac{1}{n}$
Now, the harmonic mean is the reciprocal of the arithmetic mean. The Harmonic Mean can be given by the formula
$\Rightarrow$Harmonic Mean (H.M.)$= \dfrac{n}{{\sum\limits_{i = 1}^n i }}$
Using this formula, we will get our answer.

Complete step by step solution:
In this question, we are asked to find the Harmonic Mean (H.M.) of the reciprocal of first n natural numbers.
By definition harmonic mean is the reciprocal of arithmetic mean.
Suppose we are given some data ${X_1},{X_2},{X_3},.......,{X_n}$ then the harmonic mean of this ungrouped data is given by
$\Rightarrow$Harmonic Mean (H.M.)$= \dfrac{n}{{\sum\limits_{n = 1}^n {{X_n}} }}$
So, here we have to find the Harmonic Mean of the reciprocal of first n natural numbers. Now, first n natural numbers are given by
$\Rightarrow$First n natural numbers$= 1,2,3,.......,n$
And the reciprocal of first n natural numbers is given by
$\Rightarrow$Reciprocal of first n natural numbers$= 1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},........,\dfrac{1}{n}$
Now, the sum of $1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},........,\dfrac{1}{n}$ is given by
$1 + \dfrac{1}{2} + \dfrac{1}{3} + ..... + \dfrac{1}{n} = \dfrac{{n\left( {n + 1} \right)}}{2}$
Therefore, we get
$\Rightarrow$Harmonic Mean (H.M.)$= \dfrac{n}{{\sum\limits_{i = 1}^n {\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ...... + \dfrac{1}{n}} \right)} }}$
$\Rightarrow$Harmonic Mean (H.M.)$= \dfrac{n}{{\dfrac{{n\left( {n + 1} \right)}}{2}}}$
$\Rightarrow$Harmonic Mean (H.M.)$= \dfrac{2}{{n + 1}}$
Hence, the Harmonic Mean of the reciprocal of first n natural numbers is $\dfrac{2}{{n + 1}}$. So, the correct option is option (C).

Note:
> While finding the harmonic mean of all constant numbers (c), the H.M. will be also c.
> As compared to arithmetic mean and geometric mean, the Harmonic mean has the least value, that is
$\text{AM > GM > HM}$