Question

What is HM of \[1,\dfrac{1}{2},\dfrac{1}{3}......\dfrac{1}{n}?\]

Answer 1

Hint: Arithmetic progression is the progression with some common difference in the terms. Harmonic progression has the reciprocals of the terms arranged in order.
To find the Harmonic mean first we need to convert it into AP.

Complete step-by-step answer:
The given HP is \[1,\dfrac{1}{2},\dfrac{1}{3}......\dfrac{1}{n}\]
First converting it into AP by taking reciprocals of each term we get
$ 1,2,3,......,n$
We can find out the common difference of the terms and that is
$ 2 - 1 = 3 - 2 = ..... = n - (n - 1) = 1$
The common difference between the terms as found out above is ‘1’.
The mean of this AP is given by the formula:
$
   = \dfrac{1}{n}({t_1} + {t_2} + ..... + {t_n}) \\
   \Rightarrow \dfrac{1}{n}(1 + 2 + 3 + ........ + n) \;
 $
Solving further and substituting the sum of n terms with result we get,
$
   \Rightarrow \dfrac{{n(n + 1)}}{{2n}} \\
   \Rightarrow \dfrac{{(n + 1)}}{2} \;
 $
As the Terms in AP are reciprocal of HP in a similar way the mean of HP is reciprocal of the mean of AP.
We found out above that the mean of Ap is given by
  $ = \dfrac{{(n + 1)}}{2}$
Hence, the HM of the given HP is,
I.e. $ = \dfrac{2}{{n + 1}}$
Hence, the answer is HM $ = \dfrac{2}{{n + 1}}$
So, the correct answer is “$\dfrac{2}{{n + 1}}$”.

Note: The students must be clear about the difference between AP, GP, HP.
In AP, consecutive terms will have common differences.
In GP, consecutive terms will have a common ratio.
And HP is the series formed by taking reciprocals in terms of AP.