When a train approaches a stationary observer, the apparent frequency of the whistle is n’ and when the same train recedes away from the observer, the apparent frequency is n”. Then, the apparent frequency n when the observer moves with the train is-
A. $n = \sqrt {n'n''} $
B. $n = \dfrac{{n' + n''}}{2}$
C. $n = \dfrac{{2n'n''}}{{n' - n''}}$
D. $n = \dfrac{{2n'n''}}{{n' + n''}}$

Hint: We will consider two cases in this question and solve them respectively with different figures for the same. We will use the formula for apparent frequency in both the cases and then add them after simplifying as per asked by the question.
Formula used: $f' = \dfrac{{\left( {v + {v_0}} \right)}}{{\left( {v - {v_s}} \right)}}f$

Complete Step-by-Step solution:
The frequency given by the question is $f = n$.
Case 1-
As we know, the formula for apparent frequency is $f' = \dfrac{{\left( {v + {v_0}} \right)}}{{\left( {v - {v_s}} \right)}}f$. Where f is the actual frequency, f’ is the observed frequency, v is speed, ${v_0}$ is velocity of the observer and ${v_s}$ is the velocity of the source.
Here, $
  f' = n' \\
  f = n \\
  {v_0} = 0 \\
$
Thus, the equation for n’ will be-
$
   \Rightarrow n' = n\left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right) \\
    \\
   \Rightarrow n' = n\left( {\dfrac{{v + 0}}{{v - {v_s}}}} \right) \\
    \\
   \Rightarrow n' = n\left( {\dfrac{v}{{v - {v_s}}}} \right) \\
$
Let the above equation be equation 1-
$ \Rightarrow n' = n\left( {\dfrac{v}{{v - {v_s}}}} \right)$ (equation 1)
Case 2-
Here, $
  f' = n'' \\
  f = n \\
  {v_0} = 0 \\
$
Since the source is going away from the observer, the signs will be inverted.
$
   \Rightarrow n'' = n\left( {\dfrac{{v - {v_0}}}{{v + {v_s}}}} \right) \\
    \\
   \Rightarrow n'' = n\left( {\dfrac{{v - 0}}{{v + {v_s}}}} \right) \\
    \\
   \Rightarrow n'' = n\left( {\dfrac{v}{{v + {v_s}}}} \right) \\
$
Let the above equation be equation 2-
$ \Rightarrow n'' = n\left( {\dfrac{v}{{v + {v_s}}}} \right)$ (equation 2)
Now, in equation 1-
$
   \Rightarrow n' = n\left( {\dfrac{v}{{v - {v_s}}}} \right) \\
    \\
   \Rightarrow \dfrac{{n'}}{n} = \left( {\dfrac{v}{{v - {v_s}}}} \right) \\
$
Inverting the above equation, we get-
$ \Rightarrow \dfrac{n}{{n'}} = \left( {\dfrac{{v - {v_s}}}{v}} \right)$
Now, in equation -
$
   \Rightarrow n'' = n\left( {\dfrac{v}{{v + {v_s}}}} \right) \\
    \\
   \Rightarrow \dfrac{{n''}}{n} = \left( {\dfrac{v}{{v + {v_s}}}} \right) \\
$
Inverting the above equation, we get-
$ \Rightarrow \dfrac{n}{{n''}} = \left( {\dfrac{{v + {v_s}}}{v}} \right)$
Adding both the inverted equations, we get-
$
   \Rightarrow n\left( {\dfrac{1}{{n'}} + \dfrac{1}{{n''}}} \right) = \dfrac{1}{v}\left( {v - {v_s} + v + {v_s}} \right) \\
    \\
   \Rightarrow n\left( {\dfrac{1}{{n'}} + \dfrac{1}{{n''}}} \right) = \dfrac{{2v}}{v} \\
    \\
   \Rightarrow n\left( {\dfrac{1}{{n'}} + \dfrac{1}{{n''}}} \right) = 2 \\
    \\
   \Rightarrow \left( {\dfrac{1}{{n'}} + \dfrac{1}{{n''}}} \right) = \dfrac{2}{n} \\
    \\
   \Rightarrow \dfrac{2}{n} = \dfrac{{n'' + n'}}{{n' \times n''}} \\
    \\
   \Rightarrow n = \dfrac{{2n'n''}}{{n'' + n'}} \\
$
Hence, option D is the correct option.

Note: Real frequency is exactly what it sounds like: it is a wave 's true frequency, whatever external influences might be. An observer's position has no effect on the actual frequency of a wave. On the other hand, apparent frequency is the frequency which an outside observer perceives. It may match the actual frequency, or not.