For the arithmetic progression \[a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),....,\left( {a + 2nd} \right)\] the mean deviation from the mean is
A. \[\;\left[ {\dfrac{{n\left( {n + 1} \right)d}}{{\left( {2n - 1} \right)}}} \right]\]
B. \[\left[ {\dfrac{{n\left( {n + 1} \right)d}}{{\left( {2n + 1} \right)}}} \right]\]
C. \[\left[ {\dfrac{{n\left( {n - 1} \right)d}}{{\left( {2n + 1} \right)}}} \right]\]
D. \[\left[ {\dfrac{{\left( {n + 1} \right)d}}{2}} \right]\]
E. \[\left[ {\dfrac{{n\left( {n - 1} \right)d}}{{\left( {2n - 1} \right)}}} \right]\]

Hint: In this question, we have been given the arithmetic progression sequence such as \[a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),....,\left( {a + 2nd} \right)\]. We need to find the mean for this sequence first. Also, using this mean, we will calculate the mean deviation using the mentioned formulae.

Formula used: The mean of Arithmetic Progression (AP) is \[\bar x = \dfrac{{Sum{\text{ of all the terms of a sequence}}}}{{Total{\text{ }}number{\text{ }}of{\text{ terms}}}}\]
Also, the mean deviation can be calculated from the mean as \[\left( \sigma \right) = \dfrac{1}{N}\sum {\left| {{x_i} - \bar x} \right|} \]
Here, \[N\] is the total number of terms in a sequence.
\[{x_i}\] is data values in a given sequence.
\[\bar x\] is the mean.

Complete step-by-step answer:
We know that the given AP sequence is \[a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),....,\left( {a + 2nd} \right)\]
We know that the mean \[\bar x = \dfrac{{Sum{\text{ of all the terms of a sequence}}}}{{Total{\text{ number of terms}}}}\]
So, the mean is calculated by
\[
  \bar x = \dfrac{{a + \left( {a + d} \right) + \left( {a + 2d} \right) + \left( {a + 3d} \right) + .... + \left( {a + 2nd} \right)}}{{2n + 1}} \\
   \Rightarrow \bar x = \dfrac{{\dfrac{{2n + 1}}{2}\left( {a + a + 2nd} \right)}}{{2n + 1}} \\
   \Rightarrow \bar x = \dfrac{1}{{2n + 1}}\left( {\dfrac{{2n + 1}}{2}\left( {a + a + 2nd} \right)} \right) \\
   \Rightarrow \bar x = \dfrac{{\left( {2a + 2nd} \right)}}{2} \\
 \]
By simplifying it further, we get
\[
   \Rightarrow \bar x = \dfrac{{2\left( {a + nd} \right)}}{2} \\
   \Rightarrow \bar x = \left( {a + nd} \right) \\
 \]
Hence, the mean is \[\left( {a + nd} \right)\].
So, the Mean deviation from the mean \[\left( \sigma \right) = \dfrac{1}{N}\sum {\left| {{x_i} - \bar x} \right|} \]
\[
   \Rightarrow \sigma = \dfrac{1}{{2n + 1}}\sum {\left| {{x_i} - \left( {a + nd} \right)} \right|} \\
   \Rightarrow \sigma = \dfrac{1}{{2n + 1}}\sum {\left| {{x_i} - a - nd} \right|} \\
   \Rightarrow \sigma = \dfrac{1}{{2n + 1}}\left( {nd + \left( {n - 1} \right)d + \left( {n - 2} \right)d + ... + d + 0 + d + ... + nd} \right) \\
   \Rightarrow \sigma = \dfrac{{2d}}{{2n + 1}}\left( {n + \left( {n - 1} \right) + \left( {n - 2} \right) + ...1} \right) \\
 \]
By simplifying, we get
\[
   \Rightarrow \sigma = \dfrac{{2d}}{{2n + 1}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \\
   \Rightarrow \sigma = \dfrac{{n\left( {n + 1} \right)d}}{{2n + 1}} \\
 \]
Hence, for the arithmetic progression \[a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),....,\left( {a + 2nd} \right)\] the mean deviation from the mean is \[\left[ {\dfrac{{n\left( {n + 1} \right)d}}{{\left( {2n + 1} \right)}}} \right]\].
Therefore, the correct option is (B).

Additional Information: The concept of mean plays a significant role in mathematics and statistics. The mean, often known as the average, is computed by combining all of the values and dividing the amount by the number of values. The mean deviation is a mathematical metric that computes the average variation from the mean value of a given collection of data. In simple words, the mean deviation indicates how far off the measured values are from the average value. That is, the mean deviation is utilized to compute the mean of the absolute departures from the center point of the data.

Note: Many students make mistakes in the calculation part of the mean deviation. This is the only way through which we can solve this example in an easy manner.