Let ${a_n}$, $n \geqslant 1$, be an arithmetic progression with first term 2 and common difference 4. Let ${M_n}$ be the average of the first $n$ terms, Then the sum $\sum\limits_{n = 1}^{10} {{M_n}} $ is
A) 110
B) 335
C) 770
D) 1100

Hint:
Here, we will find the value of ${M_n}$ using the formula of average and sum of $n$ terms of an AP. Then, substituting the value of ${M_n}$ in $\sum\limits_{n = 1}^{10} {{M_n}} $ and using the general formula, we will be able to find the required answer.

Formula Used:
We will use the following formulas:
1) Average $ = $ Sum of observations $ \div $ Total number of observations
2) In an arithmetic progression, the sum of $n$ terms, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ and $d$ are the first term and the common difference respectively.
3) The general formula of sum of $n$ terms is $\dfrac{{n\left( {n + 1} \right)}}{2}$

Complete step by step solution:
According to the question,
The first term of this AP, $a = 2$
And, the common difference, $d = 4$
Hence, this Arithmetic progression can be written as: $2,6,10,...{a_n}$
Now, it is given that ${M_n}$ is the average of the first $n$ terms.
We know that,
Average $ = $ Sum of observations $ \div $ Total number of observations
Now, in an arithmetic progression, the sum of $n$ terms, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
And, the total number of observations $ = n$
Therefore we can write the average ${M_n}$ of the first $n$ terms as:
${M_n} = \dfrac{{{S_n}}}{n} = \dfrac{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{n}$
Here, substituting the known values, we get,
$ \Rightarrow {M_n} = \dfrac{{n\left[ {2\left( 2 \right) + \left( {n - 1} \right)\left( 4 \right)} \right]}}{{2n}}$
Cancelling out the same terms from the numerator and the denominator and solving further, we get,
$ \Rightarrow {M_n} = \dfrac{{4 + 4n - 4}}{2} = \dfrac{{4n}}{2} = 2n$
Now,
$\sum\limits_{n = 1}^{10} {{M_n}} = \sum\limits_{n = 1}^{10} {\left( {2n} \right)} $
Here, we will use the general formula of the sum of $n$ terms which is $\dfrac{{n\left( {n + 1} \right)}}{2}$.
But, here are $2n$ terms multiplying the numerator by 2 and substituting $n = 10$, we get,
$\sum\limits_{n = 1}^{10} {{M_n}} = \sum\limits_{n = 1}^{10} {\left( {2n} \right)} = \dfrac{{2\left( {10} \right)\left( {11} \right)}}{2} = 11 \times 10 = 110$
Therefore, the sum $\sum\limits_{n = 1}^{10} {{M_n}} $ is 110.

Hence, option A is the correct answer.

Note:
An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.