Let `a_n`, `n ge 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_(n=1)^10 M_n` is

Options

A. 110
B. 335
C. 770
D. 1100

Answer

To solve the problem, we need to find the sum of the averages \( M_n \) of the first \( n \) terms of an arithmetic progression (AP) defined by its first term \( a_1 = 2 \) and common difference \( d = 4 \).

### Step-by-Step Solution:

1. **Identify the first term and common difference**:
- The first term \( a_1 = 2 \)
- The common difference \( d = 4 \)

2. **Find the general term of the arithmetic progression**:
- The \( n \)-th term of an AP is given by:
\[ a_n = a_1 + (n-1)d \]
- Substituting the values:
\[ a_n = 2 + (n-1) \cdot 4 = 2 + 4n - 4 = 4n - 2 \]

3. **Calculate the sum of the first \( n \) terms**:
- The sum of the first \( n \) terms \( S_n \) of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (2a_1 + (n-1)d) \]
- Substituting the values:
\[ S_n = \frac{n}{2} \cdot (2 \cdot 2 + (n-1) \cdot 4) = \frac{n}{2} \cdot (4 + 4n - 4) = \frac{n}{2} \cdot 4n = 2n^2 \]

4. **Find the average \( M_n \)**:
- The average of the first \( n \) terms \( M_n \) is given by:
\[ M_n = \frac{S_n}{n} = \frac{2n^2}{n} = 2n \]

5. **Calculate the sum \( \sum_{n=1}^{10} M_n \)**:
- We need to find:
\[ \sum_{n=1}^{10} M_n = \sum_{n=1}^{10} 2n = 2 \sum_{n=1}^{10} n \]
- The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
- For \( n = 10 \):
\[ \sum_{n=1}^{10} n = \frac{10 \cdot 11}{2} = 55 \]
- Therefore:
\[ \sum_{n=1}^{10} M_n = 2 \cdot 55 = 110 \]

### Final Answer:
The value of \( \sum_{n=1}^{10} M_n \) is \( 110 \).