If `a_(1), a_(2), a_(3),........, a_(n)`,... are in A.P. such that `a_(4) - a_(7) + a_(10) = m`, then the sum of first 13 terms of this A.P., is:

Options

A. 15 m
B. 10 m
C. 12 m
D. 13m

Answer

To solve the problem, we need to find the sum of the first 13 terms of an arithmetic progression (A.P.) given the condition \( a_4 - a_7 + a_{10} = m \).

### Step-by-Step Solution:

1. **Understanding the A.P. Terms**:
In an arithmetic progression, the \( n \)-th term can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Expressing the Given Terms**:
We can express the terms \( a_4 \), \( a_7 \), and \( a_{10} \) using the formula for the \( n \)-th term:
- \( a_4 = a + 3d \)
- \( a_7 = a + 6d \)
- \( a_{10} = a + 9d \)

3. **Substituting into the Given Equation**:
Substitute these expressions into the equation \( a_4 - a_7 + a_{10} = m \):
\[ (a + 3d) - (a + 6d) + (a + 9d) = m \]

4. **Simplifying the Equation**:
Simplifying the left-hand side:
\[ a + 3d - a - 6d + a + 9d = m \]
This simplifies to:
\[ a + (3d - 6d + 9d) = m \]
\[ a + 6d = m \]

5. **Finding the Sum of the First 13 Terms**:
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For the first 13 terms, we have \( n = 13 \):
\[ S_{13} = \frac{13}{2} \times (2a + 12d) \]

6. **Substituting \( a \) in Terms of \( m \)**:
From the equation \( a + 6d = m \), we can express \( a \) as:
\[ a = m - 6d \]
Now substitute \( a \) into the sum formula:
\[ S_{13} = \frac{13}{2} \times (2(m - 6d) + 12d) \]
Simplifying this:
\[ S_{13} = \frac{13}{2} \times (2m - 12d + 12d) \]
\[ S_{13} = \frac{13}{2} \times 2m \]
\[ S_{13} = 13m \]

### Final Result:
The sum of the first 13 terms of the A.P. is:
\[ \boxed{13m} \]