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 that \( a_4 - a_7 + a_{10} = m \).

### Step-by-Step Solution:

1. **Understanding the Terms of A.P.**:
The general term of an A.P. 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 need to express \( a_4 \), \( a_7 \), and \( a_{10} \):
- \( a_4 = a + (4 - 1)d = a + 3d \)
- \( a_7 = a + (7 - 1)d = a + 6d \)
- \( a_{10} = a + (10 - 1)d = a + 9d \)

3. **Substituting into the Given Equation**:
Now 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:
\[ a + 3d - a - 6d + a + 9d = m \]
Combining like terms:
\[ (a - a + a) + (3d - 6d + 9d) = m \]
This simplifies to:
\[ a + 6d = m \]

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:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For \( n = 13 \):
\[ S_{13} = \frac{13}{2} \times (2a + 12d) \]

6. **Factoring Out Common Terms**:
We can factor out the 2:
\[ S_{13} = \frac{13}{2} \times 2 \times \left(a + 6d\right) = 13 \times (a + 6d) \]

7. **Substituting \( a + 6d \)**:
From our earlier equation, we know that:
\[ a + 6d = m \]
Therefore:
\[ S_{13} = 13m \]

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