If $$a_1, a_2, a_3, \ldots, a_n, \ldots$$ 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 :

Given that the sequence $$ a_1, a_2, a_3, \ldots $$ is an arithmetic progression (A.P.), let the first term be $$ a $$ and the common difference be $$ d $$. The general term of the A.P. is $$ a_n = a + (n-1)d $$.

The given condition is $$ a_4 - a_7 + a_{10} = m $$. Express each term:

• $$ a_4 = a + (4-1)d = a + 3d $$
• $$ a_7 = a + (7-1)d = a + 6d $$
• $$ a_{10} = a + (10-1)d = a + 9d $$

Substitute these into the equation:

$$ (a + 3d) - (a + 6d) + (a + 9d) = m $$

Simplify step by step:

$$ a + 3d - a - 6d + a + 9d = m $$ $$ (a - a + a) + (3d - 6d + 9d) = m $$ $$ a + (3 - 6 + 9)d = m $$ $$ a + 6d = m \quad \text{(Equation 1)} $$

The sum of the first $$ n $$ terms of an A.P. is given by $$ S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] $$. For $$ n = 13 $$:

$$ S_{13} = \frac{13}{2} \left[ 2a + (13-1)d \right] $$ $$ S_{13} = \frac{13}{2} \left[ 2a + 12d \right] $$

Factor out the 2 in the expression inside the brackets:

$$ 2a + 12d = 2(a + 6d) $$

From Equation 1, $$ a + 6d = m $$, so substitute:

$$ 2a + 12d = 2m $$

Now substitute back into the sum formula:

$$ S_{13} = \frac{13}{2} \times 2m $$

Simplify by canceling the 2:

$$ S_{13} = 13 \times m = 13m $$

Hence, the sum of the first 13 terms is $$ 13m $$. Comparing with the options:

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

So, the correct answer is Option C.