If `a_1,a_2,a_3,. . . ,a_n. . .` are in A.P. such that `a_4−a_7+a_10=m`, then sum of the first 13 terms of the A.P.is

Options

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

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the terms in A.P.
In an Arithmetic Progression (A.P.), the nth term can be expressed as:
\[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

### Step 2: Write down the given equation
We are given:
\[ a_4 - a_7 + a_{10} = m \]

### Step 3: Substitute the terms using the A.P. formula
Using the formula for the nth term:
- \( a_4 = a_1 + 3d \)
- \( a_7 = a_1 + 6d \)
- \( a_{10} = a_1 + 9d \)

Substituting these into the equation:
\[ (a_1 + 3d) - (a_1 + 6d) + (a_1 + 9d) = m \]

### Step 4: Simplify the equation
Now, simplify the left-hand side:
\[ a_1 + 3d - a_1 - 6d + a_1 + 9d = m \]
This simplifies to:
\[ (3d - 6d + 9d) + a_1 - a_1 = m \]
\[ 6d = m \]

### Step 5: Find the sum of the first 13 terms
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \left(2a_1 + (n-1)d\right) \]
For the first 13 terms, \( n = 13 \):
\[ S_{13} = \frac{13}{2} \left(2a_1 + (13-1)d\right) \]
\[ S_{13} = \frac{13}{2} \left(2a_1 + 12d\right) \]

### Step 6: Factor out the common term
Factor out 2 from the terms inside the bracket:
\[ S_{13} = \frac{13}{2} \cdot 2 \left(a_1 + 6d\right) \]
\[ S_{13} = 13(a_1 + 6d) \]

### Step 7: Substitute \( a_1 + 6d \) with \( m \)
From our earlier step, we know:
\[ 6d = m \implies a_1 + 6d = a_1 + m - a_1 = m \]
Thus, we can substitute:
\[ S_{13} = 13m \]

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