`7^(th)` term of an A.P. is 40. Then, the sum of first 13 terms is

Options

A. 520
B. 53
C. 2080
D. 1040

Explanation

To find the sum of the first 13 terms of an arithmetic progression (A.P.) given that the 7th term is 40, we can follow these steps:

### Step 1: Understand the formula for the nth term of an A.P.
The nth term (T_n) of an A.P. can be expressed as:
\[ T_n = a + (n - 1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 2: Use the information about the 7th term.
We know that the 7th term is 40:
\[ T_7 = a + (7 - 1)d = a + 6d = 40 \]
This gives us our first equation:
\[ a + 6d = 40 \quad \text{(1)} \]

### Step 3: Use the formula for the sum of the first n terms of an A.P.
The sum of the first n terms (S_n) of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
For the first 13 terms, we substitute \( n = 13 \):
\[ S_{13} = \frac{13}{2} \left(2a + (13 - 1)d\right) = \frac{13}{2} \left(2a + 12d\right) \]

### Step 4: Simplify the sum formula.
We can factor out a 2 from the expression inside the parentheses:
\[ S_{13} = \frac{13}{2} \cdot 2 \left(a + 6d\right) = 13(a + 6d) \]

### Step 5: Substitute the value of \( a + 6d \).
From equation (1), we have \( a + 6d = 40 \). Substituting this into the sum formula:
\[ S_{13} = 13 \times 40 = 520 \]

### Final Answer:
The sum of the first 13 terms of the A.P. is:
\[ S_{13} = 520 \]