If `a_(1), a_(2), a_(3)`,... are in AP such that `a_(1) + a_(7) + a_(16) = 40`, then the sum of the first 15 terms of this AP is

Options

A. 120
B. 150
C. 280
D. 200

Explanation

To solve the problem, we need to find the sum of the first 15 terms of an arithmetic progression (AP) given that \( a_1 + a_7 + a_{16} = 40 \).

### Step-by-Step Solution:

1. **Identify the General Terms of the AP**:
The \( n \)-th term of an AP can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Express the Given Terms**:
- The first term \( a_1 \) is \( a \).
- The seventh term \( a_7 \) is:
\[ a_7 = a + 6d \]
- The sixteenth term \( a_{16} \) is:
\[ a_{16} = a + 15d \]

3. **Set Up the Equation**:
According to the problem, we have:
\[ a_1 + a_7 + a_{16} = 40 \]
Substituting the expressions we found:
\[ a + (a + 6d) + (a + 15d) = 40 \]
Simplifying this gives:
\[ 3a + 21d = 40 \]
This is our first equation.

4. **Find the Sum of the First 15 Terms**:
The sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For the first 15 terms:
\[ S_{15} = \frac{15}{2} \times (2a + 14d) \]

5. **Factor Out Common Terms**:
We can simplify \( S_{15} \):
\[ S_{15} = \frac{15}{2} \times 2(a + 7d) = 15(a + 7d) \]

6. **Express \( a + 7d \) in Terms of the First Equation**:
From our first equation \( 3a + 21d = 40 \), we can divide the entire equation by 3:
\[ a + 7d = \frac{40}{3} \]

7. **Substitute Back into the Sum**:
Now substitute \( a + 7d \) into the sum:
\[ S_{15} = 15 \left(\frac{40}{3}\right) = \frac{600}{3} = 200 \]

### Final Answer:
The sum of the first 15 terms of the AP is \( \boxed{200} \).