The sum of nineteen terms of an A.P. `a_(1), a_(2),…a_(19)` given that `a_(4) + a_(8) + a_(12) + a_(16) = 224` is

Options

A. 1200
B. 1140
C. 1064
D. none

Answer

To solve the problem, we need to find the sum of the first 19 terms of an arithmetic progression (A.P.) given that \( a_4 + a_8 + a_{12} + a_{16} = 224 \).

### Step-by-Step Solution:

1. **Define the Terms of the A.P.**:
Let the first term of the A.P. be \( a \) and the common difference be \( d \). The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]

2. **Express the Given Terms**:
We can express the specific terms \( a_4, a_8, a_{12}, a_{16} \) using the formula for the \( n \)-th term:
- \( a_4 = a + 3d \)
- \( a_8 = a + 7d \)
- \( a_{12} = a + 11d \)
- \( a_{16} = a + 15d \)

3. **Set Up the Equation**:
According to the problem, we have:
\[ a_4 + a_8 + a_{12} + a_{16} = 224 \]
Substituting the expressions we found:
\[ (a + 3d) + (a + 7d) + (a + 11d) + (a + 15d) = 224 \]

4. **Combine Like Terms**:
Combine the terms on the left side:
\[ 4a + (3d + 7d + 11d + 15d) = 224 \]
Simplifying the coefficients of \( d \):
\[ 4a + 36d = 224 \]

5. **Simplify the Equation**:
Divide the entire equation by 4:
\[ a + 9d = 56 \]

6. **Find the Sum of the First 19 Terms**:
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 19 \):
\[ S_{19} = \frac{19}{2} \times (2a + 18d) \]

7. **Substitute \( a + 9d \)**:
We can express \( 2a + 18d \) in terms of \( a + 9d \):
\[ 2a + 18d = 2(a + 9d) = 2 \times 56 = 112 \]

8. **Calculate the Sum**:
Now substitute back into the sum formula:
\[ S_{19} = \frac{19}{2} \times 112 = 19 \times 56 = 1064 \]

### Final Answer:
The sum of the first 19 terms of the A.P. is \( \boxed{1064} \).