Given that `a_(4)+a_(8)+a_(12)+a_(16)=224`, the sum of the first nineteen terms of the arithmetic progression
`a_(1),a_(2),a_(3),….` is equal to

Options

A. 1540
B. 1064
C. 3125
D. 1980

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the terms of the arithmetic progression (AP)
In an arithmetic progression, the \( n \)-th term can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the specific terms given in the problem
We need to find \( a_4, a_8, a_{12}, \) and \( a_{16} \):
- \( a_4 = a + 3d \)
- \( a_8 = a + 7d \)
- \( a_{12} = a + 11d \)
- \( a_{16} = a + 15d \)

### Step 3: Set up the equation based on the given information
We know that:
\[ a_4 + a_8 + a_{12} + a_{16} = 224 \]
Substituting the expressions we derived:
\[ (a + 3d) + (a + 7d) + (a + 11d) + (a + 15d) = 224 \]

### Step 4: Simplify the equation
Combining like terms:
\[ 4a + (3d + 7d + 11d + 15d) = 224 \]
This simplifies to:
\[ 4a + 36d = 224 \]

### Step 5: Divide the entire equation by 4
To simplify further, divide the equation by 4:
\[ a + 9d = 56 \]

### Step 6: Find the sum of the first 19 terms of the AP
The formula for the sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 19 \):
\[ S_{19} = \frac{19}{2} \times (2a + 18d) \]

### Step 7: Substitute \( 2a + 18d \)
From the equation \( a + 9d = 56 \), we can express \( 2a + 18d \) as follows:
\[ 2a + 18d = 2(a + 9d) = 2 \times 56 = 112 \]

### Step 8: Calculate \( S_{19} \)
Now substituting back into the sum formula:
\[ S_{19} = \frac{19}{2} \times 112 \]
Calculating this gives:
\[ S_{19} = 19 \times 56 = 1064 \]

### Final Answer
Thus, the sum of the first 19 terms of the arithmetic progression is:
\[ \boxed{1064} \]