Let numbers `a_(1),a_(2)…a_(16)` are in AP and `a_(1)+a_(4)+a_(7)+a_(10)+a_(13)+a_(16)=114` then `a_(1)+a_(5)+a_(12)+a_(16)` is equal to

Options

A. 38
B. 98
C. 76
D. 64

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (AP).

### Step 1: Define the terms of the AP
Let the first term of the AP be \( a \) and the common difference be \( d \). The \( n \)-th term of the AP can be expressed as:
\[ a_n = a + (n - 1)d \]

### Step 2: Write the specific terms
We need to find the values of specific terms in the AP:
- \( a_1 = a \)
- \( a_4 = a + 3d \)
- \( a_7 = a + 6d \)
- \( a_{10} = a + 9d \)
- \( a_{13} = a + 12d \)
- \( a_{16} = a + 15d \)

### Step 3: Set up the equation
According to the problem, we have:
\[ a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114 \]
Substituting the expressions for these terms:
\[ a + (a + 3d) + (a + 6d) + (a + 9d) + (a + 12d) + (a + 15d) = 114 \]
This simplifies to:
\[ 6a + (3d + 6d + 9d + 12d + 15d) = 114 \]
\[ 6a + 45d = 114 \]

### Step 4: Simplify the equation
Dividing the entire equation by 3 gives:
\[ 2a + 15d = 38 \quad \text{(Equation 1)} \]

### Step 5: Find the required sum
We need to find the value of:
\[ a_1 + a_5 + a_{12} + a_{16} \]
Substituting the terms:
- \( a_5 = a + 4d \)
- \( a_{12} = a + 11d \)

So, we have:
\[ a_1 + a_5 + a_{12} + a_{16} = a + (a + 4d) + (a + 11d) + (a + 15d) \]
This simplifies to:
\[ 4a + (4d + 11d + 15d) = 4a + 30d \]

### Step 6: Relate to Equation 1
Notice that:
\[ 4a + 30d = 2(2a + 15d) \]
Using Equation 1:
\[ 2a + 15d = 38 \]
Thus,
\[ 4a + 30d = 2 \times 38 = 76 \]

### Conclusion
The value of \( a_1 + a_5 + a_{12} + a_{16} \) is:
\[ \boxed{76} \]