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. 36
B. 96
C. 76
D. 38

Answer

To solve the problem, we start with the information given about the arithmetic progression (AP) and the specific sums of its terms.

### Step 1: Understand the terms in the AP
Let the first term of the AP be \( a_1 \) and the common difference be \( d \). The terms can be expressed as:
- \( a_1 = a_1 \)
- \( a_4 = a_1 + 3d \)
- \( a_7 = a_1 + 6d \)
- \( a_{10} = a_1 + 9d \)
- \( a_{13} = a_1 + 12d \)
- \( a_{16} = a_1 + 15d \)

### Step 2: Write the equation for the given sum
We are given that:
\[ a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114 \]
Substituting the expressions for the terms, we get:
\[ a_1 + (a_1 + 3d) + (a_1 + 6d) + (a_1 + 9d) + (a_1 + 12d) + (a_1 + 15d) = 114 \]

### Step 3: Simplify the equation
Combining like terms, we have:
\[ 6a_1 + (3d + 6d + 9d + 12d + 15d) = 114 \]
The sum of the coefficients of \( d \) is:
\[ 3 + 6 + 9 + 12 + 15 = 45 \]
Thus, we can rewrite the equation as:
\[ 6a_1 + 45d = 114 \]

### Step 4: Solve for \( a_1 + a_4 \) and \( a_{13} + a_{16} \)
We can express \( a_1 + a_{16} \) and \( a_4 + a_{13} \):
\[ a_1 + a_{16} = a_1 + (a_1 + 15d) = 2a_1 + 15d \]
\[ a_4 + a_{13} = (a_1 + 3d) + (a_1 + 12d) = 2a_1 + 15d \]
This shows that:
\[ a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} \]

### Step 5: Relate the sums
Since \( a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114 \) can be grouped as:
\[ (a_1 + a_{16}) + (a_4 + a_{13}) + (a_7 + a_{10}) = 114 \]
This can be rewritten as:
\[ 3(2a_1 + 15d) = 114 \]
Dividing both sides by 3 gives:
\[ 2a_1 + 15d = 38 \]

### Step 6: Find \( a_1 + a_5 + a_{12} + a_{16} \)
Now we need to find:
\[ a_1 + a_5 + a_{12} + a_{16} \]
Expressing \( a_5 \) and \( a_{12} \):
- \( a_5 = a_1 + 4d \)
- \( a_{12} = a_1 + 11d \)

Thus, we have:
\[ a_1 + a_5 + a_{12} + a_{16} = a_1 + (a_1 + 4d) + (a_1 + 11d) + (a_1 + 15d) \]
Combining these terms gives:
\[ 4a_1 + (4d + 11d + 15d) = 4a_1 + 30d \]

### Step 7: Substitute \( 2a_1 + 15d \)
We know from the earlier step that \( 2a_1 + 15d = 38 \). Therefore:
\[ 4a_1 + 30d = 2(2a_1 + 15d) = 2 \times 38 = 76 \]

### Final Answer
Thus, the value of \( a_1 + a_5 + a_{12} + a_{16} \) is:
\[ \boxed{76} \]