The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

Answer

To solve the problem step by step, we will use the properties of Arithmetic Progression (AP) and the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Terms of AP:**
The nth term of an AP can be expressed as:
\[ A_n = A + (n-1)D \]
where \( A \) is the first term, \( D \) is the common difference, and \( n \) is the term number.

2. **Setting Up the Equations:**
We are given two conditions:
- The 8th term is half of the 2nd term:
\[ A_8 = \frac{1}{2} A_2 \]
- The 11th term exceeds one-third of the 4th term by 1:
\[ A_{11} = \frac{1}{3} A_4 + 1 \]

3. **Expressing the Terms:**
Using the formula for the nth term:
- For the 8th term:
\[ A_8 = A + 7D \]
- For the 2nd term:
\[ A_2 = A + D \]
- For the 11th term:
\[ A_{11} = A + 10D \]
- For the 4th term:
\[ A_4 = A + 3D \]

4. **Substituting into the First Condition:**
Substitute \( A_8 \) and \( A_2 \) into the first condition:
\[ A + 7D = \frac{1}{2}(A + D) \]
Multiplying through by 2 to eliminate the fraction:
\[ 2(A + 7D) = A + D \]
Simplifying:
\[ 2A + 14D = A + D \]
Rearranging gives:
\[ A + 13D = 0 \quad \text{(Equation 1)} \]

5. **Substituting into the Second Condition:**
Substitute \( A_{11} \) and \( A_4 \) into the second condition:
\[ A + 10D = \frac{1}{3}(A + 3D) + 1 \]
Multiplying through by 3:
\[ 3(A + 10D) = A + 3D + 3 \]
Simplifying:
\[ 3A + 30D = A + 3D + 3 \]
Rearranging gives:
\[ 2A + 27D = 3 \quad \text{(Equation 2)} \]

6. **Solving the Equations:**
We have two equations:
- Equation 1: \( A + 13D = 0 \)
- Equation 2: \( 2A + 27D = 3 \)

From Equation 1, we can express \( A \):
\[ A = -13D \]
Substitute \( A \) into Equation 2:
\[ 2(-13D) + 27D = 3 \]
Simplifying:
\[ -26D + 27D = 3 \]
\[ D = 3 \]

7. **Finding A:**
Substitute \( D = 3 \) back into Equation 1:
\[ A + 13(3) = 0 \]
\[ A + 39 = 0 \quad \Rightarrow \quad A = -39 \]

8. **Finding the 15th Term:**
Now we can find the 15th term:
\[ A_{15} = A + 14D \]
Substitute \( A = -39 \) and \( D = 3 \):
\[ A_{15} = -39 + 14 \times 3 \]
\[ A_{15} = -39 + 42 = 3 \]

### Final Answer:
The 15th term of the AP is \( \boxed{3} \).