Solve the following questions on the basis of following functions
`P_(n+1) = P_(n)-P_(n-1)`, P is the term of the sequence and `P_(0) = 0, P_(1) =1`
`Q_(n+1) = Q_(n) +Q_(n+1)`, Q is the term of the sequence and `Q_(0) = 0, Q_(1) = 1`
What is the `12^(th)` term of the series `P_(n+1)` starting from n =0?

Options

A. -1
B. 0
C. 1
D. none of these

Answer

To solve the problem, we need to find the 12th term of the sequence defined by the recurrence relation \( P_{n+1} = P_n - P_{n-1} \) with initial conditions \( P_0 = 0 \) and \( P_1 = 1 \).

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- \( P_0 = 0 \)
- \( P_1 = 1 \)

2. **Calculate subsequent terms using the recurrence relation**:
- **For \( n = 1 \)**:
\[ P_2 = P_1 - P_0 = 1 - 0 = 1 \]
- **For \( n = 2 \)**:
\[ P_3 = P_2 - P_1 = 1 - 1 = 0 \]
- **For \( n = 3 \)**:
\[ P_4 = P_3 - P_2 = 0 - 1 = -1 \]
- **For \( n = 4 \)**:
\[ P_5 = P_4 - P_3 = -1 - 0 = -1 \]
- **For \( n = 5 \)**:
\[ P_6 = P_5 - P_4 = -1 - (-1) = 0 \]
- **For \( n = 6 \)**:
\[ P_7 = P_6 - P_5 = 0 - (-1) = 1 \]
- **For \( n = 7 \)**:
\[ P_8 = P_7 - P_6 = 1 - 0 = 1 \]
- **For \( n = 8 \)**:
\[ P_9 = P_8 - P_7 = 1 - 1 = 0 \]
- **For \( n = 9 \)**:
\[ P_{10} = P_9 - P_8 = 0 - 1 = -1 \]
- **For \( n = 10 \)**:
\[ P_{11} = P_{10} - P_9 = -1 - 0 = -1 \]
- **For \( n = 11 \)**:
\[ P_{12} = P_{11} - P_{10} = -1 - (-1) = 0 \]

3. **Final Result**:
- The 12th term \( P_{12} \) is \( 0 \).

### Conclusion:
The 12th term of the series \( P(n+1) \) starting from \( n = 0 \) is \( \boxed{0} \).