Each term of a sequence is the sum of its preceding two terms from the third term onwards. The second term of the sequence is -1 and the 10th term is 29. The first term is `"_________"`

Answer

To solve the problem, we need to find the first term of the sequence, denoted as \( A_1 \). We know:

- \( A_2 = -1 \)
- \( A_{10} = 29 \)

From the problem statement, we know that each term from the third term onwards is the sum of the two preceding terms. Therefore, we can express the terms of the sequence as follows:

1. **Finding the terms of the sequence:**
- \( A_3 = A_1 + A_2 = A_1 - 1 \)
- \( A_4 = A_3 + A_2 = (A_1 - 1) + (-1) = A_1 - 2 \)
- \( A_5 = A_4 + A_3 = (A_1 - 2) + (A_1 - 1) = 2A_1 - 3 \)
- \( A_6 = A_5 + A_4 = (2A_1 - 3) + (A_1 - 2) = 3A_1 - 5 \)
- \( A_7 = A_6 + A_5 = (3A_1 - 5) + (2A_1 - 3) = 5A_1 - 8 \)
- \( A_8 = A_7 + A_6 = (5A_1 - 8) + (3A_1 - 5) = 8A_1 - 13 \)
- \( A_9 = A_8 + A_7 = (8A_1 - 13) + (5A_1 - 8) = 13A_1 - 21 \)
- \( A_{10} = A_9 + A_8 = (13A_1 - 21) + (8A_1 - 13) = 21A_1 - 34 \)

2. **Setting up the equation:**
Since we know \( A_{10} = 29 \), we can set up the equation:
\[ 21A_1 - 34 = 29 \]

3. **Solving for \( A_1 \):**
\[ 21A_1 = 29 + 34 \]
\[ 21A_1 = 63 \]
\[ A_1 = \frac{63}{21} = 3 \]

Thus, the first term \( A_1 \) is **3**.