The nth term of a sequence is defined as follows. Find the first four terms: (i) `T_(n)=3n+1` (ii) `T_(n)=n^(2)+5`

Explanation

To find the first four terms of the sequences defined by the nth terms \( T_n = 3n + 1 \) and \( T_n = n^2 + 5 \), we will substitute values of \( n \) from 1 to 4 into each formula.

### Part (i): \( T_n = 3n + 1 \)

1. **Calculate \( T_1 \)**:
\[ T_1 = 3(1) + 1 = 3 + 1 = 4 \]

2. **Calculate \( T_2 \)**:
\[ T_2 = 3(2) + 1 = 6 + 1 = 7 \]

3. **Calculate \( T_3 \)**:
\[ T_3 = 3(3) + 1 = 9 + 1 = 10 \]

4. **Calculate \( T_4 \)**:
\[ T_4 = 3(4) + 1 = 12 + 1 = 13 \]

Thus, the first four terms of the sequence \( T_n = 3n + 1 \) are:
\[ 4, 7, 10, 13 \]

### Part (ii): \( T_n = n^2 + 5 \)

1. **Calculate \( T_1 \)**:
\[ T_1 = (1)^2 + 5 = 1 + 5 = 6 \]

2. **Calculate \( T_2 \)**:
\[ T_2 = (2)^2 + 5 = 4 + 5 = 9 \]

3. **Calculate \( T_3 \)**:
\[ T_3 = (3)^2 + 5 = 9 + 5 = 14 \]

4. **Calculate \( T_4 \)**:
\[ T_4 = (4)^2 + 5 = 16 + 5 = 21 \]

Thus, the first four terms of the sequence \( T_n = n^2 + 5 \) are:
\[ 6, 9, 14, 21 \]

### Final Answers:
- For \( T_n = 3n + 1 \): **4, 7, 10, 13**
- For \( T_n = n^2 + 5 \): **6, 9, 14, 21**