To find the first four terms of the sequence defined by the formula: $a_n = 61 - 8(n - 1), \quad n = 1, 2, 3, \dots$

we'll substitute $n = 1, 2, 3,$ and $4$ into the formula.

1. For $n = 1$: $a_1 = 61 - 8(1 - 1) = 61 - 8 \cdot 0 = 61$

2. For $n = 2$: $a_2 = 61 - 8(2 - 1) = 61 - 8 \cdot 1 = 61 - 8 = 53$

3. For $n = 3$: $a_3 = 61 - 8(3 - 1) = 61 - 8 \cdot 2 = 61 - 16 = 45$

4. For $n = 4$: $a_4 = 61 - 8(4 - 1) = 61 - 8 \cdot 3 = 61 - 24 = 37$

So, the first four terms of the sequence are: $61, 53, 45, 37$

Would you like further details on this? Here are some related questions for deeper understanding:

1. How would you find the 10th term of this sequence?
2. Can you determine if this sequence is arithmetic? If so, what is the common difference?
3. What is the general approach to find any term in an arithmetic sequence?
4. How would you find the sum of the first four terms of this sequence?
5. Can you derive a formula for the sum of the first $n$ terms of this sequence?

Tip: In arithmetic sequences, each term differs by a constant amount, known as the common difference.