1. The problem is to find the first four terms of the sequence defined by the formula $a_n = 61 - 6(n - 1)$ for $n = 1, 2, 3, \ldots$. 2. The formula for the $n$-th term of the sequence is given by: $$a_n = 61 - 6(n - 1)$$ This is an arithmetic sequence where the first term is $a_1 = 61$ and the common difference is $-6$. 3. Calculate the first four terms by substituting $n = 1, 2, 3, 4$: - For $n=1$: $$a_1 = 61 - 6(1 - 1) = 61 - 0 = 61$$ - For $n=2$: $$a_2 = 61 - 6(2 - 1) = 61 - 6 = 55$$ - For $n=3$: $$a_3 = 61 - 6(3 - 1) = 61 - 12 = 49$$ - For $n=4$: $$a_4 = 61 - 6(4 - 1) = 61 - 18 = 43$$ 4. Therefore, the first four terms of the sequence are: $$[61, 55, 49, 43]$$ This sequence decreases by 6 each time, starting from 61.