1. The problem is to evaluate the sequence $W_n = \sqrt{3n - 43}$ for specific values of $n$: 5, 6, 7, 8, and 9. 2. The formula given is $W_n = \sqrt{3n - 43}$. This means for each $n$, multiply by 3, subtract 43, then take the square root. 3. Calculate each term step-by-step: - For $W_5$: $$W_5 = \sqrt{3(5) - 43} = \sqrt{15 - 43} = \sqrt{-28}$$ which is not a real number since the square root of a negative number is imaginary. - For $W_6$: $$W_6 = \sqrt{3(6) - 43} = \sqrt{18 - 43} = \sqrt{-25}$$ also imaginary. - For $W_7$: $$W_7 = \sqrt{3(7) - 43} = \sqrt{21 - 43} = \sqrt{-22}$$ imaginary. - For $W_8$: $$W_8 = \sqrt{3(8) - 43} = \sqrt{24 - 43} = \sqrt{-19}$$ imaginary. - For $W_9$: $$W_9 = \sqrt{3(9) - 43} = \sqrt{27 - 43} = \sqrt{-16}$$ imaginary. 4. Since all values inside the square root are negative, none of these $W_n$ values are real numbers. 5. If we consider complex numbers, the results are: - $W_5 = \sqrt{-28} = \sqrt{28}i = 2\sqrt{7}i$ - $W_6 = \sqrt{-25} = 5i$ - $W_7 = \sqrt{-22} = \sqrt{22}i$ - $W_8 = \sqrt{-19} = \sqrt{19}i$ - $W_9 = \sqrt{-16} = 4i$ where $i$ is the imaginary unit. Final answers: $W_5 = 2\sqrt{7}i$, $W_6 = 5i$, $W_7 = \sqrt{22}i$, $W_8 = \sqrt{19}i$, $W_9 = 4i$. This shows the importance of checking the domain inside the square root to determine if the result is real or complex.