1. **State the problem:** We are given the $n^\text{th}$ term of a sequence as $$a_n = \frac{2n}{5 - 9n}.$$ We need to find: a) The 10th term, $a_{10}$. b) The limiting value of the sequence as $n \to \infty$. 2. **Formula and rules:** The $n^\text{th}$ term formula is given. To find the 10th term, substitute $n=10$ into the formula. To find the limit as $n \to \infty$, analyze the behavior of the fraction when $n$ becomes very large. 3. **Find the 10th term:** Substitute $n=10$: $$a_{10} = \frac{2 \times 10}{5 - 9 \times 10} = \frac{20}{5 - 90} = \frac{20}{-85}.$$ Simplify the fraction by dividing numerator and denominator by 5: $$a_{10} = \frac{\cancel{20}^4}{\cancel{(-85)}^{-17}} = -\frac{4}{17}.$$ 4. **Find the limit as $n \to \infty$:** Consider: $$\lim_{n \to \infty} \frac{2n}{5 - 9n}.$$ Divide numerator and denominator by $n$: $$\lim_{n \to \infty} \frac{2n/n}{(5/n) - 9n/n} = \lim_{n \to \infty} \frac{2}{\frac{5}{n} - 9}.$$ As $n \to \infty$, $\frac{5}{n} \to 0$, so: $$\lim_{n \to \infty} \frac{2}{0 - 9} = \frac{2}{-9} = -\frac{2}{9}.$$ 5. **Final answers:** a) The 10th term is $-\frac{4}{17}$. b) The limiting value as $n \to \infty$ is $-\frac{2}{9}$.