1. **State the problem:** We are given the $n^{th}$ term of a sequence as $$a_n = \frac{9 - 5n}{2n + 14}$$ We need to: a) Find the value of $n$ when the term is $-\frac{1}{2}$. b) Find the limiting value of the sequence as $n \to \infty$. 2. **Part (a): Find $n$ when $a_n = -\frac{1}{2}$** Set the term equal to $-\frac{1}{2}$: $$\frac{9 - 5n}{2n + 14} = -\frac{1}{2}$$ 3. **Cross multiply to solve for $n$:** $$2(9 - 5n) = -1(2n + 14)$$ $$18 - 10n = -2n - 14$$ 4. **Bring all terms involving $n$ to one side and constants to the other:** $$18 + 14 = -2n + 10n$$ $$32 = 8n$$ 5. **Divide both sides by 8 to isolate $n$:** $$n = \frac{32}{8}$$ $$n = 4$$ 6. **Part (b): Find the limiting value as $n \to \infty$** The term is: $$a_n = \frac{9 - 5n}{2n + 14}$$ For large $n$, the highest degree terms dominate, so: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{-5n + 9}{2n + 14} = \lim_{n \to \infty} \frac{n(-5 + \frac{9}{n})}{n(2 + \frac{14}{n})}$$ 7. **Cancel $n$ in numerator and denominator:** $$= \lim_{n \to \infty} \frac{\cancel{n}(-5 + \frac{9}{n})}{\cancel{n}(2 + \frac{14}{n})} = \lim_{n \to \infty} \frac{-5 + \frac{9}{n}}{2 + \frac{14}{n}}$$ 8. **As $n \to \infty$, $\frac{9}{n} \to 0$ and $\frac{14}{n} \to 0$, so:** $$= \frac{-5 + 0}{2 + 0} = \frac{-5}{2}$$ **Final answers:** a) $n = 4$ b) Limiting value = $-\frac{5}{2}$