Two APs have sums of their first n terms in the ratio (5n + 4) : (9n + 6) for every n…

Step 1. Cancel $\dfrac{n}{2}$. $\dfrac{2a + (n - 1)d}{2a' + (n - 1)d'} = \dfrac{5n + 4}{9n + 6}$.

Step 2. Halve top and bottom. $\dfrac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'} = \dfrac{5n + 4}{9n + 6}$.

Step 3. Choose $n$. The $18$th term needs $\dfrac{n - 1}{2} = 17$, so $n = 35$.

Step 4. The ratio. $\dfrac{a_{18}}{a'_{18}} = \dfrac{5 \times 35 + 4}{9 \times 35 + 6} = \dfrac{179}{321}$.