Let $a_1, a_2$ & $d_1, d_2$ be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have $\frac{\text{sum of n terms of first A.P.}}{\text{sum of n terms of second A.P.}}$ = $\frac{5n+4}{9n+6}$ ⇒ $\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}$ = $\frac{5n+4}{9n+6}$ ⇒ $\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}$ = $\frac{5n+4}{9n+6}$ ... (i) Substituting n = 35 in (i), we get $\frac{2a_1+34d_1}{2a_2+34d_2}$ = $\frac{(5 \times 35)+4}{(9 \times 35)+6}$ ⇒ $\frac{a_1+17d_1}{a_2+17d_2}$ = $\frac{179}{321}$ ∴ $\frac{\text{18th term of first A.P.}}{\text{18th term of second A.P.}}$ = $\frac{a_1+17d_1}{a_2+17d_2}$ = $\frac{179}{321}$