Let $a_1, d_1$ be the first term and common difference of the first A.P. and $a_2, d_2$ be the first term and commmon difference of the second A.P. If $S_1$ and $S_2$ be their sums respectively. Then
$\begin{aligned}
S_1 &=\frac{n}{2}\left[2 a_1+(n-1) d_1\right] \\
S_2 &=\frac{n}{2}\left[2 a_2+(n-1) d_2\right] \\
\therefore \frac{S_1}{S_2} &=\frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}
\end{aligned}$
$=\frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2} \quad \ldots(i)$
$\frac{S_1}{S_2}=\frac{5 n+4}{9 n+6} \quad \ldots(ii)$
We have to find
$\frac{18^{\text {th }} \text { term of first A.P. }}{18^{\text {th }} \text { term of second A.P. }}=\frac{a_1+17 d_1}{a_2+17 d_2}$
Multiplying the numerator and denominator by 2
$\frac{T_{18} \text { of I A.P. }}{T_{18} \text { of II A.P. }}=\frac{2 a_1+34 d_1}{2 a_2+34 d_2} \quad \ldots(iii)$
Comparing (i) \& (iii)
$(n-1) d_1=34 d_1 \quad \therefore n=34+1=35$
Putting $\mathrm{n}=35$ in (ii)
$\begin{aligned}
&\frac{2 a_1+(35-1) d_1}{2 a_2+(35-1) d_2}=\frac{5 \times 35+4}{9 \times 35+6} \\
&\Rightarrow \frac{2 a_1+34 d_1}{2 a_2+34 d_2}=\frac{175+4}{315+6}=\frac{179}{321} \\
&\therefore \quad \frac{T_{18} \text { of } \mathrm{I}^{\mathrm{st}} \mathrm{A} . P .}{T_{18} \text { of } \mathrm{II}^{\text {nd }} \text { A.P. }}=\frac{179}{321}
\end{aligned}$