$T_{K+1}{=}^{1824}{C}_K 5^{\frac{1824-K}{6}} 7^{\frac{K}{9}}$

Where, $K=0,1,2.....,1824$
Now, $\frac{1824-K}{6}$ will be an integer if
$K=0,6,12,18,....,1824$

Also, $\frac{K}{9}$ will be an integer if $K=0,9,18,.....,1818.$
$\Rightarrow$ Both are integers if $\mathrm{K=}0,18,36,.....,1818$

If the number of terms $=n,$

$$\begin{gathered} \Rightarrow 1818=0+\left(n–1\right)18 \\ \Rightarrow n=102 \end{gathered}$$