If the number of integral terms in the expansion of

Here, $\left(3^{2}+5^{8}\right)$

$T_{r+1}={ }^{n} C_{r}(3)^{\frac{n-r}{2}}(5)^{\frac{r}{8}}$

$\because \frac{n-r}{2}$ and $\frac{r}{8}$ are integer

So, $r$ must be $0,8,16,24 \ldots \ldots$

Now $n=t_{33}=a+(n-1) d=0+32 \times 8=256$

$\Rightarrow n=256$