8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ .
8.3.

It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).

When $n=8$, $T_{r+1}=\mathrm{C}_{8}^{r}\left(\frac{1}{2}\right)^{r} x^{\frac{(16-3 r)}{4}}$, where $r=0$, $1, \cdots, 8$. $r$ must satisfy $4 \mid(16-3 r)$, so $r$ can only be $0,4,8$.