8. Expand the binomial $\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right)^{n}$ 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's not difficult to find that the coefficients of the first 3 terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these 3 numbers form an arithmetic sequence, we have $2 \cdot \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get: $n=8$ and $n=1$ (discard).
When $\pi=8$, $T_{r+1}=C_{8}^{r}\left(\frac{1}{2}\right)^{r} x^{\frac{(16-3 r)}{4}}$. Here $r=0,1, \cdots, 8$. $r$ should satisfy $41(16-3 r)$, so $r$ can only be $0,4,8$.