Because \begin{align*} &\Big(\sqrt{x}+\frac{1}{2\sqrt[4]{x}}\Big)^n\\ =&\Big(x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{1}{4}}\Big)^n\\ =&\sum_{i=0}^nC_n^ix^{\frac{n}{2}}\cdot \frac{1}{2^{n-i}}x^{-\frac{n-i}{4}}\\ =&\sum_{i=0}^n\frac{1}{2^{n-i}}C_n^ix^{\frac{n+i}{4}} \end{align*} Therefore the bigger the value $i$ is, the bigger the $x$'s power is. This means the three coefficients are $$\frac{1}{2^{n-n}}C_n^n,\quad\frac{1}{2^{n-(n-1)}}C_n^{n-1}, \quad\frac{1}{2^{n-(n-2)}}C_n^{n-2}$$ \indent Or $$1, \quad \frac{n}{2}, \quad \frac{1}{4}\times\frac{n(n-1)}{2}$$ \indent As they form an arithmetic sequence, it must hold that $$2\times \Big(\frac{n}{2}\Big) = 1+ \frac{n(n-1)}{8}\implies n = 1, 8$$ Clearly, $x>1$, hence $n=8$ is the only possibility. When $n=8$, every term will be in the following form $$\frac{1}{2^{8-i}}C_8^ix^{\frac{8+i}{4}}\quad, i = 0, 1, 2, \cdots 8$$ Hence, the power of $x$ will be an integer if and only if $i=0,4,8$ which means the answer is $\boxed{3}$.