Let $$n\in\mathbb{N}, n\ne 1$$. After arranging the terms of the expansion of $$\left(x^{\frac{1}{2}}+\frac{1}{2x^{\frac{1}{4}}}\right)^n$$ in decreasing powers of $$x$$, the first three coefficients are in arithmetic progression. Then, the number of terms where $$x$$ appears with an integer power, is:

Write the general term of the expansion $$\left(x^{\frac{1}{2}} + \frac{1}{2x^{\frac{1}{4}}}\right)^n$$:

$$T_{r+1} = \binom{n}{r} \left(x^{\frac{1}{2}}\right)^{n-r} \left(\frac{1}{2}x^{-\frac{1}{4}}\right)^r = \binom{n}{r} \left(\frac{1}{2}\right)^r x^{\frac{2n - 3r}{4}}$$

Identify the coefficients of the first three terms ($$r = 0, 1, 2$$):

• $$C_0 = \binom{n}{0} \left(\frac{1}{2}\right)^0 = 1$$
  
• $$C_1 = \binom{n}{1} \left(\frac{1}{2}\right)^1 = \frac{n}{2}$$
  
• $$C_2 = \binom{n}{2} \left(\frac{1}{2}\right)^2 = \frac{n(n-1)}{8}$$
  

Use the condition that the coefficients are in Arithmetic Progression ($$2C_1 = C_0 + C_2$$):

$$2\left(\frac{n}{2}\right) = 1 + \frac{n(n-1)}{8}$$

$$8n = 8 + n^2 - n$$

$$n^2 - 9n + 8 = 0$$

$$(n - 1)(n - 8) = 0$$

Given that $$n \neq 1$$, we select $$n = 8$$.

Examine the power of $$x$$ for the general term with $$n = 8$$:

$$\text{Power} = \frac{2(8) - 3r}{4} = 4 - \frac{3r}{4}$$

For the power to be an integer, $$\frac{3r}{4}$$ must be an integer, which requires $$r$$ to be a multiple of $$4$$.

Since $$r$$ ranges from $$0$$ to $$8$$ ($$r \in \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$$), the valid values for $$r$$ are:

$$r = 0, 4, 8$$

Thus, there are exactly $$3$$ terms where $$x$$ appears with an integer power.