Arrange the expansion of left(x^{1/2} + frac{ | vedclass

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(C) The general term of the expansion $\left(x^{1/2} + \frac{1}{2x^{1/4}}\right)^n$ is given by $T_{r+1} = { }^n C_r (x^{1/2})^{n-r} (\frac{1}{2} x^{-

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$3$

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(C) The general term of the expansion $\left(x^{1/2} + \frac{1}{2x^{1/4}}\right)^n$ is given by $T_{r+1} = { }^n C_r (x^{1/2})^{n-r} (\frac{1}{2} x^{-1/4})^r = { }^n C_r \cdot 2^{-r} \cdot x^{\frac{2n-3r}{4}}$.The coefficients of the first three terms are $T_1 = { }^n C_0$,$T_2 = \frac{{ }^n C_1}{2}$,and $T_3 = \frac{{ }^n C_2}{4}$.Given that $T_1, T_2, T_3$ are in arithmetic progression,we have $2T_2 = T_1 + T_3$.$2 \left(\frac{{ }^n C_1}{2}\right) = { }^n C_0 + \frac{{ }^n C_2}{4} \Rightarrow n = 1 + \frac{n(n-1)}{8}$.Multiplying by $8$,we get $8n = 8 + n^2 - n \Rightarrow n^2 - 9n + 8 = 0$.$(n-1)(n-8) = 0$. Since $n$ must be greater than $2$ for three terms to exist,we take $n = 8$.The power of $x$ is $\frac{2n-3r}{4} = \frac{16-3r}{4} = 4 - \frac{3r}{4}$.For the power to be an integer,$3r$ must be divisible by $4$. Since $0 \leq r \leq 8$,the possible values for $r$ are $0, 4, 8$.Thus,there are $3$ terms with integer powers of $x$.

Arrange the expansion of $\left(x^{1/2} + \frac{1}{2x^{1/4}}\right)^n$ in decreasing powers of $x$. Suppose the coefficients of the first three terms form an arithmetic progression. Then,the number of terms in the expansion having integer power of $x$ is

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