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Q. Arrange the expansion of $\left(x^{\frac{1}{2}}+\frac{1}{2 x^{\frac{1}{4}}}\right)^n$ in decreasing powers of $x$. Suppose the coefficient of the first three terms (taken in that order) form an arithmetic progression. Then the number of terms in the expansion having integral powers of $x$, is

Binomial Theorem

Solution:

$T _{ r +1}={ }^{ n } C _{ r }\left( x ^{\frac{1}{2}}\right)^{ n - r }\left(\frac{1}{2 x ^{\frac{1}{4}}}\right)^{ r }=\frac{{ }^{ n } C _{ r }}{2^{ r }} x ^{\frac{2 n -3 r }{4}}$
As, $T_1, T_2, T_3$ (in A.P.)
$\Rightarrow 2\left(\frac{{ }^{ n } C _1}{2}\right)={ }^{ n } C _0+\frac{{ }^{ n } C _2}{2^2} \Rightarrow n =8 $
$\therefore \frac{16-3 r }{4}=\text { integer, when } r =0,4,8$