Question

Arrange the expansion of ${\left(x^{\frac{1}{2}}+\frac{1}{2x^{\frac{1}{4}}}\right)}^n$ 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

• A. 1
• B. 2
• C. 3
• D. more than 3

Answer 1

Answer: (C)

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