A) 5th term
B) 6th term
C) 11 th term
D) no term
Correct Answer: D
Solution :
Given expansion is \[{{\left( \frac{2\sqrt{x}}{5}-\frac{1}{2x\sqrt{x}} \right)}^{11}}\] General term \[{{T}_{r+1}}={{(-1)}^{r}}{{\,}^{11}}{{C}_{r}}{{\left( \frac{2\sqrt{x}}{5} \right)}^{11-r}}{{\left( \frac{1}{2{{x}^{3/2}}} \right)}^{r}}\] \[=\frac{{{2}^{11-2r}}}{{{5}^{11-r}}}{{(-1)}^{r}}\,{{\,}^{11}}{{C}_{r}}\,{{x}^{\frac{11-r}{2}-\frac{3r}{2}}}\] For term independent of x put \[\frac{11-r}{2}-\frac{3r}{2}=0\] \[\Rightarrow \] \[\frac{11-4r}{2}=0\] \[\Rightarrow \] \[r=\frac{11}{4}\in N\] \[\therefore \] There is no term which is independent of x.
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