A) \[a=1/2\]and \[n=6\]
B) \[a=1/3\] and \[n=5\]
C) \[a=2\]and \[n=3\]
D) \[a=1/4\]and \[n=1\]
Correct Answer: A
Solution :
Now, \[{{T}_{4}}{{=}^{n}}{{C}_{3}}{{(ax)}^{n-3}}{{\left( \frac{1}{x} \right)}^{3}}=\frac{5}{2}\] (given) \[\Rightarrow \] \[^{n}{{C}_{3}}{{a}^{n-3}}{{x}^{n-6}}=5/2\] \[\Rightarrow \] \[n-6=0\] (\[\because \] RHS is independent of x) \[\Rightarrow \] \[n=6\] On putting n = 6 in Eq. (i), we get \[^{6}{{C}_{3}}\,{{a}^{3}}=\frac{5}{2}\Rightarrow {{a}^{3}}=\frac{1}{8}\Rightarrow a=\frac{1}{2}\]
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