A) 18
B) 16
C) 12
D) - 10
Correct Answer: D
Solution :
\[{{T}_{3}}={{\,}^{n}}{{C}_{2}}{{(x)}^{n-2}}{{\left( -\frac{1}{2x} \right)}^{2}}\] and \[{{T}_{4}}={{\,}^{n}}{{C}_{3}}{{(x)}^{n-3}}{{\left( -\frac{1}{2x} \right)}^{3}}\] But according to the condition, \[\frac{-\,n(n-1)\times 3\times 2\times 1\times 8}{n(n-1)(n-2)\times 2\times 1\times 4}=\frac{1}{2}\Rightarrow n=-10\]You need to login to perform this action.
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