A) \[\frac{n-4}{5}\]
B) \[\frac{2(n-4)}{5}\]
C) \[\frac{5}{n-4}\]
D) \[\frac{5}{2(n-4)}\]
Correct Answer: D
Solution :
Option is not correct \[{{T}_{4}}={{\,}^{n}}{{C}_{3}}{{(a)}^{n-3}}{{(-2b)}^{3}}\] and \[{{T}_{5}}={{\,}^{n}}{{C}_{4}}{{(a)}^{n-4}}{{(-2b)}^{4}}\] \[\because \]\[{{T}_{4}}+{{T}_{5}}=0\] (given) \[\therefore \]\[{{\,}^{n}}{{C}_{3}}{{(a)}^{n-3}}{{(-2b)}^{3}}+{{\,}^{n}}{{C}_{4}}{{(a)}^{n-4}}{{(-2b)}^{4}}=0\] \[\Rightarrow \]\[{{(a)}^{n-4}}{{(-2b)}^{3}}[a{{\,}^{n}}\,{{C}_{3}}+{{\,}^{n}}{{C}_{4}}(-2b)]=0\] \[\Rightarrow \] \[\frac{a}{b}=\frac{2.{{\,}^{n}}{{C}_{4}}}{{{\,}^{n}}{{C}_{3}}}\] \[=2.\frac{n(n-1)(n-2)(n-3)}{4.3.2.1}\times \frac{3.2.1}{n(n-1)(n-2)}\] \[=\frac{n-3}{2}\]
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