A) \[\frac{5}{n-4}\]
B) \[\frac{1}{5(n-4)}\]
C) \[\frac{n-5}{6}\]
D) \[\frac{n-4}{5}\]
Correct Answer: D
Solution :
\[{{T}_{5}}+{{T}_{6}}=0,\] \[{{\,}^{n}}{{C}_{4}}{{a}^{n-4}}{{(-b)}^{4}}+{{\,}^{n}}{{C}_{5}}{{a}^{n-5}}{{(-b)}^{5}}=0\]\[{{\,}^{n}}{{C}_{4}}{{a}^{n-4}}{{b}^{4}}={{\,}^{n}}{{C}_{5}}{{a}^{n-5}}{{b}^{5}}\] \[{{\,}^{n}}{{C}_{4}}\frac{{{a}^{n}}}{{{a}^{4}}}.{{b}^{4}}={{\,}^{n}}{{C}_{5}}\frac{{{a}^{n}}}{{{a}^{5}}}.{{b}^{5}}\Rightarrow \frac{a}{b}=\frac{{{\,}^{n}}{{C}_{5}}}{^{n}{{C}_{4}}}\] \[\frac{a}{b}=\frac{n!4!n-4!}{5!n-5!n!}=\frac{n!4!n-5!n-4!}{5\,4!n!n!-5!}\] \[\frac{a}{b}=\frac{n-4}{5}\]
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