A) 15
B) 20
C) 10
D) 5
Correct Answer: D
Solution :
\[{{T}_{2}}=n\,{{(x)}^{n-1}}{{(a)}^{1}}=240\] .....(i) \[{{T}_{3}}=\frac{n\,(n-1)}{1.2}{{x}^{n-2}}{{a}^{2}}=720\] ?..(ii) \[{{T}_{4}}=\frac{n\,(n-1)(n-2)}{1.2.3}{{x}^{n-3}}{{a}^{3}}=1080\] ?..(iii) To eliminate x, \[\frac{{{T}_{2}}\,.\,{{T}_{4}}}{{{T}_{3}}^{2}}=\frac{240\,.\,1080}{720\,.\,720}=\frac{1}{2}\] Þ \[\frac{{{T}_{2}}}{{{T}_{3}}}\,\,.\,\,\frac{{{T}_{4}}}{{{T}_{3}}}=\frac{1}{2}\] Now, \[\frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{{}^{n}{{C}_{r}}}{{}^{n}{{C}_{r-1}}}=\frac{n-r+1}{r}\] Putting r = 3 and 2 in above expression, we get \[\Rightarrow \frac{n-2}{3}\,.\,\frac{2}{n-1}=\frac{1}{2}\]\[\Rightarrow n=5\].
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