A) 12
B) 14
C) 9
D) 11
Correct Answer: B
Solution :
Since, \[^{n}{{C}_{4}},{{\,}^{n}}{{C}_{5}}\]and \[{{\,}^{n}}{{C}_{6}}\]are in A.P. \[\therefore \]\[2{{\times }^{n}}{{C}_{5}}{{=}^{n}}{{C}_{4}}{{+}^{n}}{{C}_{6}}\] \[\Rightarrow \]\[2\times \frac{\left| \!{\nderline {\, n \,}} \right. }{\left| \!{\nderline {\, 5 \,}} \right. \,\left| \!{\nderline {\, n-5 \,}} \right. }=\frac{\left| \!{\nderline {\, n \,}} \right. }{\left| \!{\nderline {\, 4 \,}} \right. \,\,\left| \!{\nderline {\, n-4 \,}} \right. }+\frac{\left| \!{\nderline {\, n \,}} \right. }{\left| \!{\nderline {\, 6 \,}} \right. \,\,\left| \!{\nderline {\, n-6 \,}} \right. }\] \[\Rightarrow \]\[\frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}\] \[\Rightarrow \]\[\frac{2}{5}(n-4)=1+\frac{(n-4)(n-5)}{6\times 5}\] \[\Rightarrow \]\[12(n-4)=30+{{n}^{2}}-9n+20\] \[\Rightarrow \]\[21n-{{n}^{2}}-98=0\Rightarrow {{n}^{2}}-21n+98=0\] \[\Rightarrow \]\[(n-14)(n-7)=0\Rightarrow n=7,14\] But only n = 14 lies in the options.
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