A) 3
B) 4
C) 5
D) 6
E) 7
Correct Answer: E
Solution :
\[\therefore \] \[{{T}_{2}}{{=}^{n}}{{C}_{1}}x\] and \[{{T}_{3}}{{=}^{n}}{{C}_{2}}{{x}^{2}},{{T}_{4}}{{=}^{n}}{{C}_{3}}{{x}^{3}}\] Since,\[{{T}_{2}},{{T}_{3}},{{T}_{4}}\]are in AP \[\Rightarrow \] \[\frac{^{n}{{C}_{1}}{{+}^{n}}{{C}_{3}}}{2}{{=}^{n}}{{C}_{2}}\] \[^{n}{{C}_{1}}{{+}^{n}}{{C}_{3}}={{2}^{n}}{{C}_{2}}\] \[\Rightarrow \]\[\frac{n!}{(n-1)!(1)!}+\frac{n!}{(n-3)!3!}=\frac{2n!}{2!(n-2)!}\] \[\Rightarrow \] \[\frac{1}{(n-1)(n-2)}+\frac{1}{3!}=\frac{1}{(n-2)}\] \[\Rightarrow \] \[1+\frac{(n-1)(n-2)}{6}=(n-1)\] \[\Rightarrow \] \[6+{{n}^{2}}-3n+2=6n-6\] \[\Rightarrow \] \[{{n}^{2}}-3n-6n+8+6=0\] \[\Rightarrow \] \[{{n}^{2}}-9n+14=0\] \[\Rightarrow \] \[(n-7)(n-2)=0\] \[\Rightarrow \] \[n=7or\,2\]
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