A) 4
B) 5
C) 6
D) 7
E) 8
Correct Answer: D
Solution :
It is given that,\[{{a}_{1}}{{=}^{n}}{{C}_{1}},{{a}_{2}}{{=}^{n}}{{C}_{2}}\]and\[{{a}_{3}}{{=}^{n}}{{C}_{3}}\]are in AP. \[\Rightarrow \] \[{{2}^{n}}{{C}_{2}}{{=}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{3}}\] \[\Rightarrow \] \[\frac{2n(n-1)}{2!}=n+\frac{n(n-1)(n-2)}{3!}\] \[\Rightarrow \] \[n(n-1)=n+\frac{n(n-1)(n-2)}{6}\] \[\Rightarrow \] \[6n-6=6+{{n}^{2}}-3n+2\] \[\Rightarrow \] \[{{n}^{2}}-9n+14=0\] \[\Rightarrow \] \[n=7,2\] If\[n=2,\]then there are only three terms in the expansion of\[{{(1+x)}^{n}}\]. Therefore;\[n=7\].You need to login to perform this action.
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