Thank you for reporting, we will resolve it shortly
Q.
Let $ {{(1+x)}^{n}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{n}}{{x}^{n}} $ .If $ {{a}_{1}},{{a}_{2}} $ and $ {{a}_{3}} $ are in $AP$, then the value of $n$ is
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 $ .