Question

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

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

7

Answer 2

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; $n=7$ .