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
• E. 8

Answer 1

Answer: (D)

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; $n=7$ .