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 $\mathrm{AP}$, then the value of $\mathrm{n}$ is

• A. $4$
• B. $5$
• C. $6$
• D. $7$
• E. $8$

Answer 1

The correct option is D

$7$

Explanation for correct option:

Determining the value of $\mathrm{n}$:

Given ${\left(1+x\right)}^n=1+a_{1}x+a_2{x}^2+...+a_n{x}^n$

Also $a_1,a_2,a_3$ are in $\mathrm{AP}$.

We know that,

${\left(1+x\right)}^n={}^{n}C_0+{}^{n}C_{1}x+{}^{n}C_2{x}^2+....+{}^{n}C_n{x}^n$

Comparing with the given expression we get,

$$\begin{gathered} {}^{n}C_0=1 \\ {}^{n}C_1=a_1 \\ {}^{n}C_2=a_2 \\ {}^{n}C_3=a_3 \end{gathered}$$

Now we are given that, $a_1,a_2,a_3$ are in $\mathrm{AP}$, hence,

$$\begin{gathered} 2a_2=a_1+a_3 \\ \Rightarrow 2{}^{n}C_2={}^{n}C_1+{}^{n}C_3 \\ \Rightarrow 2\left(\frac{n\left(n-1\right)}{2}\right)=n+\frac{n\left(n-1\right)\left(n-2\right)}{6} \\ \Rightarrow 6\left(n^2-n\right)=6n+n\left(n^2-3n+2\right) \\ \Rightarrow 6n^2-6n=6n+n^3-3n^2+2n \\ \Rightarrow n^3-9n^2+14n=0 \\ \Rightarrow n\left(n^2-9n+14\right)=0 \\ \Rightarrow n\left(n-2\right)\left(n-7\right)=0 \\ \Rightarrow n=0,2,7 \end{gathered}$$

Reject $n=0,2$ because the given expansion has more than three terms. Therefore we have, $n=7$

Hence, the correct answer is option (D).