Question

If the coefficients of second, third and fourth terms in the expansion of $(1+x{)}^n$ are in AP, then $n$ is equal to:

• A. $7$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$7$

Coefficient of second,third and fourth terms in the expansion of

$(1+x{)}^n$ are ${}^n{C}_1{,}^n{C}_2$ and ${}^n{C}_3$

Since these are in A.P. $\Rightarrow 2{\cdot }^n{C}_2{=}^n{C}_1{+}^n{C}_3$

$\Rightarrow 2\cdot \frac{n!}{2!(n-2)!}=\frac{n!}{(n-1)!1!}+\frac{n!}{(n-3)!3!}$

$\Rightarrow \frac{6(n-1)!}{(n-2)!}=\frac{6(n-1)!}{(n-1)!}+\frac{(n-1)!}{(n-3)!}$

$\Rightarrow 6(n-1)=6+(n-1)(n-2)=n^2-3n+8$

$\Rightarrow n^2-9n+14=0$

$\Rightarrow (n-7)(n-2)=0\Rightarrow n=7,2$ (but $n\ge 3$)

Hence $n=7$ is the answer