Question

If coefficients of 2, 3 and 4 terms in the binomial expansion of $(1+x{)}^n$ are in A.P, then $n^2-9n$ is equal to

• A. -7
• B. 7
• C. 14
• D. -14

Answer 1

The correct option is D -14

Coefficients of 2, 3 and 4 terms are respectively ${}^n{C}_1{,}^n{C}_2$ and ${}^n{C}_3$ are in A.P.
$\Rightarrow {2.}^n{C}_2{=}^n{C}_1{+}^n{C}_3$
$\Rightarrow \frac{2n!}{2!(n-2)!}=\frac{n!}{(n-1)!}+\frac{n!}{3!(n-3)!}$
On solving, $n^2-9n+14=0\Rightarrow n^2-9n=-14$.