Question

If the coefficients of $x^9,x^{10}$ and $x^{11}$ in the expansion of $(1+x{)}^n$ are in arithmetic progression, then $n^2-41n$ is equal to

• A. 399
• B. 298
• C. -398
• D. 198

Answer 1

Answer: (C)

Given that, $x^9,x^{10}$ and $x^{11}$ are in AP.
So, ${}^n{C}_9,{}^n{C}_{10},{}^n{C}_{11}$ are in $AP$
$\therefore 2^n{C}_{10}={}^n{C}_9+{}^n{C}_{11}$
$\Rightarrow 2=\frac{{}^n{C}_9}{{}^n{C}_{10}}+\frac{{}^n{C}_{11}}{{}^n{C}_{10}}$
$\Rightarrow 2=\frac{10}{n-9}+\frac{n-10}{11}$
$\Rightarrow 2=\frac{110+(n-10)(n-9)}{11(n-9)}$
$\Rightarrow 2=\frac{110+n^2-9n-10n+90}{11n-99}$
$\Rightarrow 22n-198=200+n^2-19n$
$\Rightarrow n^2-41n=-398$