Question

If the coefficients of $x^{9}, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^{n}$ are in arithmetic progression, then $n^{2}-41 n$ 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 $AP$
$\therefore 2^{n} C_{10}={ }^{n} C_{9}+{ }^{n} C_{11}$
$\Rightarrow 2=\frac{{ }^{n} C_{9}}{{ }^{n} C_{10}}+\frac{{ }^{n} C_{11}}{{ }^{n} C_{10}}$
$\Rightarrow 2=\frac{10}{n-9}+\frac{n-10}{11}$
$\Rightarrow 2=\frac{110+(n-10)(n-9)}{11(n-9)}$
$\Rightarrow 2=\frac{110+n^{2}-9 n-10 n+90}{11 n-99}$
$\Rightarrow 22 n-198=200+n^{2}-19 n$
$\Rightarrow n^{2}-41 n=-398$