Question

If the coefficeints of $x^9,x^{10},x^{11}$ in the expansion of ${(1+x)}^n$ are in arithmetic progression, then $n^2-41n=$

• A. $398$
• B. $298$
• C. $-398$
• D. $198$

Answer 1

The correct option is C $-398$

${}^n{C}_9{,}^n{C}_{10}{,}^n{C}_{11}$ are in A.P.

${\Rightarrow }^{2n}{C}_{10}{=}^n{C}_9{+}^n{C}_{10}$

$\Rightarrow 2=\frac{{}^n{C}_9}{{}^n{C}_{10}}+\frac{{}^n{C}_{11}}{{}^n{C}_{10}}$

$\Rightarrow 2=\frac{10}{n-9}+\frac{n-10}{11}$

$\Rightarrow 2=\frac{110+n^2-19n+90}{11n-99}$

$\Rightarrow n^2-41n=-398$