Question

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

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

Answer 1

The correct option is A $398$

Coefficients of $x^9,x^{10},x^{11}$ in ${(1+x)}^n$ are in A.P.

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

We know that,

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

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

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

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

$\Rightarrow n^2-41n=398$ is the answer.