The coefficient of $9 t h, 10 t h$ and $11 t h$ terms in the expansion of $(1 + x)^{n}$ are in $A . P .$ then $n =$

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Solution

According to the question, $^{n} C_{8} +^{n} C_{10} = {2.}^{n} C_{9}$
$= \frac{n!}{(n - 8)! 8!} + \frac{n!}{(n - 10)! 10!} = \frac{2 n!}{(n - 9)! 9!}$
$\Rightarrow \frac{1}{(n - 8) (n - 9)} + \frac{1}{10 - 9} = \frac{2}{(n - 9) - 9}$
$\Rightarrow \frac{90 + (n - 8) (n - 9)}{90 (n - 8) (n - 9)} = \frac{2}{(n - 9) 9}$
$\Rightarrow \frac{90 + n^{2} - 17 n + 72}{10 (n - 8)} = 2$
$\Rightarrow 90 + n^{2} - 17 n + 72 = 20 n - 160$
$\Rightarrow n^{2} - 37 n + 322 = 0$
$∴ n = 23$
$n = 14$

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