If coefficient of 9th, 10th and 11th terms of (1 + x)n are in A.P., then value of n can be

A
15
B
32
C
23
D
17

Solution:

$\begin{array}{l} 2 (^{n} C_{9}) =^{n} C_{8} +^{n} C_{10} \\ \Rightarrow 2 = \frac{9}{n - 8} + \frac{n - 9}{10} \\ \Rightarrow 20 (n - 8) = 90 + (n - 8) (n - 9) \\ \Rightarrow n^{2} - 37 n + 322 = 0 \end{array}$

$\Rightarrow (n - 14) (n - 23) = 0$

Thus, $n$ can be 23

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