If coefficients of 2, 3 and 4 terms in the binomial expansion of $(1 + x)^{n}$ are in A.P, then $n^{2} - 9 n$ is equal to

-7

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-14

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Solution

The correct option is D -14
Coefficients of 2, 3 and 4 terms are respectively $^{n} C_{1},^{n} C_{2}$ and $^{n} C_{3}$ are in A.P.
$\Rightarrow {2.}^{n} C_{2} =^{n} C_{1} +^{n} C_{3}$
$\Rightarrow \frac{2 n!}{2! (n - 2)!} = \frac{n!}{(n - 1)!} + \frac{n!}{3! (n - 3)!}$
On solving, $n^{2} - 9 n + 14 = 0 \Rightarrow n^{2} - 9 n = - 14$ .

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