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(D) The coefficients of the $2^{nd}$,$3^{rd}$,and $4^{th}$ terms in the expansion of $(1 + x)^n$ are $^nC_1$,$^nC_2$,and $^nC_3$ respectively.Since th

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$-14$

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(D) The coefficients of the $2^{nd}$,$3^{rd}$,and $4^{th}$ terms in the expansion of $(1 + x)^n$ are $^nC_1$,$^nC_2$,and $^nC_3$ respectively.Since these are in $A.P.$,we have $2(^nC_2) = ^nC_1 + ^nC_3$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:$2 	imes \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.Dividing by $n$ (assuming $n 
eq 0$):$(n-1) = 1 + \frac{(n-1)(n-2)}{6}$.$6(n-1) = 6 + (n^2 - 3n + 2)$.$6n - 6 = 6 + n^2 - 3n + 2$.$n^2 - 9n + 14 = 0$.Therefore,$n^2 - 9n = -14$.

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

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