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(A) In the expansion of $(1 + x)^n$,the coefficients of the second,third,and fourth terms are $^nC_1, ^nC_2$,and $^nC_3$ respectively.Given that $^nC_

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$7$

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(A) In the expansion of $(1 + x)^n$,the coefficients of the second,third,and fourth terms are $^nC_1, ^nC_2$,and $^nC_3$ respectively.Given that $^nC_1, ^nC_2, ^nC_3$ are in $A.P.$,we have:$2(^nC_2) = ^nC_1 + ^nC_3$$2 	imes \frac{n(n - 1)}{2} = n + \frac{n(n - 1)(n - 2)}{6}$Dividing by $n$ (since $n 
eq 0$):$n - 1 = 1 + \frac{(n - 1)(n - 2)}{6}$$6(n - 1) = 6 + (n^2 - 3n + 2)$$6n - 6 = n^2 - 3n + 8$$n^2 - 9n + 14 = 0$$(n - 7)(n - 2) = 0$So,$n = 7$ or $n = 2$.For $n = 2$,the expansion $(1 + x)^2$ has only three terms,so the fourth term does not exist. Thus,$n = 7$ is the only valid solution.

If the coefficients of the second,third,and fourth terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n$ is equal to

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