If the coefficients of the 5^{th},6^{th},and | vedclass

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(C) The coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms are $^nC_4$,$^nC_5$,and $^nC_6$ respectively.Given that these are in $A.P.$,we have $

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$7$ or $14$

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(C) The coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms are $^nC_4$,$^nC_5$,and $^nC_6$ respectively.Given that these are in $A.P.$,we have $2(^nC_5) = ^nC_4 + ^nC_6$.Dividing by $^nC_5$,we get $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$.Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ and $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$.So,$2 = \frac{5}{n-4} + \frac{n-5}{6}$.Multiplying by $6(n-4)$,we get $12(n-4) = 30 + (n-5)(n-4)$.$12n - 48 = 30 + n^2 - 9n + 20$.$n^2 - 21n + 98 = 0$.$(n-7)(n-14) = 0$.Thus,$n = 7$ or $n = 14$.

If the coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n =$

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