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(D) The general term in the expansion of $(x + a)^n$ is $T_{r+1} = {}^nC_r x^{n-r} a^r$.Given $T_2 = {}^nC_1 x^{n-1} a = 240$ $(i)$,$T_3 = {}^nC_2 x^{

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$5$

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(D) The general term in the expansion of $(x + a)^n$ is $T_{r+1} = {}^nC_r x^{n-r} a^r$.Given $T_2 = {}^nC_1 x^{n-1} a = 240$ $(i)$,$T_3 = {}^nC_2 x^{n-2} a^2 = 720$ $(ii)$,$T_4 = {}^nC_3 x^{n-3} a^3 = 1080$ $(iii)$.Dividing $(ii)$ by $(i)$,we get $\frac{T_3}{T_2} = \frac{{}^nC_2 x^{n-2} a^2}{{}^nC_1 x^{n-1} a} = \frac{n-1}{2} \cdot \frac{a}{x} = \frac{720}{240} = 3$ $\Rightarrow \frac{a}{x} = \frac{6}{n-1}$.Dividing $(iii)$ by $(ii)$,we get $\frac{T_4}{T_3} = \frac{{}^nC_3 x^{n-3} a^3}{{}^nC_2 x^{n-2} a^2} = \frac{n-2}{3} \cdot \frac{a}{x} = \frac{1080}{720} = \frac{3}{2}$ $\Rightarrow \frac{a}{x} = \frac{9}{2(n-2)}$.Equating the two expressions for $\frac{a}{x}$:$\frac{6}{n-1} = \frac{9}{2(n-2)}$$12(n-2) = 9(n-1)$$12n - 24 = 9n - 9$$3n = 15 \Rightarrow n = 5$.

If the second,third,and fourth terms in the expansion of $(x + a)^n$ are $240, 720,$ and $1080$ respectively,then the value of $n$ is

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