If the 2^{text{nd}},3^{text{rd}},and 4^{text{ | vedclass

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(A) The terms are given by the binomial expansion formula $T_{r+1} = {}^{n}C_{r} x^{n-r} a^{r}$.$T_2 = {}^{n}C_1 x^{n-1} a = 96$ $(i)$$T_3 = {}^{n}C_2

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$n+1$

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(A) The terms are given by the binomial expansion formula $T_{r+1} = {}^{n}C_{r} x^{n-r} a^{r}$.$T_2 = {}^{n}C_1 x^{n-1} a = 96$ $(i)$$T_3 = {}^{n}C_2 x^{n-2} a^2 = 216$ $(ii)$$T_4 = {}^{n}C_3 x^{n-3} a^3 = 216$ $(iii)$Dividing $(i)$ by $(ii)$:$\frac{{}^{n}C_1 x^{n-1} a}{{}^{n}C_2 x^{n-2} a^2} = \frac{96}{216}$ $\Rightarrow \frac{n x}{\frac{n(n-1)}{2} a} = \frac{4}{9}$ $\Rightarrow \frac{2x}{(n-1)a} = \frac{4}{9}$ $\Rightarrow 9x = 2(n-1)a$.Dividing $(ii)$ by $(iii)$:$\frac{{}^{n}C_2 x^{n-2} a^2}{{}^{n}C_3 x^{n-3} a^3} = \frac{216}{216}$ $\Rightarrow \frac{\frac{n(n-1)}{2} x}{\frac{n(n-1)(n-2)}{6} a} = 1$ $\Rightarrow \frac{3x}{(n-2)a} = 1$ $\Rightarrow 3x = (n-2)a$.From the two equations:$\frac{9x}{3x} = \frac{2(n-1)a}{(n-2)a}$ $\Rightarrow 3 = \frac{2(n-1)}{n-2}$ $\Rightarrow 3n-6 = 2n-2$ $\Rightarrow n=4$.Substituting $n=4$ into $3x = (n-2)a$ gives $3x = 2a$,so $a = \frac{3}{2}x$.Substituting into $(i)$:$4 x^3 (\frac{3}{2}x) = 96$ $\Rightarrow 6x^4 = 96$ $\Rightarrow x^4 = 16$ $\Rightarrow x=2$.Then $a = \frac{3}{2}(2) = 3$.Thus,$a+x = 3+2 = 5$. Since $n=4$,$a+x = n+1$.

If the $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(x+a)^{n}$ are $96, 216$,and $216$ respectively,and $n$ is a positive integer,then $a+x=$

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