If the coefficients of x^2 and x^3 in the exp | vedclass

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(D) The general term in the expansion of $(3+ax)^9$ is given by $T_{r+1} = {^9C_r} (3)^{9-r} (ax)^r = {^9C_r} (3)^{9-r} a^r x^r$. For the coefficient

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$\frac{9}{7}$

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(D) The general term in the expansion of $(3+ax)^9$ is given by $T_{r+1} = {^9C_r} (3)^{9-r} (ax)^r = {^9C_r} (3)^{9-r} a^r x^r$. For the coefficient of $x^2$, we set $r=2$: Coefficient of $x^2 = {^9C_2} (3)^{9-2} a^2 = {^9C_2} (3)^7 a^2$. For the coefficient of $x^3$, we set $r=3$: Coefficient of $x^3 = {^9C_3} (3)^{9-3} a^3 = {^9C_3} (3)^6 a^3$. Given that these coefficients are equal: ${^9C_2} (3)^7 a^2 = {^9C_3} (3)^6 a^3$. Using ${^9C_2} = \frac{9 	imes 8}{2} = 36$ and ${^9C_3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$: $36 	imes 3^7 	imes a^2 = 84 	imes 3^6 	imes a^3$. Dividing both sides by $3^6 a^2$ (assuming $a 
eq 0$): $36 	imes 3 = 84 	imes a$. $108 = 84a$. $a = \frac{108}{84} = \frac{9}{7}$.

If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+ax)^9$ are the same, then the value of $a$ is

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