Find $a$ if the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+a x)^{9}$ are equal.

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by

${T_{r + 1}} = {\,^n}{C_r}{a^{n - r}}{b^r}$

Assuming that $x^{2}$ occurs in the $(r+1)^{\text {th }}$ term in the expansion of $(3+a x)^{9}$, we obtain

${T_{r + 1}} = {\,^9}{C_r}{(3)^{9 - r}}{(ax)^r} = {\,^9}{C_r}{(3)^{2 - r}}{a^r}{x^r}$

Comparing the indices of $x$ in $x^{2}$ and in $T_{r+1},$ we obtain

$r=2$

Thus, the coefficient of $x^{2}$ is

${\,^9}{C_2}{(3)^{9 - 2}}{a^2} = \frac{{9!}}{{2!7!}}{(3)^7}{a^2} = 36{(3)^7}{a^2}$

Assuming that $x^{3}$ occurs in the $(k+1)^{\text {th }}$ term in the expansion of $(3+a x)^{9}$, we obtain

${T_{k + 1}} = {\,^9}{C_k}{(3)^{9 - k}}{(ax)^k} = {\,^9}{C_k}{(3)^{9 - k}}{a^k}{x^k}$

Comparing the indices of $x$ in $x^{3}$ and in $T_{k+1},$ we obtain $k=3$

Thus, the coefficient of $x^{3}$ is

${\,^9}{C_3}{(3)^{9 - 3}}{a^3} = \frac{{9!}}{{3!6!}}{(3)^6}{a^3} = 84{(3)^6}{a^3}$

It is given that the coefficient of $x^{2}$ and $x^{3}$ are the same.

$84(3)^{6} a^{3}=36(3)^{7} a^{2}$

$\Rightarrow 84 a=36 \times 3$

$\Rightarrow a=\frac{36 \times 3}{84}=\frac{104}{84}$

$\Rightarrow a=\frac{9}{7}$

Thus, the required value of $a$ is $9 / 7$