Let $x^2$ occur in the (r + 1)th term of the expansion $(3 + ax)^9$ . Then $T_{r+1}$ = $\,\!{}^{9}C_r (3)^{9-r} (ax)^r$ = $\,\!{}^{9}C_r (3)^{9-r} a^r x^r$ On comparing power of x in $x^2$ and $T_{r + 1}$, we get r = 2 ⇒ $T_{2+1}$ = ${}^{9}C_2 (3)^{9-2} a^2x^2$ ∴ Coefficient of $x^2$ in $T_3$ is $\,\!{}^{9}C_2 (3)^7 a^2$ = 36 $(3)^7 a^2$ Now let $x^3$ occurs in the (r + 1)th term of the expansion $(3+ax)^9$ ∴ $T_{r+1}$ = $\,\!{}^{9}C_r (3)^{9-r} (ax)^r$ = ${}^{9}C_r (3)^{9-r} a^r x^r$ On comparing power of x in $x^3$ and $T_{r + 1}$, we get r = 3 ⇒ $T_{3+1}$ = $\,\!{}^{9}C_3(3)^{9-3}(ax)^3$ = $\,\!{}^{9}C_3(3)^6 a^3x^3$ Coefficient of $x^3$ in $T_4$ is ${}^{9}C_3 3^6 a^3$ = $84(3)^6 a^3$ We are given coefficient of $x^2$ = coefficient of $x^3$ ∴ 36 $(3))^7 a^2$ = 84 $(3)^6 a^3$ ⇒ $\frac{a^3}{a^2}$ = $\frac{36(3)^7}{84(3)^6}$ = $\frac{9}{7}$ ⇒ a = $\frac{9}{7}$