Let the coefficients of three consecutive terms T_r, T_r+1 and T_r+2 in the binomial expansion of (a + b)¹²  be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4}{)}^{12}$. Then p + q

$\begin{array}{l}(\mathrm{a}+\mathrm{b}{)}^{\frac{1}{2}}\\ {\mathrm{T}}_{\mathrm{r}},{\mathrm{T}}_{\mathrm{r}+1},{\mathrm{T}}_{\mathrm{r}+2}\to \mathrm{GP}\\ \text{ So, }\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}}=\frac{{\mathrm{T}}_{\mathrm{r}+2}}{{\mathrm{T}}_{\mathrm{r}+1}}\\ \frac{{ }^{12}{\mathrm{C}}_{\mathrm{r}}}{{ }^{12}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{{ }^{12}{\mathrm{C}}_{\mathrm{r}+1}}{{ }^{12}{\mathrm{C}}_{\mathrm{r}}}\\ \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1}\\ (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})\left(\mathrm{r}\right)\\ -\mathrm{r}+12\mathrm{r}+13=12\mathrm{r}-{\mathrm{r}}^2\\ 13=0\end{array}$

No value of r possible
So P = 0

${\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)}^{12}=\sum ^{12} {\mathrm{C}}_{\mathrm{r}}{\left(3^{\frac{1}{4}}\right)}^{12-\mathrm{r}}{\left(4^{\frac{1}{3}}\right)}^{\mathrm{r}}$

$\begin{array}{r}\text{ Exponent of }\left(3^{\frac{1}{4}}\right)\text{ exponent of }\left(4^{\frac{1}{3}}\right)\text{ term }\\ \begin{array}{ccr}12& 0& 27\\ 0& 12& 256\\ \mathrm{q}=27+256=283& & \\ \mathrm{p}+\mathrm{q}=0+283=283& & \end{array}\end{array}$