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 $p + q$ is equal to:

<p>$$\begin{aligned} & (\mathrm{a}+\mathrm{b})^{\frac{1}{2}} \\ & \mathrm{~T}_{\mathrm{r}}, \mathrm{~T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP} \\ & \text { So, } \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{\mathrm{T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}} \\ & \frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}} \\ & \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1} \\ & (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})(\mathrm{r}) \\ & -\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^2 \\ & 13=0 \end{aligned}$$</p> <p>No value of r possible</p> <p>So $\mathrm{P}=0$</p> <p>$$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_r\left(3^{\frac{1}{4}}\right)^{12-\mathrm{r}}\left(4^{\frac{1}{3}}\right)^{\mathrm{r}}$$</p> <p>Exponent of $\left(3^{\frac{1}{4}}\right)$ exponent of $\left(4^{\frac{1}{3}}\right)$ term</p> <p>$$\begin{array}{ccc} 12 & 0 & 27 \\ 0 & 12 & 256 \\ \mathrm{q}=27+256=283 & & \\ \mathrm{p}+\mathrm{q}=0+283=283 & & \end{array}$$</p>