If the coefficients of rth, (r + 1)th and (r +2)th terms in the expansion of (1 + x)¹⁴ are in AP, then r is /are

Coefficient of rth, (r + 1) th and (r +2)th terms in (1 +x)¹⁴ are ${ }^{14}{\mathrm{C}}_{\mathrm{r}-1}{,}^{14}{\mathrm{C}}_{\mathrm{r}}{,}^{14}{\mathrm{C}}_{\mathrm{r}+1}$ respectively
Now, according to the question, $2{(}^{14}{\mathrm{C}}_{\mathrm{r}}){=}^{14}{\mathrm{C}}_{\mathrm{r}-1}{+}^{14}{{\mathrm{C}}_{\mathrm{r}}}_{+1}$
On dividing both sides by ${ }^{14}{\mathrm{C}}_{\mathrm{r}}$, we get

$2=\frac{{ }^{14}{\mathrm{C}}_{\mathrm{r}-1}}{{ }^{14}{\mathrm{C}}_{\mathrm{r}}}+\frac{{ }^{14}{\mathrm{C}}_{\mathrm{r}+1}}{{ }^{14}{\mathrm{C}}_{\mathrm{r}}}$

$\begin{array}{rl} & \Rightarrow 2=\frac{\mathrm{r}}{14-\mathrm{r}+1}+\frac{14-(\mathrm{r}+1)+1}{\mathrm{r}+1}\\ & \Rightarrow 2=\frac{\mathrm{r}}{15-\mathrm{r}}+\frac{14-\mathrm{r}}{\mathrm{r}+1}\\ & \Rightarrow 2(15-\mathrm{r})(\mathrm{r}+1)=\mathrm{r}(\mathrm{r}+1)+(15-\mathrm{r})(14-\mathrm{r})\\ & \Rightarrow -2{\mathrm{r}}^2+28\mathrm{r}+30=2{\mathrm{r}}^2-28\mathrm{r}+210\\ & \Rightarrow 4{\mathrm{r}}^2-56\mathrm{r}+180=0\Rightarrow {\mathrm{r}}^2-14\mathrm{r}+45=0\\ & \Rightarrow (\mathrm{r}-9)(\mathrm{r}-5)=0\\ & \Rightarrow \mathrm{r}=5,9\end{array}$