$$\begin{aligned} & \frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) 6)=(\mathrm{n}-2) \cdot 180^{\circ} \\ & \mathrm{an}+3 \mathrm{n}^2-3 \mathrm{n}=(\mathrm{n}-2) \cdot 180^{\circ}\quad\text{.... (1)} \end{aligned}$$

Now according to question

$$\begin{aligned} & a+(n-1) 6^{\circ}=219^{\circ} \\ & \Rightarrow a=225^{\circ}-6 n^{\circ}\quad\text{.... (2)} \end{aligned}$$

Putting value of a from equation (2) in (1)

We get

$$\begin{aligned} & \left(225 \mathrm{n}-6 \mathrm{n}^2\right)+3 \mathrm{n}^2-3 \mathrm{n}=180 \mathrm{n}-360 \\ & \Rightarrow 2 \mathrm{n}^2-42 \mathrm{n}-360=0 \\ & \Rightarrow \mathrm{n} 2-14 \mathrm{n}-120=0 \\ & \mathrm{n}=20,-6 \text { (rejected) } \end{aligned}$$