Mathematics - Sequence & Series Question with Solution | TestHub

$\frac{n}{2}[2a+(n-1)6]=(n-2)180^{\circ }$ and $an+3n^2-3n=3n(n-2)180^{\circ }$...(i) $\begin{array}{rl}& \therefore \text{ Given }a+(n-1)6^{\circ }=219^{\circ }\\ & \Rightarrow a=225^{\circ }-6n^{\circ }\end{array}$ Putting value of $a$ in (i) $\begin{array}{rl}& \text{ We get }\left(225-6n^2\right)+3n^2-3n=180n-360^{\circ }\\ & \Rightarrow 2n^2-42n-360=0\\ & \Rightarrow n^2-14n-120=0\\ & \Rightarrow (n-20)(n+6)=0\\ & \Rightarrow n=20,-6\text{ (Rejected) }\\ & \therefore n=20\end{array}$