JEE Main 2025 — 28 January, Evening Shift — Question 23

$\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) 6)=(\mathrm{n}-2) .180^{\circ}$

an $+3 n^{2}-3 n=(n-2) .180^{\circ}$

Now according to question

$\mathrm{a}+(\mathrm{n}-1) 6^{\circ}=219^{\circ}$

$\Rightarrow \mathrm{a}=225^{\circ}-6 \mathrm{n}^{\circ}$

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

We get

$\left(225 n-6 n^{2}\right)+3 n^{2}-3 n=180 n-360$

$\Rightarrow 3 \mathrm{n}^{2}-42 \mathrm{n}-360=0$

$\Rightarrow \mathrm{n}^{2}-14 \mathrm{n}-120=0$

$\mathrm{n}=20,-6$ (rejected)