The angles of a quadrilateral are in AP, and the greatest angle is double

Let the smallest term be x, and the largest term be 2x

Then AP formed= x, ?, ?, 2x

So

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+(\mathrm{a}+(\mathrm{n}-1) \mathrm{d})]=\frac{\mathrm{n}}{2}[$ First term $+($ Last term $)]$

$360^{\circ}=4 / 2[x+2 x] \ldots . .[$ We know that $\rightarrow a+(n-1) d=$ last term $=2 x]$

$\Rightarrow 180^{\circ}=3 x$

$\Rightarrow x=60^{\circ}$

Now, $60^{\circ}$ is least angle.

$=60^{\circ}=\pi / 180^{\circ} \times 60^{\circ}$

$\Rightarrow 60^{\circ}=\pi / 3 \mathrm{rad}$