The angles of a quadrilateral are in A.P. The greatest angle is thrice the least angle. Find the greatest angle.

The correct option is D: ${135}^{\circ }$

Let the angles which are in A.P be $x,x+d,x+2d,x+3d$

Greatest angle $=3\times$ least angle [Given]

$\Rightarrow x+3d=3x$

$\Rightarrow d=\frac{2x}{3}\dots \dots \left(i\right)$

Now by property of quadrilaterals, sum of all angles is ${360}^{\circ }$, therefore

$x+(x+d)+(x+2d)+(x+3d)={360}^{\circ }$

$\Rightarrow 4x+6d={360}^{\circ }$

$\Rightarrow 2x+3d={180}^{\circ }\dots \dots \left(ii\right)$

Now, $d=\frac{2x}{3}$

$\therefore 2x+3\times \frac{2x}{3}={180}^{\circ }$

$\Rightarrow 4x={180}^{\circ }$

$\Rightarrow x={45}^{\circ }$

Now, put the value of $x$ in $eq.\left(i\right),$ we get

$d=\frac{2x}{3}=\frac{2\times 45}{3}=30$

So, the greatest angle $=x+3d=45+3\times 30={135}^{\circ }$