In a triangle with one angle $2\pi /3$ , the lengths of the sides form an A.P. If the length of the greatest side is 7 cm, the radius of the circumcircle of the triangle is

Let the sides of the triangle be

7, 7 – d, 7 – 2d.

Since the given angle is the greatest (being obtuse) angle of the triangle, it is opposite to the greatest side of the triangle and we have,

$\begin{array}{l}{7}^2=(7-d{)}^2+(7-2d{)}^2\\ -2(7-d)(7-2d)\cos 2\pi /3\end{array}$

$\begin{array}{l} \\ \Rightarrow 7^2=2\times 7^2-42d+5d^2\\ -2\left(7^2-21d+2d^2\right)(-1/2\\ \Rightarrow d^2-9d+14=0\\ \Rightarrow (d-7)(d-2)=0\\ \Rightarrow d=2(d=7\text{ is not possible })\end{array}$

Therefore, the sides of the triangle are 7 cm, 5 cm, 3 cm

Area of the triangle $\mathrm{\Delta }=\frac{1}{2}\times 5\times 3\times \sin \frac{2\pi }{3}=\frac{15\sqrt{3}}{4}$

${\mathrm{cm}}^2$ and the radius of the circumcircle $R=\frac{7\times 5\times 3}{4(15\sqrt{3}/4)}$

$=\frac{7\times 5\times 3}{15\sqrt{3}}=\frac{7\sqrt{3}}{3}\mathrm{cm}$.