The interior angles of a polygon with n sides, are in an A.P. with common difference $$6^{\circ}$$ . If the largest interior angle of the polygon is $$219^{\circ}$$, then n is equal to

The interior angles of a polygon with $$n$$ sides are in A.P. with common difference $$6°$$ and the largest angle is $$219°$$.

We start by letting the smallest angle be $$a$$. The angles in A.P. are: $$a, a+6, a+12, \ldots, a+(n-1)\times 6$$. Since the largest angle is $$219°$$, we have $$a + (n-1)\times 6 = 219$$, which gives $$a = 219 - 6(n-1)\quad\cdots(1)$$.

Next, the sum of interior angles of an $$n$$-sided polygon is $$(n-2)\times 180°$$. The sum of the A.P. can be written as $$\frac{n}{2}[2a + (n-1)\times 6] = (n-2)\times 180$$. Noting that the first term is $$a$$ and the last term is $$219$$, this becomes $$\frac{n}{2}[a + 219] = (n-2)\times 180$$.

Substituting $$a$$ from equation (1) into the sum yields $$\frac{n}{2}[219 - 6(n-1) + 219] = (n-2)\times 180$$ $$\frac{n}{2}[438 - 6n + 6] = 180(n-2)$$ $$\frac{n}{2}[444 - 6n] = 180(n-2)$$ $$n(444 - 6n) = 360(n-2)$$ $$444n - 6n^2 = 360n - 720$$ $$-6n^2 + 84n + 720 = 0$$ $$6n^2 - 84n - 720 = 0$$ $$n^2 - 14n - 120 = 0$$.

Solving the quadratic gives $$n = \frac{14 \pm \sqrt{196 + 480}}{2} = \frac{14 \pm \sqrt{676}}{2} = \frac{14 \pm 26}{2}$$, so $$n = \frac{40}{2} = 20$$ or $$n = \frac{-12}{2} = -6$$. We reject $$n = -6$$ since $$n$$ must be positive.

Therefore, n = 20.