Question ID: #499
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:
Solution:
Let the sides of the polygon be $n$.
Sum of interior angles $S_n = (n-2) \times 180^\circ$.
The angles are in A.P. with $d = 6$.
The largest angle (last term) $l = 219$.
We know $l = a + (n-1)d$.
$$219 = a + (n-1)6 \Rightarrow a = 219 – 6(n-1) = 225 – 6n$$
Sum of A.P. $S_n = \frac{n}{2}(a + l)$.
Substituting values:
$$(n-2)180 = \frac{n}{2}(225 – 6n + 219)$$
$$(n-2)180 = \frac{n}{2}(444 – 6n)$$
$$(n-2)180 = n(222 – 3n)$$
$$180n – 360 = 222n – 3n^2$$
$$3n^2 – 42n – 360 = 0$$
Dividing by 3:
$$n^2 – 14n – 120 = 0$$
Factorizing:
$$(n – 20)(n + 6) = 0$$
Since $n$ must be positive, $n = 20$.
Ans. (20)
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