Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
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 \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: Sum of interior angles of an $n$-gon is $(n-2)\cdot 180^{\circ}$. Largest angle $=a+(n-1)d$. Two equations $\Rightarrow$ a quadratic in $n$.
Let first angle be $a^{\circ}$. Largest: $a+(n-1)\cdot 6=219\Rightarrow a=225-6n.$
Sum: $\dfrac{n}{2}\bigl[2a+(n-1)\cdot 6\bigr]=(n-2)\cdot 180.$ Substitute:
$$\dfrac{n}{2}\bigl[2(225-6n)+6(n-1)\bigr]=180(n-2)\Rightarrow\dfrac{n}{2}(444-6n)=180n-360.$$
$$n(222-3n)=180n-360\Rightarrow 3n^{2}-42n-360+180n-222n=0$$
$$\Rightarrow 2n^{2}-42n-360=0\Rightarrow n^{2}-14n-120=0\Rightarrow(n-20)(n+6)=0.$$
Reject $n=-6$; $n=20$ (and smallest angle $=225-120=105^{\circ}>0$ checks out).
Correct Answer: 20