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8. Sequences and Series
hard
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______
A$10$
B$30$
C$20$
D$50$
(JEE MAIN-2025)
Solution
$\frac{n}{2}(2 a+(n-1) 6)=(n-2) \cdot 180^{\circ}$
$an+3 n^2-3 n=(n-2) \cdot 180^{\circ}$
Now according to question
$a+(n-1) 6^{\circ}=219^{\circ}$
$\Rightarrow a=225^{\circ}-6 n^{\circ}$
Putting value of a from equation $(2)$ in $(1)$
We get
$\left(225 n-6 n^2\right)+3 n^2-3 n=180 n-360$
$\Rightarrow 2 n^2-42 n-360=0$
$\Rightarrow n 2-14 n-120=0$
$n=20,-6 \text { (rejected) }$
$an+3 n^2-3 n=(n-2) \cdot 180^{\circ}$
Now according to question
$a+(n-1) 6^{\circ}=219^{\circ}$
$\Rightarrow a=225^{\circ}-6 n^{\circ}$
Putting value of a from equation $(2)$ in $(1)$
We get
$\left(225 n-6 n^2\right)+3 n^2-3 n=180 n-360$
$\Rightarrow 2 n^2-42 n-360=0$
$\Rightarrow n 2-14 n-120=0$
$n=20,-6 \text { (rejected) }$
Standard 11
Mathematics
