The interior angles of a polygon are in A.P. The smallest angle isn 120^° and the common difference is 5^°. Find then number of sides of the polygon.

Let the number of sides of the polygon be nnThen the sum of all the interior angles =(n x 180-360)nSum of the interior anglesn=120+125+130+ * s to n termsn=(n / 2)[240+(n-1) 5] Therefore (n / 2)[240+(n-1) 5]=n x 180-360n => n^2-25 n+144=0 => (n-9)(n-16)=0 => n=9 or 16nBut when n=16, the greatest interior angle is 120^°+(16-1) 5^°=195^° which is not possible, for interior angle is $

The interior angles of a polygon are in A.P. The smalle... | Sequence & Series (Mathematics)

Let the number of sides of the polygon be

n

\nThen the sum of all the interior angles

=(n \times 180-360)

\nSum of the interior angles\n

=120+125+130+\cdots

n

terms\n

=\frac{n}{2}[240+(n-1) 5] \therefore \frac{n}{2}[240+(n-1) 5]=n \times 180-360

\Rightarrow n^2-25 n+144=0 \Rightarrow(n-9)(n-16)=0 \Rightarrow n=9

or 16\nBut when

n=16

, the greatest interior angle is

120^{\circ}+(16-1) 5^{\circ}=195^{\circ}

which is not possible, for interior angle is

<180^{\circ}

.\nHence the number of sides

=9

Explanation

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