Arithmetic progression

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If the interior angles of a polygon are in $A P$ . with common difference 5° and smallest angle is 120°, then the number of sides of the polygon is

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Solution

Let there be $n$ sides of the polygon. Then, the sum
of all interior angles is $(2 n - 4)$ right angles.
$\begin{array}{l} \frac{n}{2} {240 + (n - 1) 5} = (2 n - 4) 90 . \\ n^{2} - 25 n + 144 = 0 \Rightarrow n = 9, 16 \end{array}$
But, for $n = 16$ , the largest angle $= 120 + (16 - 1) 5 = 195^{0}$ which is impossible.
Hence, $n = 9 .$

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Arithmetic progression

Remember concepts with our Masterclasses.

If the interior angles of a polygon are in AP. with common difference 5° and smallest angle is 120°, then the number of sides of the polygon is

Let there be n sides of the polygon. Then, the sumof all interior angles is (2n - 4) right angles.n2{240+(n-1)5}=(2n-4)90.n2-25n+144=0⇒n=9,16But, for n=16, the largest angle =120+(16-1)5=1950which is impossible.Hence, n=9.

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If the interior angles of a polygon are in $A P$ . with common difference 5° and smallest angle is 120°, then the number of sides of the polygon is