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`        The angles of a convex n-sided polygon form an arithmetic progression whoose common difference (in degrees) is an non-zero integer. Find the largest possible value of n for which this is possible. (A polygon is convex if it's interior angles are all less than 180°.)`
5 months ago

```							hello utsav this is a wonderful question. we will use the fact that sum of internal angles of all polygons (convex as well as concave) is equal to 180(n-2), n being the number of sides.let a1, a2, a3, …......an be the angles in increasing order. obviously a1 is greater than zero and an is less than 180.now, sum of angles cn also be writen as n/2*(2a1+(n-1)d)= 180(n-2) or a1= 180(1 –  2/n) – (n – 1)d/2an= a1+(n-1)d= 180(1 –  2/n) – (n – 1)d/2+(n-1)d= 180(1 –  2/n) + (n – 1)d/2 must be less than 180or (n – 1)d/2 must be less than 360/nor 720/n(n – 1) must be greater than dbut d is greater than equal to 1.so 720/n(n – 1) must be greater than 1or 720 must be greater than n(n – 1)or n^2 – n – 720 less than 0by wavy curve, you get n less than equal to 27.33 but since n is an integer,nmax= 27kindly approve :)
```
5 months ago
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