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The interior angles of a polygon are in arithmetic progression. The smallest angle is 120°, and the common difference is 5°, Find the number of sides of the polygon.

The interior angles of a polygon are in arithmetic progression. The smallest angle is 120°, and the common difference is 5°, Find the number of sides of the polygon.

Grade:11

3 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Hello Student,
Please find the answer to your question
Let there be in n sides in the polygon.
Then by geometry, sum of all n interior angles of polygon = (n – 2) * 180°
Also the angles are in A. P. with the smallest angle = 120° , common difference = 5°
∴ Sum of all interior angles of polygon
= n/2[2 * 120 + ( n – 1) * 5
Thus we should have
n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180
⇒ n/2 [5n + 235] = (n – 2 ) * 180
⇒ 5n2 + 235n = 360n – 720
⇒ 5n2 – 125n + 720 = 0 ⇒ n2 – 25n + 144 = 0
⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9
Also if n = 16 then 16th angle = 120 + 15 * 5 = 195° > 180°
∴ not possible.
Hence n = 9.

Thanks
Aditi Chauhan
askIITians Faculty
Vivek Kumar
19 Points
6 years ago
Smallest angle=120degreeCommon difference=5A P is 120, 125, 130,……..The sum of interior angles of a polygon= (n-2)180Hence Sum of n terms of an A P = (n-2)180n/2 {2.120+(n-1)5} = 180(n-2)5n^2 -125n +720 = 0n^2 -25n +144=0n=9 or 16hence number of sides can be 9 or 16
John michael
15 Points
4 years ago
Angles of polygon =(n-2)180°
Sum of first n terms s=n/2(2a+(n-1)d)
Let substitute the data of suitable place ,
And equate the relation
Angle of polygon = sum of nth terms of sequence

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