Question

the angles of apolygon are in AP with common difference 5 if the smallest angle is 120degree find the number of sides of the polygon.?​

Answer 1

Concepts used:

☞ Sum of interior angles of a polygon = (n - 2) × 180°

☞ Sum of n terms = (n ÷ 2) [2a + (n - 1)d]

Solution:

It is given that:

• First term = a = 120°
• Common difference = d = 5

According to the question,

Sum of interior angles = Sum of n terms

⇒ (n - 2) × 180 = (n ÷ 2) [2a + (n - 1)d]

⇒ (n - 2) × 180 = (n ÷ 2) [2 × 120 + (n - 1)5]

⇒ (n - 2) × 180 = (n ÷ 2) [240 + 5n - 5]

⇒ (n - 2) × 180 = (n ÷ 2) [5n + 235]

⇒ (n - 2) × 180 = [n (5n + 235)] ÷ 2

⇒ 180n - 360 = (5n² + 235n) ÷ 2

⇒ 2(180n - 360) = 5n² + 235n

⇒ 360n - 720 = 5n² + 235n

⇒ 5n² + 235n = 360n - 720

⇒ 5n² + 720 = 360n - 235n

⇒ 5n² + 720 = 125n

⇒ 5n² - 125n + 720 = 0

Dividing by 5 on both sides,

➥ n² - 25n + 144 = 0

Factorise the above quadratic equation by splitting the middle term.

n² - 16n - 9n + 144 = 0

➝ n (n - 16) -9 (n - 16) = 0

➝ (n - 9) (n - 16) = 0

∴ n = 16 or 9

If n = 16

16th angle

= a + (n - 1)d

= 120 + (16 - 1)5

= 120 + 15(5)

= 120 + 75

= 195°

This is not possible because an angle in a polygon cannot be greater than 180°

∴ n = 9