ty divided into 'n'sectors in such a way that the angles at the centre

1.	ty divided into 'n'sectors in such a way that the angles at the centre of circle are inA.P. If the smallest angle is 8 and the largest is 72', caleulate n and the angle of the 3"d sector taking thesmallest angle as first.
Answer» First angle, a = 8°nth angle, an = 72°let the common difference = d sum of angles = 360°⇒ (n/2)[a + an] = 360⇒(n/2)[8+72] = 360⇒(n/2)[80] = 360⇒ 40n = 360⇒ n = 360/40⇒ n = 9 72 = 8+(n-1)d⇒ 72 - 8= (9-1)d⇒ 64 = 8d⇒ d = 64/8⇒ d = 8° a₃ = a + (n-1)d⇒a₃ = 8 + (3-1)×8⇒a₃ = 24 n = 9angle of third sector = 24°

ty divided into 'n'sectors in such a way that the angles at the centre of circle are inA.P. If the smallest angle is 8 and the largest is 72', caleulate n and the angle of the 3"d sector taking thesmallest angle as first.

Answer»

First angle, a = 8°nth angle, an = 72°let the common difference = d

sum of angles = 360°⇒ (n/2)[a + an] = 360⇒(n/2)[8+72] = 360⇒(n/2)[80] = 360⇒ 40n = 360⇒ n = 360/40⇒ n = 9

72 = 8+(n-1)d⇒ 72 - 8= (9-1)d⇒ 64 = 8d⇒ d = 64/8⇒ d = 8°

a₃ = a + (n-1)d⇒a₃ = 8 + (3-1)×8⇒a₃ = 24

n = 9angle of third sector = 24°