Question

A circle is divided into 18 sectors such that the angles
subtended at the centre form an anthmetic
progression. Given that the
angle of the smallest sector is 11.5 , find the angle of the biggest
sector​

Answer 1

Answer:Let the common difference be d.

Let Tr​ denote the angle of the rth sector.

Using AP formula:

Tr​=8+(r−1)d

72=8+(n−1)d

⟹(n−1)d=64            (1)

Sum of all angles: 360o

Sum of first n terms of AP is given by:2n​(2a+(n−1)d)

2n​(16+64)=360

⟹n(40)=360⟹n=9

Substituing n=9 in (1):

d=8

Therefore, angle of fifth sector = T5​=8+4⋅8=40o​