i) Two least angles are 3° & 5° and all angles are in AP
ii) So the series is: 2, 5, 7, 9, .......
Sum of all these angles = (n/2){2*3 + (n - 1)*(2)}, since a = 2 and d = 2
Sum all angles = 360°, since it is a circle
Equating both above, (n/2)(6 + 2n - 2) = 360
==> n² + 2n - 360 = 0
Factorizing, (n + 20)(n - 18) = 0
So either n = 18 or -20
But counts can't be < 0; so n = 18
Thus there are 18 sectors.
ii) So the series is: 2, 5, 7, 9, .......
Sum of all these angles = (n/2){2*3 + (n - 1)*(2)}, since a = 2 and d = 2
Sum all angles = 360°, since it is a circle
Equating both above, (n/2)(6 + 2n - 2) = 360
==> n² + 2n - 360 = 0
Factorizing, (n + 20)(n - 18) = 0
So either n = 18 or -20
But counts can't be < 0; so n = 18
Thus there are 18 sectors.