Question

(a) A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic
progression. The angle of the largest sector is 4 times the angle of the smallest sector.
Giventhat the radius of the circle is 5 cm, find the perimeter of the smallest sector. [6]

(b) The first, second and third terms of a geometric progression are 2k + 3, k + 6 and k, respectively.
Given that all the terms of the geometric progression are positive, calculate
(i) the value of the constant k, [3]
(ii) the sum to infinity of the progression. [2]

Answer 1

okay do you know what an arithmetic progression looks like

Answer 2

yea

Answer 3

okay so if the 6th sector is 4 times bigger than the first sector what must be the progression factor?

Answer 4

for the 6 sectors
lets say the first sector was a angles
a,a+r,a+2r,a+3r....,
now
a+a+r+a+2r+a+3r...a+5r=360
and u know that
a+5r=4a

Answer 5

let the angels be=a, a+b, a+2b,a+3b,a+4b and a+5b
now a+5b=4a
=>b=3a/5;

now total angle in circle = 360
so, 6a+15b = 360
a=24
or a = 2pi/15;

now perimeter = 2r+ar/pi
=2r+2r/15
=32r/15
=96cm

Answer 6

2 equations 2 unknowns solve for a and r, and the he rest