A circle is divided into 18 sectors such that the angle subtended at the centre form an arithmetic progression.Given the angle of the smallest sector is 115 degree.Find
a)the common difference
b)the angle of the biggest sector.
If I understand the problem correctly then the smallest angle cannot be 115, could you mean 11.5 instead?
If so then you basically have 18 angles starting at 11.5 and increasing by a value n. All angles total 360 therefore:
11.5 + (11.5 + n) + (11.5 + 2n) + ... + (11.5 + 17n) = 360
n works out nicely actually if 11.5 was what you meant. Can you solve it from here or do you need more help?
a) $\displaystyle S_n = \frac{n}{2}(2a+(n-1)d)$
We know that
- a = 11.5
- n = 18
$\displaystyle S_{18} = \frac{18}{2}(23+17d) = 2\pi$
Solve for d.
NB: if you're using degrees set $\displaystyle S_{18}$ equal to $\displaystyle 360^{\circ}$
b) The biggest sector is the last (18th) one
$\displaystyle U_{18} = a + (n-1)d = 11.5 + 17d$