A circle is completely divided into n sectors in such a way that the angles
Question:

A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 80 and the largest is 720 , calculate n and the angle in the fifth sector.

Solution:

A circle is divided into n sectors.

Given,

Angles are in A.P

Smallest angle $=a=8^{\circ}$

Largest angle $=1=72^{\circ}$

Final term of last term of an A.P series is l = a + (n – 1)×d

So,

$72^{\circ}=8^{\circ}+(n-1) \times d$

$(n-1) \times d=64^{\circ} \longrightarrow(1)$

Sum of all angles of all divided sectors is $360^{\circ}$

Sum of $n$ terms of A.P whose first term and the last term are known is $\frac{n}{2}\{a+1\}$

Where n is the number of terms in A.P.

So,

$\frac{\mathrm{n}}{2}\left\{8^{\circ}+72^{\circ}\right\}=360^{\circ}$

$n\left(40^{\circ}\right)=360^{\circ}$

$n=\frac{360^{\circ}}{40^{\circ}}$

$n=9 \rightarrow(2)$

From equations (1) & (2) we get,

$(9-1) \times d=64^{\circ}$

$8 \times d=64^{\circ}$

$\mathrm{d}=\frac{64^{\circ}}{8}$

$d=8^{\circ}$

The circle is divided into nine sectors whose angles are in A.P with a common difference of 8°.

Angle in fifth sector is a $+(5-1) \times d=40^{\circ}$

∴ n = 9

The angle in the fifth sector = 40°.