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 $8^o$ and the largest is ${72}^o$ then the angle in the fifth sector is ?

• A. ${40}^o$
• B. ${35}^o$
• C. ${42}^o$
• D. ${43}^o$

Answer 1

The correct option is A ${40}^o$

Let the common difference be $d$.
Let $T_r$ denote the angle of the $r^{\text{th}}$ sector.
Using AP formula:
$T_r=8+(r-1)d$
$72=8+(n-1)d$
$⟹(n-1)d=64$ (1)
Sum of all angles: ${360}^o$
Sum of first n terms of AP is given by:$\frac{n}{2}(2a+(n-1\left)d\right)$
$\frac{n}{2}(16+64)=360$
$⟹n\left(40\right)=360⟹n=9$
Substituing $n=9$ in (1):
$d=8$
Therefore, angle of fifth sector = $T_5=8+4\cdot 8={40}^o$