Acircle 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. given that the radius of the circle is 5cm, find the perimeter of the smallest sector.

It is a AP => a₁, a₂, a₃, a₄, a₅, a₆. (a being the angles in degree

a₆ =4a₁ (given) and  a₁ + a₂ + a₃ + a₄ + a₅ + a₆ =360°(given & R=5 cm (given)
and n, the number of terms = 6

Moreover a₁ is the smallest angle and a₆ the largest

We have to calculate a₁: In a AP, the sum S= (a₁ +a₆)n/2 ==>
S=(a₁ + 4a₁)6/2 = 360° ==> (5a₁)(3) =360° ==> a₁ =24°
Now let's find the length of the sector which central angle = 24°
Perimeter of a circle =2πR = 2π.5 =10π (or 10x180°)
The central angle =24°, so the length of the arc = (10x180)/24 = 75 cm
To the length of the arc we have to add the 2 Radius (5x2=10) ,
Hence the PERIMETER of the sector =75+10 =85 cm

It’s going to be d. 4x^2-4x-19’