A circular cake of radius $r$ is cut into 22 sectors. The areas of the sectors form an increasing arithmetic progression. The area of the eighth sector is twice the area of the smallest sector. Find, in terms of $\pi$, the angle of the largest sector. [Area of Sector = $\tfrac{1}{2} r^2 \theta$ where $\theta$ is the angle of the sector]

1. Establish a relationship between the area of the smallest sector and the common difference.

Let the area of the smallest sector be $a$ and the common difference of the arithmetic progression be $d$.
Given that the area of the eighth sector is twice the area of the smallest sector:
$A_8 = 2A_1$
$a + 7d = 2a$
$a = 7d$

2. Set up an equation for the sum of all 22 sectors and solve for $d$.

The sum of the areas of all 22 sectors is the total area of the circular cake, which is $\pi r^2$.
$S_{22} = \frac{22}{2}[2a + (22 - 1)d] = \pi r^2$
$11[2(7d) + 21d] = \pi r^2$
$11[14d + 21d] = \pi r^2$
$11[35d] = \pi r^2$
$385d = \pi r^2$
$d = \frac{\pi r^2}{385}$

3. Calculate the area of the largest sector ($A_{22}$).

The largest sector is the 22nd sector.
$A_{22} = a + 21d$
$A_{22} = 7d + 21d = 28d$
$A_{22} = 28 \left( \frac{\pi r^2}{385} \right)$
$A_{22} = \frac{4}{55} \pi r^2$

4. Find the angle of the largest sector.

Let the angle of the largest sector be $\theta$.
Using the area of a sector formula:
$\frac{1}{2} r^2 \theta = \frac{4}{55} \pi r^2$
$\frac{1}{2} \theta = \frac{4}{55} \pi$
$\theta = \frac{8\pi}{55}$