A disc is cut into 12 sectors with areas in arithmetic progression. The largest angle is twice the smallest. Find the smallest angle. I was given a question which states -

A circular disc is cut into twelve sectors whose areas are in an
  arithmetic sequence. The angle of the largest sector is twice the
  angle of the smallest sector.
Find the size of the angle of the smallest sector.

Here's what I have done so far,
$\Rightarrow U_1 = w$
$\Rightarrow U_n = 2w$
$\Rightarrow w + (n - 1)d = 2w$ 
$\Rightarrow n = 12$
$\therefore\ d = \frac{w}{11}$
And now I am clueless about where to go from here...

You got $d = a$1$/11$.
($a$1 is angle of the smallest sector)
We can write: $11d = a$1   ........(1)
Now, we know:
Area of sector = ${\frac{1}2}.{r^2}.\theta$
And,
Sum of area of all sectors = Area of circle
(${\frac{1}2}.{r^2}.a$1) + (${\frac{1}2}.{r^2}.a$2) + ..... + (${\frac{1}2}.{r^2}.a$12) = $\pi.r^2$
${\frac{1}2}.r^2$.{ $a$1 + $a$2 + ..... + $a$12 } = $\pi.r^2$
$a$1 + $a$2 + ..... + $a$12 = 2$\pi$
Applying the formula for the sum of $n$ terms of an AP:
${\frac{12}2}.${ $2a$1 + $(12-1).d$ } = 2$\pi$
$6$.{ $2a$1 + $11d$ } = 2$\pi$
Putting value of $11d$ from (1):
$6$.{ $2a$1 + $a$1 } = 2$\pi$
$a$1 = $\frac{\pi}9$
