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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...

Blue
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JonDoe
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    In the last equation, surely you mean $$ d= \frac{w}{11} $$ so now $d$ is the increment between the different terms. The next step is to write an expression for the sum of the sectors. On the other hand, the total angle is $360^{\circ}$ (or $2\pi$), and on the other hand it's the sum of the arithmetic sequence. Please write this as an equation. – Matti P. Aug 12 '19 at 06:17
  • @MattiP. - I'm so sorry, you're right. Must have made that mistake in a hurry. Thanks! – JonDoe Aug 12 '19 at 06:20

3 Answers3

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Recall that the sum of twelve sectors is the whole disc, $$360^{\circ}=\sum_{k=1}^{12} (w+(k-1)d)=12w+d\sum_{k=1}^{12}(k-1)=12w+66d$$ which, together with $w + (12 - 1)d = 2w$, i.e. $11d=w$ give you all you need to find $w$, that is the angle (in degrees) of the smallest sector: $$360^{\circ}=12w+66d=12w+6w\implies w=\frac{360^\circ}{18}=20^\circ.$$

Robert Z
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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 = $\frac{\pi}9$

Astitva S.
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You could also use the formula -

$\frac{n}{2}(U_1 + U_n)$

Like this,

$\Rightarrow\ \frac{12}{2}(w^{\circ} + 2w^{\circ}) = 360^{\circ}$

$\Rightarrow\ 6(3w^{\circ}) = 360^{\circ}$

$\Rightarrow\ w^{\circ} = \frac{360^{\circ}}{18}$

$\therefore\ w^{\circ} = 20^{\circ}$

Community
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JonDoe
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    This is the right answer: ignore the differences and jump straight to the sum. – Neil Aug 12 '19 at 20:51
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    @Justin I realise now that you accepted your own answer... I thought that this was not possible here, but I was wrong. https://math.meta.stackexchange.com/questions/21221/accepting-your-own-answer – Robert Z Sep 10 '19 at 12:16