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### Circle in n Sectors

```
Date: 7/8/96 at 4:25:1
From: Wong Cheong Siong
Subject: Circle in Sectors

Dear Dr. Math,

A circle is completely divided into n sectors in such a way that
the angles of the sectors are in arithmetic progression. If the
smallest of these angles is 8 degrees, and the largest is 52 degrees,
calculate n.

Best Regards,
Wong Cheong Siong
```

```
Date: 7/8/96 at 5:53:55
From: Doctor Anthony
Subject: Re: Circle in n Sectors

The sum of the angles must equal 360 degrees, and we use the formula
Sum of n terms = (n/2)(first term + last term)
(n/2)(8 + 52) = 360
(n/2)*60 = 360
n/2 = 6   and so n = 12
```

```
Date: 7/8/96 at 11:24:12
From: Doctor Brian
Subject: Re: Circle in n Sectors

Let n = the number of intermediate terms.
Let d = the common difference between terms.
We have 8 + (8 + d) + (8 + 2d) + ... + (8 + nd) + 52 = 360.
We also have that (8 + (n+1)d) = 52, because that would be the next
term in the arithmetic progression.

For the first equation, subtract the first 8 and the 52 to get that
(8 + d) + ... + (8 + nd) = 300.

Use the fact that the sum from 1 to n is 1/2 * n * (n+1), and use the
fact that there are n 8's to get the equation
8n + 1/2 * n * (n+1) * d = 300.

From the second equation you can subtract the 8 to get that (n+1)d =
44. This equation can be substituted verbatim into the other equation
to get

8n + 1/2 * n * 44 = 300.  This gives 8n + 22n = 300, 30n = 300,
n = 10, and substituting back, that d = 4.  The n = 10 represents the
intermediate steps, so there are a total of 12 arcs....8, 12, 16, 20,
... , 52.
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Sequences, Series

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