1. ## Arthimetic Series 3

A number of interlocking rings each 1 cm thick are hanging from a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance from the top of the top ring and the bottom of the bottom ring?

Here is a picture of what this thing looks like:

So, this is my attempt:

t1 = 20 cm d = -1 cm tn = 3 cm n = ? Sn = ? (We are trying to figure out Sn?)

First, I try to solve for n:

tn = t1 + (n-1)d

3 = 20 + (n-1)-1

3 = 20 + (n+1)

-17 = n + 1

n = -18

Now, I try to solve for Sn and plugging in the tn I found:

Sn = n/2 (t1 + tn)

Sn = -18/2 (20 + 3)

Sn = -207

The number shouldn't be positive...

The answer is actually 173 cm... I'm not sure where I went wrong... :/

2. ## Re: Arthimetic Series 3

You are taking circumference whereas we have to consider diameter to find the distance from top to bottom

3. ## Re: Arthimetic Series 3

I'm not sure how to do this question... Could you elaborate more?

4. ## Re: Arthimetic Series 3

You are going on the right track. Firstly it is the diameter which is given so you are right. In calculation n cannot be negative: You have gone wrong in opening bracket.
3 = 20 + (n-1)(-1)
3 = 20 -n +1
That will give n = 18
Now the sum of these terms is also correct i.e., 207
Now out of total 18 rings we have double length ( overlapping 1 cm each so we lose 32 cm on that count and 1 cm at top and bottom. taht makes a total of 34 cm . So the total distance would be 207-34 = 173 cm

### a number of interlocking rings each 1 cm thick are hanging from a peg

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