
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:
http://i.snag.gy/IJ7kX.jpg
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 + (n1)d
3 = 20 + (n1)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... :/

Re: Arthimetic Series 3
You are taking circumference whereas we have to consider diameter to find the distance from top to bottom

Re: Arthimetic Series 3
I'm not sure how to do this question... Could you elaborate more?

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 + (n1)(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 20734 = 173 cm