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Math Help - Arthimetic Series 3

  1. #1
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    Exclamation 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... :/
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  2. #2
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    Re: Arthimetic Series 3

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

    I'm not sure how to do this question... Could you elaborate more?
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  4. #4
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    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
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