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a convergent geometric series with a=100 and r=1/10

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This problem is adapted from the famous Greek paradox "Achilles and the Tortoise": suppose a rabbit and a turtle are in a race. The rabbit is ten times as fast as the turtle, and the turtle is given a 100-foot head start. The paradox claims that the turtle will always be ahead of the rabbit since once the rabbit has covered the 100 feet, the turtle is now 10 feet ahead, and once the rabbit covers that 10 feet, the turtle is still 1 foot ahead and so forth. in reality, this is not a paradox (it is not a self-referencing, self-contradicting statement). It can be solved by a series. So... how far must the rabbit run to catch up to the turtle?

https://brainmass.com/math/algebraic-geometry/illustrating-convergent-geometric-series-431163

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The rabbit runs 100 feet the tortoise moves 10 feet. 10 = 1/10(100)
The rabbit runs 10 feet the ...

Solution Summary

This solution demonstrates a convergent geometric series with a=100 and r=1/10.

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