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This is a geometric sequence with terms 100,50,25,12.5,...
The sum of a geometric sequence is S(n) = a(1-r^n)/(1-r) where r=common ratio and a=the 1st term
Here the common ratio is 50/100 = 1/2
For the ball to 10 yds from the hole, it has covered a distance of 190 yds,
so S(n) = 190
190 = 100(1-(1/2)^n)(1-(1/2))
Solve for n using logarithms -> n = 4.32
So after the 5th shot, the ball is less than 10 yds from hole
For 5 yds from the hole, solve the above with S(n) = 195
This gives n = 5.32
So after the 6th shot, the ball is < 5 yds away
In theory, the ball never covers the full 200 yds, since the remaining distance, while getting very small, is always non-zero.