See attachment for illustration
Let a1 be 2/3(50) = 100/3
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Since this is in infinite geometric progression then we'll use
[tex]S_n = \frac{a_1}{1-r} [/tex]
where r is the common ratio
[tex]S_n=\frac{100/3}{1- \frac{2}{3} }[/tex]
[tex]S_n = 100[/tex]
Since the distance traveled by the ball is twice 2/3 of the previous then we'll have Sn multiplied by 2
[tex]S_n = 100(2)[/tex]
[tex]S_n = 200 ft[/tex]
But remember that we have not included the initial 50ft distance traveled since a1 is 2/3 of 50.
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Total distance traveled by the ball then is
[tex]d_{total} = 200 + 50[/tex]
[tex]d_{total} = 250ft[/tex]
