Question 765572
<pre>
The ball bounces to a 1/3 its initial height every time, which means the series of heights forms a geometric progression, with common ratio as 1/3
First term a = Initial height = 12
2nd term = 1st rebound = 12*1/3 = 4
3rd term = 2nd rebound = 4*1/3 = 4/3

So the height of the 8th rebound is given by the 9th term in the series.
9th term = a*r^(n-1) = 12*(1/3)^8 = 12/6561 = 4/2187

Height of rebound for 8th time = 4/2187.


2) Total distance travelled at the 10th time it hits ground is the sum of  the 
upward travel and the downward travels. Both are geometric progressions.

Downward travel:
1st term = 12
Common ratio = 1/3
Sum of 10 terms = 
{{{a*(1-r^10)/(1-r)}}}

{{{12*(1 - 1/59049)/(2/3)}}}

= {{{18*(59048/59049)}}}

= {{{2*59048/6561 = 118096/6561 = 17.999}}}

2) Upward (rebound) travel

1st term = 4
Common ratio = 1/3
Sum of 9 terms = (since it rebounds 9 times)
{{{a*(1-r^9)/(1-r)}}}

{{{4*(1 - 1/19683)/(2/3)}}}

= {{{(6)*(19682/19683)}}}

= {{{(118092/19683) = 5.999}}}

So the total travel = 17.999 + 5.999 = 23.998 approximately

As you would see, the sum is already approaching the "sum to infinity" of the
series.


:)