Bouncing ball geometric sequence question When a ball falls vertically off a table, it rebounds 75% of its height after each bounce. If it travels a total distance of 490 cm, how high was the table top above the floor?
The trouble I am having is that the ball will never actually stop bouncing because it keeps re
bouncing 75% of the previous height each time. So how will I solve this?
 A: Here is a method which avoids summing a geometric progression (at least it hides a method for computing the sum). 
Call the height of the table $h=4a$ and the total distance travelled $d$. To the top of the first bounce the ball travels down $4a$ and back up $3a$. The remainder of the track of the ball is exactly as if it had fallen off a table of height $3a$ - and simply by scaling we see that this would be $\frac 34d$. So we have $7a+\frac34d=d$ and it is easy to solve from there.
A: Let $h_n$ be the height of the ball after the $n$-th bounce, and $h>0$ the height of the table. Then we have
$$
h_0=h,\ h_n=\frac34 h_{n-1} \quad \forall n \ge 1,
$$
i.e.
$$
h_n=\left(\frac34\right)^nh_0=\left(\frac34\right)^nh.
$$
The total distance that the ball travels after its fall is
$$
d=h_0+2\sum_{n=1}^\infty h_n=\left[-1+2\sum_{n=0}^\infty\left(\frac34\right)^n\right]h=\left(-1+\frac{2}{1-\frac34}\right)h=7h.
$$
Thus
$$
h=\frac17d=70 \text{ cm }.
$$
A: Using $S= a/(1-r)$ and allowing $x$ to be the original height you have the following equation. (Note:  the original drop is only added to the series because it is the drop distance.)
$$ x+\frac{ 2x(.75)}{ 1- .75 }=490$$
We double the $x$ in the numerator which is equal to (a) because after the original drop the ball goes up then comes back down, this makes us double the distance.
Solving we get: $x+6x=490$
 $$7x=490$$
Answer:             x=70cm
