Logarithm/Geometric sequence word problem help.

[Bird]

Suppose that you drop a ball from a window 25 meters above the ground. The ball bounces to 75% of its previous height with each bounce. Find the total number of meters the ball travels between the time it is dropped and the 10th bounce.
A) 133.7 b)163.7 c) 185.0 d) 188.7 e) 196.0

If it helps I'm in math 12. I have a sheet with formulas ,but they are just basic geometric sequence formulas.

 

[wjm11]

Suppose that you drop a ball from a window 25 meters above the ground. The ball bounces to 75% of its previous height with each bounce. Find the total number of meters the ball travels between the time it is dropped and the 10th bounce.
A) 133.7 b)163.7 c) 185.0 d) 188.7 e) 196.0

If it helps I'm in math 12. I have a sheet with formulas ,but they are just basic geometric sequence formulas.

Draw yourself a sketch of the ball bouncing 10 times.
Generate the sequence, writing the height the ball bounces to each time on your sketch.

So, we start at 25, then bounce, up to 18.75 and back down again, then bounce,... and so on. Stop at the 10th bounce. The calculations should now become obvious.

 

[Bird]

I worked it out the long way (calculating each bounce and adding them up) and got 116.3 meters which unfortunatly isn't one of the answers.
I tried using a formula (Sn= a(1-r^n)/1-r) which gave me an impossible answer of 69.8. There's also 3 other questions like this on the assignment so there must be a way to do this question.
I also noticed that using percents tends to mess up the answers I get from the formulas ,making the supposed sum less than the 'a' value.

 

[soroban]

Hello, Bird!

It took a few tries to get the answer . . .

Suppose you drop a ball from a window 25 meters above the ground.
The ball bounces to 75% of its previous height with each bounce.
Find the total distance the ball travels from the time it is dropped to the 10th bounce.

. . (A) 133.7. . (B) 163.7. . (C) 185.0. . (D) 188.7. . (E) 196.0

I assume that "to the 10th bounce" means when the ball strikes the ground for the 10th time.

First, write 75% as \(\displaystyle \tfrac{3}{4}\)


Let's chart the movement of the ball . . .

. . \(\displaystyle \begin{array}{ccc} 25 & \downarrow & 1\\ \\ (\frac{3}{4})(25) & \uparrow \\ \\ (\frac{3}{4})(25) & \downarrow & 2\\ \\ (\frac{3}{4})^2(25) & \uparrow \\ \\ (\frac{3}{4})^2(25) & \downarrow & 3\\ \\ \vdots \\ \\ (\frac{3}{4})^9(25) & \uparrow \\ \\ (\frac{3}{4})^9(25) & \downarrow & 10\end{array}\)


The total distance is:
.. \(\displaystyle D \;=\;25 + 2\bigg[(\frac{3}{4})(25) + (\frac{3}{4})^2(25) + \hdots + (\frac{3}{4})^9(25)\bigg] \)

. . \(\displaystyle D \;=\;25 + 2(\frac{3}{4})(25)\underbrace{\bigg[ 1 + \tfrac{3}{4} + (\tfrac{3}{4})^2 + \hdots + (\tfrac{3}{4})^8\bigg]}_{\text{geometric series}}\)

The geometric series has the sum: .\(\displaystyle \dfrac{1 - (\frac{3}{4})^9}{1-\frac{3}{4}} \:=\:\dfrac{242,\!461}{65,\!536}\)


\(\displaystyle \text{Therefore: }\ \;=\;25 + \dfrac{75}{2}\!\cdot\!\dfrac{242,\!461}{65,\!536} \;=\;163.7372971 \;\hdots\text{ answer (B)}\)