1. ## Geometric progression

Each stroke of a pump removes 8.2% of the remaining air from a container. What percent of the air remains after 50 strokes?

I originally thought this was simple, use the formula $\displaystyle a_n=a_1r^{n-1}$
where
$\displaystyle a_1=1$
$\displaystyle r=0.918$
and $\displaystyle n=50$

substituting in these values i get
$\displaystyle a_{50}=1*0.918 ^{50-1}$
$\displaystyle a_{50}=0.015111$
and therefore ~1.51%
however, the solution in my textbook shows the intermediary step to be
$\displaystyle a_{50}=1*0.918 ^{50}$
and final solution as
$\displaystyle a_{50}=0.01387$

Which brings me to my main question: Could somebody explain to me why the textbook's solution shows $\displaystyle 0.918^{50}$ instead of $\displaystyle 0.918^{50-1}$? Is it something to do with $\displaystyle a_{1}=1$? a mistake in the book? or something else?

2. Hello, mattty!

You miscounted . . .

Each stroke of a pump removes 8.2% of the remaining air from a container.
What percent of the air remains after 50 strokes?

I originally thought this was simple, use the formula $\displaystyle a_n\:=\:a_1r^{n-1}$ . . . . no
where: .$\displaystyle a_1=1,\;r=0.918,\;n=50$

You should be using: .$\displaystyle a_n \:=\:a_1r^{{\color{red}n}}$

Check out the initial condition.

When $\displaystyle n = 0,\;a_o \:=\:a_1r^0 \quad\Rightarrow\quad a_o = a_1$

When no air has been pumped out, the amount is the initial amount, $\displaystyle a_1$

3. that's a forehead smack moment if i ever saw one...
yeah, i can see now, when $\displaystyle n=1$ 100% of the air has been pumped out
thanks for pointing out the obvious.... i need that kind of help a lot -_-