geometric progression ?

**dugongster** · December 5th 2010, 10:09 PM

"If one-fourth of the air in a tank is removed by each stroke of an air pump, find the fractional part of the air remaining after seven strokes of the pump ?"

I'm getting 0.5 after subtracting the sum of (1/4 + 3/16 + 3/64 ... 3/16384). Am I doing something wrong ? For sure 0.5 is not the correct answer as the textbook says.

Thank you once again !

**HallsofIvy** · December 6th 2010, 03:32 AM

Your "1/4" and "3/16" are correct but you have the wrong pattern after that. After removing 1/4+ 3/16= 7/16 of the you have 9/16 left and so the third stroke remove 1/4(9/16)= 9/64, not 3/64.

I think it is simpler to calculate how much is left after each pump. After the first pump, 1/4 of the air is removed leaving 3/4, 1/4 of that is 3/16 leaving 9/16, 1/4 of that is 9/64 leaving 27/64, etc. You should be able to see that you are getting powers of 3/4. After n pumps, $\left(\frac{3}{4}\right)^n$ of the air is left.

**Archie Meade** · December 6th 2010, 04:58 AM

Yes, after each stroke, 0.75 of the amount of air before the stroke will remain.

After 1 stroke, $\frac{3}{4}$ of the air remains.

After 2 strokes, $\frac{3}{4}\left[\frac{3}{4}\right]$ of the air remains....

and on.

Working with the amount removed is not as compact...

After 1 stroke, the remaining fraction of the original amount of air is

$1-\frac{1}{4}$

After 2 strokes, the fraction that remains is

$\left(1-\frac{1}{4}\right)-\frac{1}{4}\left[1-\frac{1}{4}\right]$

After 3 strokes, the fraction that remains is

$\left(1-\frac{1}{4}\right)-\frac{1}{4}\left(1-\frac{1}{4}\right)-\frac{1}{4}\left[\left(1-\frac{1}{4}\right)-\frac{1}{4}\left(1-\frac{1}{4}\right)\right]$

and on.

This gives a pattern of...

1 stroke...... $\frac{3}{4}$

2 strokes.... $\frac{3}{4}-\frac{1}{4}\;\frac{3}{4}=\frac{3}{4}\left(1-\frac{1}{4}\right)=\frac{3}{4}\;\frac{3}{4}$

3 strokes.... $\frac{3}{4}-\frac{1}{4}\;\frac{3}{4}-\frac{1}{4}\left(\frac{3}{4}-\frac{1}{4}\;\frac{3}{4}\right)=\frac{3}{4}\left[1-\frac{1}{4}-\left(\frac{1}{4}-\frac{1}{4}\;\frac{1}{4}\right)\right]$

$=\frac{3}{4}\left[\frac{3}{4}-\frac{1}{4}\left(1-\frac{1}{4}\right)\right]=\frac{3}{4}\left[\frac{3}{4}-\frac{1}{4}\;\frac{3}{4}\right]=\frac{3}{4}\left(\frac{3}{4}\right)\left[1-\frac{1}{4}\right]$

shows the pattern

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