Question

The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

☝️☝️In this situation does the list of numbers involved make an arithmetic progression, and why☝️☝️



Plz answer fast with steps
Steps involved will get the brainliest answer.....

Answer 1

Answer:

no

it is not an arithmetic progression

Step-by-step explanation:

Let a be the amount of air initially Amount of air remaining after 1st pump =a –(a/4)=3a/4

Amount of air remaining after 2nd  pump= (3a/4) – (1/4)(3a/4)=9a/16

Amount of air remaining after 3rd  pump= (9a/16) – (1/4)(9a/16)=27a/64

So the series is like

a,3a/4,9a/4,27a/64……..

Difference Ist and second term=-a/4

Difference between Second and Third term= -3a/16

So difference is not constant

So it is not Arithmetic Progression

Answer 2

Answer:

Step-by-step explanation:

Let the amount of air present in the cylinder X units.

1/4×x=3x/4 units.

1/4×3x/4=3x/16units.

air remaining=3x/4-3x/16.

=12x-3x/16.

=9x/16.

x,3x/4,9x/16..........

Common difference (d)=9x/16-3x/4

=9x-12x/16

=-3x/16

Common difference (d)=3x/4 -x

=3x-4x/4

=-x/4.

List of number are not torming A.P.

Common difference is not same..