Question

the amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

Answer 1

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

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Answer 2

Let the volume of air in a cylinder, initially, be V litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3…and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.