(ii) The amount of air present in a cylinder when a vacuum pump remove

1.	(ii) The amount of air present in a cylinder when a vacuum pump removes of the4air remaining in the cylinder at a time.
Answer» Let a be the amount of air initially Amount of air remaining after 1stpump =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 likea,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 constantSo it is not Arithmetic Progression

(ii) The amount of air present in a cylinder when a vacuum pump removes of the4air remaining in the cylinder at a time.

Answer»

Let a be the amount of air initially Amount of air remaining after 1stpump =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 likea,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 constantSo it is not Arithmetic Progression