#### Questions

In which of the following situations, does the list of numbers involved make an arithmetic progression and why?

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 which of the following situations, the sequence of numbers formed will form an A.P.?

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

#### Solution 1

let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes `1/4` of air remaining in the cylinder at a time.

In other words, after every stroke, only `1 - 1/4 = 3/4` part of air will remain.

Therefore, volumes will be V, `3/4V , (3/4)^2V , (3/4)^3V`...

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

#### Solution 2

Here, let us take the initial amount of air present in the cylinder as 100 units.

So,

Amount left after vacuum pump removes air for 1^{st} time= `100 - (1/4) 100`

= 100 - 25

= 75

Amount left after vacuum pump removes air for 2^{nd} time= `75 - (1/4)75`

= 75 - 18.75

= 56.25

Amount left after vacuum pump removes air for 3^{rd} time= `56.25 - (1/4) 56.25`

= 56.25 - 14.06

= 42.19

Thus, the amount left in the cylinder at various stages is 100, 75, 56, 25, 42, 19

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here

`a_1 - a = 75 - 100`

= -25

Also

`a_2 - a_1 = 56.25 - 75`

= -18.75

Since `a_1- a != a_2 - a_1`

The sequence is not an A.P.