Question

In the following situation, 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 the following situation, the sequence of number 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.

Answer 1

Let the quantity of air in the cylinder be 1.

T₁ = 1

Air removed = `1/4`

`"T"_2 = 1 - 1/4 = (4 - 1)/4 = 3/4`

Air removed = `3/4 xx 1/4 = 3/16`

`"T"_3 = 3/4 - 3/16 = (12 - 3)/16 = 9/16`

Air removed = `9/16 xx 1/4 = 9/64`

`"T"_4 = 9/16 - 9/64 = (36 - 9)/64 = 27/64`

Series: `1, 3/4, 9/16, 27/64`

`"d"_1 = 3/4 - 1 = (-1)/4`

`"d"_2 = 9/16 - 3/4 = (-3)/16`

Here the common difference is not the same hence it is A.P. Not there.

Answer 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.