In Which of the Following Situations, Does the List of Numbers Involved Make as 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. - Mathematics

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 1st time= 100 - (1/4) 100

= 100 - 25

= 75

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

= 75 - 18.75

= 56.25

Amount left after vacuum pump removes air for 3rd 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.

Concept: Arithmetic Progression
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APPEARS IN

NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.1 | Q 1.2 | Page 99
RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.3 | Q 3.2 | Page 11