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

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(ii) 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.

Solution (i):

The given condition can be written as;

Fare of taxi for 1 km = 15

Fare of taxi for first 2 kms = 15+8 = 23

Fare of taxi for first 3 kms = 23+8 = 31

Fare of taxi for first 4 kms = 31+8 = 39

And so on……

As a result, 15, 23, 31, 39… creates an A.P. because each subsequent phrase is 8 more than the previous one.

Solution(ii):

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

The vacuum pump removes 1/4th of the air in the cylinder at a time with each stroke. Or, to put it another way, 1-1/4 = 3/4th of the air will remain after each stroke.

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

Clearly, as we can see, the neighboring terms in this series do not have a common difference. As a result, this is not an A.P.