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

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

(ii) The amount of air present in a cylinder when a vacuum pump removes $\dfrac{1}{4}$ of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

(i) It can be observed that

Taxi fare for 1st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 + 8 = 31

Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 …… forms an A.P. because every term is 8 more than the preceding term.

Hence, the numbers involved forms an A.P. with common difference = 8.

(ii) Let the initial volume of air in a cylinder be V litre. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of air remaining in the cylinder at a time. In other words, after every stroke, only $1 - \dfrac{1}{4} = \dfrac{3}{4}$ part of air will remain.

Therefore, volumes will be $V, \dfrac{3}{4}V, \dfrac{3}{4}V \times \dfrac{3}{4}, \dfrac{3}{4}V \times \dfrac{3}{4} \times \dfrac{3}{4},$ ………..

⇒ $V, \dfrac{3}{4}V, \Big(\dfrac{3}{4}\Big)^2V, \Big(\dfrac{3}{4}\Big)^3V, ………..$

⇒ $V, \dfrac{3}{4}V, \dfrac{9}{16}V, \dfrac{27}{64}V, …….$

Difference between 2nd and 1st term :

$\dfrac{3}{4}V - V = \dfrac{3 - 4}{4}V = -\dfrac{1}{4}V$

Difference between 3rd and 2nd term :

$\dfrac{9}{16}V - \dfrac{3}{4}V = \dfrac{9 - 12}{16}V = -\dfrac{3}{16}V$

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

Hence, numbers involved do not form an A.P.

(iii) Cost of digging for first metre = ₹150

Cost of digging for first 2 metres = ₹150 + ₹50 = ₹200

Cost of digging for first 3 metres = ₹200 + ₹50 = ₹250

Cost of digging for first 4 metres = ₹250 + ₹50 = ₹300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

Hence, the numbers involved forms an A.P. with common difference = 50.

(iv) We know that,

If ₹P is deposited at r% compound interest per annum for n years, money will be $P\Big(1 + \dfrac{r}{100}\Big)^n$ after n years.

After every year, money will be :

$10000\Big(1 + \dfrac{8}{100}\Big), 10000\Big(1 + \dfrac{8}{100}\Big)^2, 10000\Big(1 + \dfrac{8}{100}\Big)^3, ……….$

Clearly, adjacent terms of this series do not have the same difference between them.

Hence, numbers involved do not form an A.P.