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

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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 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.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum.

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

Solution: Fare for 1st km = Rs. 15

Fare for 2nd km = Rs. 15 + 8 = Rs 23

Fare for 3rd km = Rs. 23 + 8 = 31

Here, each subsequent term is obtained by adding a fixed number (8) to the previous term.

Hence, it is an AP

(ii) The amount of air present in the cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

Solution: Let us assume, initial quantity of air = 1

Therefore, quantity removed in first step = ¼

Remaining quantity after first step

=1-1/4 = 3/4

Quantity removed in second step

= 3/4x1/4 = 3/16

Remaining quantity after second step

= 3/4 - 3/16 = 9/16

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

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

Solution: Cost of digging of 1st meter = 150

Cost of digging of 2nd meter = 150 + 50 = 200

Cost of digging of 3rd meter = 200 + 50 = 250

Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.

Hence, it is an AP.

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

Solution: Amount in the beginning = Rs. 10000

Interest at the end of 1st year @ 8% = 10000 x 8% = 800

Thus, amount at the end of 1st year = 10000 + 800 = 10800

Interest at the end of 2nd year @ 8% = 10800 x 8% = 864

Thus, amount at the end of 2nd year = 10800 + 864 = 11664

Since, each subsequent term is not obtained by adding a fixed number to the previous term; hence, it is not an AP.