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Question

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 14 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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Solution

(i) Fare for first km = Rs.15

Fare for second km = Rs.(15+8)= Rs.23

Fare for third km = Rs.(23+8)=Rs.31

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

Hence, it is in A.P.

(ii) Let us assume, initial quantity of air =1(i)

Therefore, quantity removed in first step =14

Remaining quantity after first step =114=34(ii)

Quantity removed in second step = 34×14=316

Remaining quantity after second step = 34316=916(iii)

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

Hence, it is not an AP.

(iii) 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) Amount in the beginning = Rs. 10000

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

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

Interest at the end of 2nd year @ 8% = 10800×8%=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.


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