Question

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.

Answer 1

Answer: We can write the given condition as;

Taxi fare for 1 km = 15

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

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

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

And so on……

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

Answer 2

Step-by-step explanation:

Fare for first km = Rs. 15

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

Fare for third km = Rs. 23+8=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 .....1)

Therefore, quantity removed in first step = 41

Remaining quantity after first step1− 41 = 43 ...Quantity removed in second step= 43× 41=163

Remaining quantity after second step= 43− 163 = 169 ....3)

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 1

st

meter =150

Cost of digging of 2

nd

meter =150+50=200

Cost of digging of 3

rd

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 1

st

year @ 8% = 10000×8%=800

Thus, amount at the end of 1

st

year =10000+800=10800

Interest at the end of 2

nd

year @ 8% = 10800×8%=864

Thus, amount at the end of 2

nd

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.