Q) In which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) Th
e taxi fare after each km when the fare is Rs 15 for the first km and Rs. 8 for each additional km.

Ans: Let’s first start making a sequence and then check if the sequence makes an AP or not..

It is given that the taxi fare for the 1st km is Rs 15. clearly this becomes our 1st term,

Next, it is given that the fare is Rs. 8 for each additional km.

It means if we go 2 kms, then fare for 1st km (Rs. 15) will be counted PLUS fare for 2nd  km (Rs. 8) will be counted.

Hence, Fare for 2 kms will become 15 + 8 = 23

Next, if we go 3 kms, then fare for 1st km (Rs. 15) will be counted PLUS fare for 2nd km (Rs. 8) will be counted + PLUS fare for 3rd

Hence, Fare for 3 kms will become = 15 + 8 + 8

or we can write it as: 15 + 2 x 8 = 15 + 16 = 31

This becomes 3rd term of sequence

Similarly, if we go 4 kms, then fare for 1st km (Rs. 15) will be counted PLUS fare for 2nd km (Rs. 8) will be counted + PLUS fare for 3rd km (Rs. 8) will be counted + PLUS fare for 4th km (Rs. 8) will be counted

Hence, Fare for 3 kms will become = 15 + 8 + 8 + 8

or we can write it as: 15 + 3 x 8 = 15 + 24 = 39

This becomes 4th term of sequence

Thus, the sequence starts to emerge as 15, 23, 31, 39,……

Here the difference between any two consecutive terms will be 8 because we have added 8 for each additional km.

Therefore, the difference will be common across the sequence

and since the common difference is same across any two consecutive terms, the sequence formed is an AP.

Note: This is very commonly used situation in practical life and after understanding this, now you can calculate fare for any distance you travel using formula for nth term

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