Q) 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. 

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 1^st km is Rs 15. clearly this becomes our 1^st 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 2^nd  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 1^st km (Rs. 15) will be counted PLUS fare for 2^nd km (Rs. 8) will be counted + PLUS fare for 3^rd

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 1^st km (Rs. 15) will be counted PLUS fare for 2^nd km (Rs. 8) will be counted + PLUS fare for 3^rd km (Rs. 8) will be counted + PLUS fare for 4^th 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 4^th 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 n^th term

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