**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 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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