In which of the following situations , does the list of numbers of involved make an arithmetic progression , and why ?
The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km .

Answer

To determine whether the taxi fare after each kilometer forms an arithmetic progression (AP), we can follow these steps:

### Step 1: Identify the fare structure
The taxi fare is structured as follows:
- For the first kilometer, the fare is Rs. 15.
- For each additional kilometer, Rs. 8 is added.

### Step 2: Write the fare for each kilometer
Let’s denote the fare for each kilometer:
- For the 1st kilometer (A1): Rs. 15
- For the 2nd kilometer (A2): Rs. 15 + Rs. 8 = Rs. 23
- For the 3rd kilometer (A3): Rs. 15 + 2 * Rs. 8 = Rs. 31
- For the 4th kilometer (A4): Rs. 15 + 3 * Rs. 8 = Rs. 39

Thus, the sequence of fares is:
- A1 = 15
- A2 = 23
- A3 = 31
- A4 = 39

### Step 3: Calculate the common difference
Now, we need to find the difference between consecutive terms:
- Difference between A2 and A1:
\( A2 - A1 = 23 - 15 = 8 \)

- Difference between A3 and A2:
\( A3 - A2 = 31 - 23 = 8 \)

- Difference between A4 and A3:
\( A4 - A3 = 39 - 31 = 8 \)

### Step 4: Check if the differences are equal
Since the difference between each pair of consecutive terms is the same (8), we can conclude that the fare forms an arithmetic progression.

### Conclusion
The taxi fare after each kilometer does indeed form an arithmetic progression because the difference between consecutive fares is constant.

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