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In which of the following situations , d...

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 .

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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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