📚 Question
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 ₹15 for the first km and ₹8 for each additional km.
📌 Given Information
- Fare for the first kilometer: ₹15
- Fare for each additional kilometer: ₹8
- We need to check if the cumulative fares form an AP
🎯 To Find
Whether the taxi fares form an Arithmetic Progression (AP) and justify the answer.
💡 Key Concept
Arithmetic Progression (AP): A sequence of numbers where the difference between consecutive terms is constant.
If \(a_1, a_2, a_3, \ldots\) is an AP, then:
\[a_2 – a_1 = a_3 – a_2 = a_4 – a_3 = d\]
where \(d\) is the common difference.
✍️ Step-by-Step Solution
Step 1: Calculate the fare for each kilometer
Let’s find the cumulative taxi fare after traveling 1 km, 2 km, 3 km, and so on:
- After 1 km: ₹15
- After 2 km: ₹15 + ₹8 = ₹23
- After 3 km: ₹23 + ₹8 = ₹31
- After 4 km: ₹31 + ₹8 = ₹39
- After 5 km: ₹39 + ₹8 = ₹47
So, the sequence of fares is: \[15, 23, 31, 39, 47, \ldots\]
Step 2: Check if the differences are constant
Calculate the difference between consecutive terms:
\(a_2 – a_1 = 23 – 15 = 8\)
\(a_3 – a_2 = 31 – 23 = 8\)
\(a_4 – a_3 = 39 – 31 = 8\)
\(a_5 – a_4 = 47 – 39 = 8\)
Observation: The difference between consecutive terms is constant and equals 8.
Step 3: Conclusion
Since the difference between consecutive terms is constant (\(d = 8\)), the taxi fares form an Arithmetic Progression.
The AP can be written as:
\[15, 23, 31, 39, 47, \ldots\]
with first term \(a = 15\) and common difference \(d = 8\).
✅ Final Answer
Yes, the taxi fares form an Arithmetic Progression.
The sequence is: 15, 23, 31, 39, 47, … with common difference \(d = 8\).
🔄 Alternative Method: Using nth Term Formula
We can also verify this by deriving a general formula for the fare after \(n\) kilometers:
For the first kilometer: ₹15
For additional \((n-1)\) kilometers: ₹8 × \((n-1)\)
Total fare after \(n\) km:
\[a_n = 15 + 8(n-1) = 15 + 8n – 8 = 7 + 8n\]
This is in the form \(a_n = a + (n-1)d\) where \(a = 15\) and \(d = 8\), confirming it’s an AP.
⚠️ Common Mistakes to Avoid
- Confusing per-km fare with cumulative fare: Remember, we’re checking if the total cumulative fares form an AP, not the per-kilometer charges.
- Not checking all differences: Always verify that all consecutive differences are equal, not just the first two.
- Forgetting the first term: The first kilometer costs ₹15, not ₹8. Don’t start the sequence from ₹8.
- Math notation errors: Ensure proper use of parentheses when writing differences like \(a_2 – a_1\).
📝 Practice Problems
- An auto-rickshaw charges ₹20 for the first km and ₹12 for each additional km. Do the fares form an AP? If yes, find the first five terms.
- A parking lot charges ₹30 for the first hour and ₹15 for each subsequent hour. Write the sequence of charges for 6 hours and verify if it’s an AP.
- If a taxi charges ₹\(a\) for the first km and ₹\(b\) for each additional km, prove that the fares form an AP with common difference \(b\).
🔗 Related Topics
Written by Farhan Mansuri
M.Sc. Mathematics | B.Ed. | 15+ Years Teaching Experience
Farhan Mansuri is a dedicated mathematics educator with over 15 years of experience teaching CBSE curriculum. He specializes in making complex mathematical concepts accessible to Class 10 students.
🔗 Connect on LinkedIn