📚 Question
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
📌 Given Information
- Cost for digging the first metre: ₹150
- Cost increases by ₹50 for each subsequent metre
- We need to check if the cumulative costs form an AP
🎯 To Find
Whether the cost of digging forms 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 cost for each metre of digging
Let’s find the cumulative cost after digging 1m, 2m, 3m, and so on:
- 1st metre: ₹150
- 2nd metre: ₹150 + ₹50 = ₹200
- 3rd metre: ₹200 + ₹50 = ₹250
- 4th metre: ₹250 + ₹50 = ₹300
- 5th metre: ₹300 + ₹50 = ₹350
So, the sequence of costs is: \[150, 200, 250, 300, 350, \ldots\]
Step 2: Check if the differences are constant
Calculate the difference between consecutive terms:
\(a_2 – a_1 = 200 – 150 = 50\)
\(a_3 – a_2 = 250 – 200 = 50\)
\(a_4 – a_3 = 300 – 250 = 50\)
\(a_5 – a_4 = 350 – 300 = 50\)
Observation: The difference between consecutive terms is constant and equals 50.
Step 3: Conclusion
Since the difference between consecutive terms is constant (\(d = 50\)), the costs of digging form an Arithmetic Progression.
The AP can be written as:
\[150, 200, 250, 300, 350, \ldots\]
with first term \(a = 150\) and common difference \(d = 50\).
✅ Final Answer
Yes, the cost of digging forms an Arithmetic Progression.
The sequence is: 150, 200, 250, 300, 350, … with common difference d = 50.
🔄 Alternative Method: Using nth Term Formula
We can also verify this by deriving a general formula for the cost of digging the \(n\)th metre:
For the first metre: ₹150
For each additional metre: Cost increases by ₹50
Cost for the \(n\)th metre:
\[a_n = 150 + (n-1) \times 50 = 150 + 50n – 50 = 100 + 50n\]
This is in the form \(a_n = a + (n-1)d\) where \(a = 150\) and \(d = 50\), confirming it’s an AP.
📊 Verification Table
Conclusion: All differences are constant (50), confirming this is an AP.
🏗️ Real-Life Application
Why does the cost increase?
As we dig deeper:
- More effort is required to remove soil from greater depths
- Additional safety measures are needed
- Harder rock layers may be encountered
- More time and labor are required
This uniform increase in cost per metre makes it a perfect example of an Arithmetic Progression in real life!
⚠️ Common Mistakes to Avoid
- Confusing per-metre cost with cumulative cost: We’re looking at the cost for each metre, not the total cumulative cost.
- Misreading “rises by ₹50”: The cost increases by ₹50, so the 2nd metre costs ₹200, not ₹50.
- Starting from zero: The sequence starts from ₹150 (first metre), not ₹0.
- Not verifying all differences: Always check multiple consecutive differences to confirm the pattern.
📝 Practice Problems
- A construction company charges ₹5000 for the first floor and ₹3000 more for each additional floor. Do the charges form an AP? Find the cost for the 10th floor.
- If digging costs ₹\(a\) for the first metre and increases by ₹\(d\) for each subsequent metre, write the general formula for the cost of the \(n\)th metre.
- A well is 20 metres deep. Using the given cost structure (₹150 for first metre, increasing by ₹50), find the total cost of digging the entire well.
- Compare this scenario with Question 1(ii) (vacuum pump). Why is one an AP and the other a GP?
🔗 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.
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