📚 Question
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(iv) The amount of money in the account every year, when ₹10,000 is deposited at compound interest at 8% per annum.
📌 Given Information
- Principal amount: ₹10,000
- Interest rate: 8% per annum (compound interest)
- We need to check if the amounts form an AP
🎯 To Find
Whether the amount of money in the account forms an Arithmetic Progression (AP) and justify the answer.
💡 Key Concept
Compound Interest Formula:
\[A = P\left(1 + \frac{r}{100}\right)^n\]
where \(P\) = Principal, \(r\) = Rate, \(n\) = Time period
For AP: Difference between consecutive terms must be constant.
\[a_2 – a_1 = a_3 – a_2 = d\]
✍️ Step-by-Step Solution
Step 1: Calculate amount after each year
Using compound interest formula with \(P = 10000\), \(r = 8\%\):
Initially (Year 0): \(a_1 = 10000\)
After 1 year: \(a_2 = 10000 \times \left(1 + \frac{8}{100}\right) = 10000 \times 1.08 = 10800\)
After 2 years: \(a_3 = 10000 \times (1.08)^2 = 10000 \times 1.1664 = 11664\)
After 3 years: \(a_4 = 10000 \times (1.08)^3 = 10000 \times 1.259712 = 12597.12\)
After 4 years: \(a_5 = 10000 \times (1.08)^4 = 10000 \times 1.36048896 = 13604.89\)
Sequence: \[10000, 10800, 11664, 12597.12, 13604.89, \ldots\]
Step 2: Check if differences are constant (AP test)
Calculate the differences between consecutive terms:
\(a_2 – a_1 = 10800 – 10000 = 800\)
\(a_3 – a_2 = 11664 – 10800 = 864\)
\(a_4 – a_3 = 12597.12 – 11664 = 933.12\)
\(a_5 – a_4 = 13604.89 – 12597.12 = 1007.77\)
Observation: The differences are NOT constant: \(800 \neq 864 \neq 933.12 \neq 1007.77\)
Step 3: Check if ratios are constant (GP test)
Calculate the ratios between consecutive terms:
\(\frac{a_2}{a_1} = \frac{10800}{10000} = 1.08\)
\(\frac{a_3}{a_2} = \frac{11664}{10800} = 1.08\)
\(\frac{a_4}{a_3} = \frac{12597.12}{11664} = 1.08\)
\(\frac{a_5}{a_4} = \frac{13604.89}{12597.12} = 1.08\)
Observation: The ratios ARE constant: \(1.08\). This is a Geometric Progression (GP), not an AP!
Step 4: Conclusion
Since the difference between consecutive terms is NOT constant, the amounts do NOT form an Arithmetic Progression.
However, it forms a Geometric Progression with common ratio \(r = 1.08\).
✅ Final Answer
No, the amount of money does NOT form an Arithmetic Progression.
The differences between consecutive terms are not constant. Instead, it forms a Geometric Progression with common ratio r = 1.08.
🔍 Why It’s NOT an AP
Compound Interest vs Simple Interest:
- Simple Interest: Fixed amount added each year → Forms an AP
- Compound Interest: Interest on interest → Forms a GP
In compound interest, the interest amount itself grows each year because it’s calculated on an increasing principal. This creates a multiplicative pattern (GP), not an additive pattern (AP).
📊 Comparison: Simple vs Compound Interest
Note: Simple Interest would form an AP with d = 800, but compound interest forms a GP with r = 1.08.
⚠️ Common Mistakes to Avoid
- Confusing simple and compound interest: Simple interest creates AP, compound interest creates GP.
- Assuming all financial growth is AP: Only linear growth (fixed additions) forms AP.
- Not calculating multiple differences: Always verify with at least 3-4 consecutive differences.
- Rounding errors: Use precise calculations for compound interest to see the pattern clearly.
📝 Practice Problems
- If ₹10,000 is deposited at simple interest at 8% per annum, would the amounts form an AP? Find the first 5 terms.
- A population grows by 5% each year. Does the population form an AP or GP over the years?
- Compare the total amounts after 10 years for simple interest vs compound interest at 8% on ₹10,000.
- If a bank offers 10% compound interest, what is the common ratio of the GP formed by yearly amounts?
🔗 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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