The Logic Behind the Math: The GP Derivation

Let's derive the CI formula using the logic of sequences. If you have a Principal $P$ and an annual rate $r$ (in decimal form):

1. End of Year 1: $A_1 = P + Pr = P(1+r)$
2. End of Year 2: $A_2 = A_1 + A_1r = A_1(1+r) = P(1+r)(1+r) = P(1+r)^2$
3. End of Year $n$: $A_n = P(1+r)^n$

This is precisely the $n^{th}$ term of a Geometric Progression where the first term is $P(1+r)$ and the common ratio is $(1+r)$.

Step-by-Step Solved Example

Problem: Find the Compound Interest on ₹10,000 at 10% per annum for 3 years, compounded annually.

• Step 1: Identify Variables. $P = 10,000$, $R = 10\%$, $n = 3$.
• Step 2: Convert Rate. $r = 10/100 = 0.1$.
• Step 3: Apply the Growth Factor. The factor is $(1 + 0.1) = 1.1$.
• Step 4: Calculate Total Amount. $A = 10,000 * (1.1)^3$.
• Step 5: Solve Powers. $1.1 * 1.1 * 1.1 = 1.331$. So, $A = 10,000 * 1.331 = 13,310$.
• Step 6: Extract Interest. $CI = A - P = 13,310 - 10,000 = 3,310$.

Alternative Methods: The Successive Percentage Trick

For small values of $n$ (like 2 or 3 years), use the effective rate formula: $x + y + (xy/100)$. For 10% over 2 years: $10 + 10 + (100/100) = 21\%$. This is much faster than powers for MCQ-based exams like JEE Mains.

Exam Trap Alert: The "Compounding Frequency" Pitfall

A common mistake in NEET/JEE is ignoring the word "Semi-annually" or "Quarterly."

If compounding is half-yearly:

• New Rate ($R'$) = $R/2$
• New Time ($n'$) = $2n$

Students often divide the rate but forget to double the time, leading to an answer that is usually an option (A) or (B) designed to trick you!

Practice Problem (JEE Level)

Question: A sum of money triples itself in 5 years at a certain rate of compound interest. In how many years will it become 81 times itself?

Hint: Use the property $3P = P(1+r)^5$. Then find $t$ such that $81P = P(1+r)^t$. Notice that $81 = 3^4$.