Compound interest formula

Simple interest pays you the same amount every year, because it is always calculated on the original principal. Compound interest is different — each year's interest gets added back to the principal, so the next year's interest is calculated on a slightly larger sum. Do this year after year and the growth accelerates: interest starts earning interest. Working this out year by year is slow and error-prone for anything beyond two or three years, which is exactly why the compound interest formula exists — a single expression that jumps straight to the final amount, no matter how many years or how the compounding is spread across the year. In ICSE Class 9, this formula is the entire content of the Commercial Mathematics unit, and it is tested both forwards (find the amount) and backwards (find the principal, rate, or time from a given amount).

The core insight

The compound interest formula $A = P(1+r/100)^n$ compresses year-by-year repeated simple interest into one calculation, and adjusting the rate and period together handles compounding more often than once a year.

The core idea

From repeated simple interest to one formula. If you compute simple interest for year 1, add it to the principal, compute simple interest again on that new total for year 2, and repeat, you eventually notice a pattern: each year multiplies the current amount by the same factor, $\left(1+\dfrac{r}{100}\right)$ . Multiplying by the same factor $n$ times is exponentiation, which is exactly what the formula captures. For a principal $P$ invested at rate $r\%$ per annum for $n$ years, compounded annually, the amount at the end of $n$ years is:

A = P\left(1+\dfrac{r}{100}\right)^n

where: $A$ is the amount (principal plus all accumulated interest), $P$ is the principal (initial sum invested), $r$ is the rate of interest per annum as a percentage, and $n$ is the number of years (compounding periods). The compound interest itself is obtained as a separate final step:

CI = A - P

Compounding more than once a year. When interest is compounded half-yearly instead of annually, the year is split into two compounding periods, so two changes must be made together: the rate used per period is halved to $\dfrac{r}{2}\%$ , and the number of periods is doubled to $2n$ . The formula becomes:

A = P\left(1+\dfrac{r/2}{100}\right)^{2n}

This is not a new formula — it is the same $A = P(1+\text{rate})^{\text{periods}}$ pattern, just with "rate" and "periods" redefined to match how often interest is actually credited. Forgetting to change both together is the single most common source of errors in this topic.

Reading the formula both ways. Because $A = P\left(1+\dfrac{r}{100}\right)^n$ links five quantities — $A$ , $P$ , $r$ , $n$ , $CI$ — an exam question can hand you any three of $A$ , $P$ , $r$ , $n$ and ask for the fourth. Solving for the principal only needs division:

P = \dfrac{A}{\left(1+\dfrac{r}{100}\right)^n}

Solving for $r$ or $n$ needs the same idea in reverse: divide $A$ by $P$ to isolate $\left(1+\dfrac{r}{100}\right)^n$ , then either take an $n$ -th root (to find $r$ ) or match the base to a known power (to find $n$ ). ICSE-level reverse problems are deliberately built so the numbers land on a recognisable perfect power, like $1.1^2$ or $1.05^3$ , so careful trial or factoring finds the answer without logarithms.

Figure

A staircase diagram showing a principal growing year by year under compound interest, each step taller than the last because interest is earned on interest

Derivation ADeriving the compound interest formula from repeated simple interest

The goal, which every ICSE student should be able to reproduce from first principles rather than merely quote, is to show that adding simple interest onto a growing principal year after year, for $n$ years running, produces exactly the single closed formula $A = P(1+r/100)^n$ , so that the shortcut can be trusted instead of treated as a rule handed down without justification.

Step 1. Find the amount after year 1. Simple interest for one year on principal $P$ at rate $r\%$ is $\dfrac{P \times r \times 1}{100}$ , so the amount at the end of year 1 is $A_1 = P + \dfrac{Pr}{100} = P\left(1+\dfrac{r}{100}\right)$ . This is just the principal multiplied by the growth factor $\left(1+\dfrac{r}{100}\right)$ .

Step 2. Find the amount after year 2, treating $A_1$ as the new principal. The interest for year 2 is calculated on $A_1$ , not on the original $P$ — that is the defining feature of compound interest. So $A_2 = A_1\left(1+\dfrac{r}{100}\right) = P\left(1+\dfrac{r}{100}\right)\left(1+\dfrac{r}{100}\right) = P\left(1+\dfrac{r}{100}\right)^2$ .

Step 3. Spot the pattern and generalise. Each year multiplies the current amount by the same factor $\left(1+\dfrac{r}{100}\right)$ , because the interest rate and the growth mechanism do not change from year to year. After year 3, $A_3 = P\left(1+\dfrac{r}{100}\right)^3$ , and after $n$ years, $A_n = P\left(1+\dfrac{r}{100}\right)^n$ . This is the compound interest formula: $A = P\left(1+\dfrac{r}{100}\right)^n$ .

Step 4. Verify with a tiny case. Take $P = 100$ , $r = 10\%$ , $n = 2$ . Year-by-year: $A_1 = 100 + 10 = 110$ ; $A_2 = 110 + 11 = 121$ (interest on $110$ , not on $100$ ). The formula gives $A = 100(1.1)^2 = 100 \times 1.21 = 121$ — matching exactly, confirming the shortcut reproduces the repeated-interest calculation.

Worked example

Find the amount and the compound interest on ₹15,000 for 2 years at 8% per annum, compounded annually. Then extend the same principal and rate to half-yearly compounding for 1 year, to see how the adjustment works in practice.

Step 1. Apply the annual formula directly. $A = P\left(1+\dfrac{r}{100}\right)^n = 15000\left(1+\dfrac{8}{100}\right)^2 = 15000(1.08)^2$ .

Step 2. Compute the power carefully. $(1.08)^2 = 1.1664$ , so $A = 15000 \times 1.1664 = ₹17{,}496$ . The compound interest is $CI = A - P = 17496 - 15000 = ₹2{,}496$ . So the annual-compounding amount is ₹17,496 and the compound interest is ₹2,496.

Step 3. Now compound half-yearly instead, over 1 year at the same 8% per annum. Halve the rate and double the periods: rate per period $= \dfrac{8}{2} = 4\%$ , periods $= 2 \times 1 = 2$ . $A = 15000\left(1+\dfrac{4}{100}\right)^2 = 15000(1.04)^2 = 15000 \times 1.0816 = ₹16{,}224$ , giving $CI = 16224 - 15000 = ₹1{,}224$ .

Step 4. Sanity-check by reversing the first calculation. If told the amount after 2 years at 8% per annum was ₹17,496, recovering the principal should give back ₹15,000: $P = \dfrac{17496}{(1.08)^2} = \dfrac{17496}{1.1664} = ₹15{,}000$ . This confirms the amount ₹17,496 and compound interest ₹2,496 for the annual case, and ₹16,224 with compound interest ₹1,224 for the one-year half-yearly case at the same nominal rate.

Common traps

Halving the rate without doubling the time (or vice versa) for half-yearly compounding — both adjustments must be made together, or the answer is wrong.
Confusing $A$ and $CI$ . The formula gives the amount $A$ directly; the compound interest is the separate final step $CI = A - P$ , easy to forget under exam pressure.
Misreading the exponent $n$ . $n$ must equal the actual number of compounding periods, not simply the number of years stated, whenever compounding is more frequent than annual.
Rounding powers like $(1.08)^2$ too early. Exam answers depend on exact decimal values, so expand these fully rather than approximating.
Writing $\dfrac{r}{100}$ incorrectly, such as $0.8$ instead of $0.08$ for $r = 8$ — always double-check this conversion before raising it to a power.

Ask the tutor: If rearranging the formula to find $r$ , $n$ , or $P$ is confusing, or you want more practice comparing annual versus half-yearly compounding, ask the AI tutor to walk through a few problems with you.

Why does compound interest grow faster than simple interest even at the same rate?

Simple interest always computes that year's interest on the fixed original principal, so each year adds exactly the same rupee amount. Compound interest computes each year's interest on the previous year's amount, which already includes earlier interest — so the base keeps growing, and each year's interest is slightly larger than the last. Over many years this "interest on interest" effect compounds noticeably, which is why compound interest overtakes simple interest by a growing margin as $n$ increases.

If a question just says compounded, without saying annually or half-yearly, what should I assume?

Assume annual compounding unless the question explicitly states otherwise. ICSE problems that intend half-yearly (or, rarely, quarterly) compounding will always say so directly, usually with the phrase "compounded half-yearly" or "interest is paid every six months." If no such phrase appears, use $n$ years and rate $r\%$ directly in the standard formula without any halving or doubling.

Do I need logarithms to find the rate or time when they are unknown?

No, not at this level. ICSE reverse problems are set up so that after dividing $A$ by $P$ , the resulting power of $\left(1+\dfrac{r}{100}\right)$ matches a recognisable perfect power, such as $1.1^2 = 1.21$ or $1.05^3 = 1.157625$ . Recognising these common values, or factoring carefully, lets you read off $r$ or $n$ directly. Logarithms are a university-level tool for cases where the numbers do not work out neatly.

Compound interest formula

The core idea

Worked example

Common traps

Before this

Where this leads