Compound interest, growth and depreciation

A bank does not treat your savings the same way every year. In the first year it pays you interest on what you deposited. But from the second year onward, it pays you interest on your deposit plus all the interest you already earned — interest earning interest. This snowball effect is compound interest, and it is the reason a loan left unpaid for years can balloon far beyond what simple interest would predict. The same year-on-year percentage idea shows up everywhere outside banking too. A town's population grows by a fixed percentage each year, and a car or machine depreciates — loses value — by a fixed percentage each year. All three situations share one arithmetic pattern: take last year's amount, add or subtract a percentage of it, and repeat. This article builds that pattern step by step, then compresses it into a formula you can use to check your work quickly.

The core insight

Compound interest, growth and depreciation all repeat the same step — add or subtract $R\%$ of the previous year's amount, not the original amount — which is why after $n$ years the amount becomes $A = P(1 \pm R/100)^n$ .

The core idea

Simple interest vs compound interest. In simple interest, every year's interest is calculated on the same original principal $P$ , so the interest added each year is identical. In compound interest, each year's interest is calculated on the amount at the start of that year — principal plus every year's interest added so far. Because that base keeps growing, the rupee amount of interest added grows a little each year too.

Building it up year by year. For a principal $P$ at rate $R\%$ per annum, the pattern for the first two years is:

\text{Amount after Year 1} = P + \frac{R}{100}\times P

\text{Amount after Year 2} = (\text{Amount after Year 1}) + \frac{R}{100}\times(\text{Amount after Year 1})

where: $P$ is the principal (starting amount, in rupees), $R$ is the rate of interest per year (as a percentage), and each year's interest is a percentage of the previous year's amount, not of $P$ . At the Class 8 level, this step-by-step build-up is the expected method for 1 to 3 years — it keeps every step visible and stops you from silently sliding back into the simple-interest habit.

The compact formula. Repeating the step above $n$ times is exactly equivalent to a single compact expression:

A = P\left(1 + \frac{R}{100}\right)^n

where: $A$ is the amount after $n$ years, and the compound interest itself is $CI = A - P$ . This formula is a useful shorthand for describing the pattern or for a quick check, but for Class 8 problems the year-on-year build-up remains the method you show in full.

Growth and depreciation use the same shape. A quantity that increases by $R\%$ each year (population, for instance) follows the same rising pattern as compound interest. A quantity that decreases by $R\%$ each year (the resale value of a machine, for instance) follows the mirror pattern, subtracting instead of adding:

A = P\left(1 - \frac{R}{100}\right)^n

where: the symbols mean the same as above, except now $R\%$ of the previous year's amount is taken away rather than added. The figure below shows both patterns starting from the same value.

Figure

Two curves starting from the same value one rising for growth and one falling for depreciation over three years

Derivation

Derivation AThe compound amount formula A equals P times one plus R over 100 to the power n

The goal is to show why repeating the same year-on-year percentage step $n$ times, each time on the newly grown amount rather than the original principal, collapses into the single compact expression $A = P(1+R/100)^n$ , so that the step-by-step method and the formula are guaranteed to give the identical answer.

Step 1. Write the multiplying factor for one year. Adding $R\%$ of an amount $X$ to itself means $X + \frac{R}{100}X$ , and factoring out $X$ gives $X\left(1+\frac{R}{100}\right)$ . So going from any year's amount to the next year's amount always means multiplying by the same fixed factor $\left(1+\frac{R}{100}\right)$ .

Step 2. Apply this factor for Year 1. Starting from the principal $P$ , the amount after Year 1 is $A_1 = P\left(1+\frac{R}{100}\right)$ .

Step 3. Apply the same factor again for Year 2, this time to $A_1$ , not to $P$ . $A_2 = A_1\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)^2$ . Each further year multiplies by the same factor once more, so after $n$ years the principal has been multiplied by $\left(1+\frac{R}{100}\right)$ a total of $n$ times.

Step 4. Collect the repeated multiplication into a power. Multiplying by the same factor $n$ times is exactly $\left(1+\frac{R}{100}\right)^n$ , giving $A = P\left(1+\frac{R}{100}\right)^n$ . Replacing the plus sign with a minus sign models the same reasoning for a quantity shrinking by $R\%$ each year, giving $A = P\left(1-\frac{R}{100}\right)^n$ for depreciation.

Worked example

Find the compound interest on ₹8,000 for 3 years at 10% per annum, compounded annually, extending the 2-year case, and verify the answer using the compact formula.

Step 1. Amount after Year 1: interest is 10% of ₹8,000. $\frac{10}{100}\times 8000 = 800$ , so the amount is $8000+800=8800$ .

Step 2. Amount after Year 2: interest is 10% of the new amount, ₹8,800, not ₹8,000. $\frac{10}{100}\times 8800 = 880$ , so the amount is $8800+880=9680$ .

Step 3. Amount after Year 3: interest is 10% of ₹9,680. $\frac{10}{100}\times 9680 = 968$ , so the amount is $9680+968=10648$ .

Step 4. The compound interest over 3 years is $CI = A - P = 10648 - 8000 = 2648$ .

Step 5. Check with the compact formula. $A = 8000\left(1+\frac{10}{100}\right)^3 = 8000\times(1.1)^3 = 8000\times1.331 = 10648$ , matching Step 3 exactly, so $CI = 10648-8000 = 2648$ again. So the compound interest over 3 years is ₹2,648, compared with a simple interest of $\frac{8000\times10\times3}{100}=2400$ over the same 3 years — the gap between them, ₹248, is entirely due to interest earning interest in Years 2 and 3.

Competitive stretch

A sum of ₹8,000 compounded annually for 2 years becomes ₹9,680. Find the rate of interest per annum, using the compact formula as a check on trial values.

Step 1. Write the compact formula with the known values filled in. $9680 = 8000\left(1+\frac{R}{100}\right)^2$ .

Step 2. Isolate the squared factor by dividing both sides by 8000. $\left(1+\frac{R}{100}\right)^2 = \frac{9680}{8000} = 1.21$ .

Step 3. Take the square root of both sides. $1+\frac{R}{100} = \sqrt{1.21} = 1.1$ , so $\frac{R}{100} = 0.1$ , giving $R = 10\%$ per annum, matching the rate used in the worked example above and confirming the two approaches agree.

Common traps

Calculating every year's interest on the original principal instead of the previous year's amount — this silently turns compound interest into simple interest.
Confusing growth (add each year) with depreciation (subtract each year) when both use the same-looking percentage rate and the same formula shape.
Applying the compact formula $A = P(1\pm R/100)^n$ without first checking $n$ is a whole number of years compounded annually — the formula assumes the same rate applies at the end of every year.
Reporting the total amount $A$ when the question asks for the compound interest $CI = A-P$ , or the reverse.

Ask the tutor: Want to see this worked out for 4 years, or compare compound interest side-by-side with simple interest on a table for the same numbers?

Why is compound interest for the first year the same as simple interest

Because in Year 1 there is no earlier interest yet to earn interest on — the amount you are calculating interest from is still just the original principal $P$ . The two methods only start to differ from Year 2 onward, once compound interest begins calculating on principal plus the interest already added, while simple interest keeps calculating on the same untouched $P$ every year. That is why in the worked example the Year 1 interest of ₹800 is identical either way, but Year 2 and Year 3 pull apart.

Can I always use the formula A equals P times 1 plus R over 100 to the power n instead of the year-by-year method

For a whole number of years compounded annually at a constant rate, yes — the derivation above shows the formula is mathematically identical to repeating the year-on-year step $n$ times. At the Class 8 level, though, you are usually expected to show the step-by-step build-up as your working, because it demonstrates you understand why the interest grows each year rather than just plugging numbers into a formula. Use the formula as a quick check on your final answer, not as a replacement for the steps.

Is population growth exactly the same calculation as compound interest

The arithmetic is identical — both add a fixed percentage of the previous year's value each year — but the real-world assumption behind it is different. Compound interest grows at a rate fixed by a bank or loan agreement, which genuinely stays constant. Population growth rates and depreciation rates are estimates that can change year to year in reality; when a problem gives you a fixed percentage for several years, it is asking you to apply the same year-on-year percentage method, not claiming the rate is guaranteed to stay that way forever.