Growth and Depreciation: Concepts, Formulas and Examples

Growth means a quantity increases by a fixed percentage every time period, and depreciation means it decreases by a fixed percentage. Unlike simple interest, where the same amount is added each year, compound growth applies the percentage to the new, bigger (or smaller) value each time, so the change itself keeps changing.

This idea matters everywhere money and nature are involved. Banks use it for compound interest, companies use it to predict how machines lose value, and scientists use it to describe how populations grow. If you cannot do growth and depreciation calculations, you will misread loan statements, investment returns and even biology facts.

Key Concepts

• Principal (P): the starting value in rupees, kilograms, or any unit.
• Rate (r): the percentage change per time period, written as a decimal in formulas (for example 10% becomes 0.10).
• Time (n): the number of periods (usually years).
• Compound growth: the new amount A = P(1 + r)^n.
• Depreciation: the reduced value V = P(1 − r)^n.
• Appreciation: another name for growth when talking about assets like land.

Formulas

• Amount after growth: A = P(1 + r)^n, where P is the initial value, r is the rate per period as a decimal, and n is the number of periods. Assumes the rate stays constant and growth happens once per period.
• Value after depreciation: V = P(1 − r)^n, with the same meanings. Assumes a constant rate of decline.
• Total growth: A − P. Total depreciation: P − V.
• Rate from two values: r = (A/P)^(1/n) − 1, used when you know the start and end values.

Worked Example

A car bought for ₹4,00,000 depreciates at 10% per year. Find its value after 3 years.

1. Identify: P = 4,00,000, r = 0.10, n = 3.
2. Formula: V = P(1 − r)^n = 4,00,000 × (0.90)^3.
3. Compute (0.90)^3: 0.90 × 0.90 = 0.81; 0.81 × 0.90 = 0.729.
4. Multiply: 4,00,000 × 0.729 = 2,91,600.

The car is worth ₹2,91,600 after 3 years. Check year by year: after year 1 it loses ₹40,000 (value ₹3,60,000), after year 2 it loses ₹36,000 (value ₹3,24,000), after year 3 it loses ₹32,400 (value ₹2,91,600). The yearly loss shrinks because the percentage applies to a smaller value each year.

Common Misconceptions

• "10% for 3 years means 30% total." Wrong for compound change. 10% growth for 3 years gives (1.10)^3 = 1.331, a 33.1% increase, not 30%. The extra comes from growth on growth.
• "Depreciation is the same every year." No — only the percentage is fixed. In the car example the loss fell from ₹40,000 to ₹32,400. Equal yearly loss would be the straight-line (simple) method, which is a different formula.
• "Growth always means money." Growth also describes bacteria doubling or a city's population rising — the same formula works wherever a constant percentage applies per period.

Real-World Applications

• Banks and mutual funds quote compound returns on savings and investments.
• Companies depreciate machinery and vehicles in their accounts to plan replacements.
• Governments project population growth for planning schools, hospitals and roads.
• Scientists model the growth of bacteria or the decay of medicines in the body.

Growth and depreciation are the same idea in two directions: apply a fixed percentage to a changing base, period after period. Master A = P(1 + r)^n and V = P(1 − r)^n, always convert the rate to a decimal, and check your answer by stepping through year by year.