Compound Interest

Module 16 — Lesson 0016 · In Depth · 40 min

Simple interest earns on principal only. Compound interest earns interest on interest — so money grows like a GP, not an AP. Master the amount-first formula and the three frequency shifts; everything else is a variation.

1. The Core Variables

Symbol	Meaning
`P`	Principal (original amount)
`R`	Annual rate %
`n` / `t`	Time in years (or number of conversion periods)
`A`	Amount (final value) = P + CI
`CI`	Compound interest = A − P
`k`	Compounding frequency per year (1=annual, 2=half-yearly, 4=quarterly)

2. The Fundamental Formula — Amount First

In SI you find interest first. In CI you find amount first — because interest is calculated on a growing base.

A = P × [1 + R/100]ⁿ

Then CI = A − P = P{[1 + R/100]ⁿ − 1}

Why the power? Year 1: P(1+R/100). Year 2: that entire amount earns again → P(1+R/100)². Each year multiplies by the same factor — a GP with ratio (1+R/100). SI adds the same interest each year (AP); CI multiplies.

Worked: P=1000, R=10%, n=2 → A=1000×1.1²=1210, CI=210. SI for same would be 1000×10%×2=200 — CI is 10 more (interest on first year's 100).

3. Compounding Frequencies (Non-Annual)

Interest is not always annual. Divide the rate by k, multiply periods by k:

A = P × [1 + R/(100·k)]^k·n

Frequency	k	Rate used	Periods	Formula
Annual	1	R	n	`P[1+R/100]ⁿ`
Half-yearly	2	R/2	2n	`P[1+R/200]²ⁿ`
Quarterly	4	R/4	4n	`P[1+R/400]⁴ⁿ`
Monthly	12	R/12	12n	`P[1+R/1200]¹²ⁿ`

Worked: Half-yearly vs annual — P=8000, R=10%, n=1 year.
Annual: A=8000×1.1=8800, CI=800.
Half-yearly: R/2=5%, periods=2 → A=8000×1.05²=8000×1.1025=8820, CI=820. More frequent compounding → slightly more interest because the first half-year's interest earns in the second half-year.
Exam tip: If they say "compounded half-yearly for 1.5 years" → n=1.5, k=2 → periods=3, rate R/2.

4. Advanced Concepts

A. Difference between CI and SI

Very common placement question — same P, R, n.

For 2 years: CI − SI = P × (R/100)²
For 3 years: CI − SI = P × (R/100)² × (R/100 + 3) — which is P×(R/100)² × (3 + R/100)

Derivation for 2 years: CI = P[1+R/100]²−P = P[2R/100 + (R/100)²], SI = P×2R/100. Subtract → P(R/100)². That leftover is exactly "interest on interest" for one period.

Worked: P=1000, R=10% → SI(2y)=200, CI=210, diff=10. Formula: 1000×0.1²=10 ✓.
For 3 years: CI =1000×1.331−1000=331, SI=300, diff=31. Formula: 1000×0.01×3.1=31 ✓.

B. Multiplier Concept (Doubling & Rule of 72)

If money doubles in x years at CI, then: 4× in 2x years, 8× in 3x years, 16× in 4x years — a GP 2, 4, 8, 16…

Why: doubling means factor 2 per x years. After k cycles, factor = 2^k.

Rule of 72 (estimate): Doubling time ≈ 72 / R years. At 8% → doubles in ~9 years. Useful for quick MCQs when they ask "in how many years will it double?"

C. Successive (Different) Rates

If R₁ for year 1, R₂ for year 2, …: just chain the factors.

A = P × (1 + R₁/100) × (1 + R₂/100) × …

Worked: P=5000, R₁=10%, R₂=12% → A=5000×1.10×1.12=6160. This is the same as non-uniform compounding — treat each year as its own factor. Depreciation would use minus: (1 − R/100).

D. The Tree Method (Year-by-Year Visualization)

Instead of the formula, break interest year by year — especially handy for fractional years (e.g., 1.5 years) or to see where the extra CI comes from.

Tree: P=1000, R=10%, 2 years
Year 1: 10% of 1000 = 100
Year 2: interest on principal 10% of 1000 = 100 + interest on Year 1's interest 10% of 100 = 10 → total Year 2 = 110
Total CI = 100 + 110 = 210 (matches formula).
Visual: each year's interest sprouts its own sub-interest — that's the compounding tree.

E. Growth & Depreciation

Same formula with +/−:

Population / appreciation after n years: P(1 + R/100)ⁿ
Depreciation (asset loses value): P(1 − R/100)ⁿ

Example: Car worth 100000 depreciates 10% per year → after 2 years 100000×0.9²=81000.

5. Summary — Choose Your Tool

Need	Use
Amount after n years (uniform R)	`P(1+R/100)ⁿ`
Half-yearly / quarterly	`P(1+R/(100k))^kn`
Different rates per year	Chain: `P∏(1+Rᵢ/100)`
CI−SI (2y)	`P(R/100)²`
CI−SI (3y)	`P(R/100)²(R/100+3)`
Doubling → 4×, 8×	Multiply time by cycles: 2x, 3x

Amount in CI: A = ?

P=1000, R=10%, 2 years — CI?

Half-yearly for 1 year at 10%: rate and periods?

Quarterly for 1 year at 8%: A = ?

CI−SI for 2 years, same P,R is?

Money doubles in 6 years at CI. When does it become 8×?

P with R₁=10% then R₂=12%: A = ?

Depreciation 10% per year for 2 years on 100000?

Notes

CI :- Basics :-

CI ⇒ interest on interest ; SI ⇒ on principal only

A ⇒ P(1+R/100)ⁿ ; CI = A−P

Why power ⇒ each year multiplies by (1+R/100) → GP

Frequencies :-

General ⇒ P[1+R/(100k)]^kn ; k=1 annual, 2 half, 4 quarterly

Half ⇒ R/2, 2n ; Quarterly ⇒ R/4, 4n

(eg P=8000,10%,1y half→5%×2→8820 vs 8800 annual)

Advanced :-

CI−SI (2y) ⇒ P(R/100)² ; (3y) ⇒ P(R/100)²(R/100+3)

Doubling ⇒ 2→4 (2x), →8 (3x), →16 (4x) ; Rule 72≈72/R years

Successive rates ⇒ P(1+R₁/100)(1+R₂/100)… ; depreciation (1−R/100)ⁿ

Tree method ⇒ Year1 on P, Year2 on P + interest-on-interest

(eg 1000,10%,2y: Y1 100, Y2 100+10=110 → CI 210)

Variables :-

A ⇒ amount ; P ⇒ principal ; R ⇒ rate ; n/k ⇒ periods/frequency

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