Compound interest and successive accumulation

Learning outcome

Build compound amounts period by period and distinguish total compound interest from interest earned in one particular year.

Concepts and assumptions

Compound interest adds each completed period’s interest to the balance. That enlarged balance becomes the next period’s base. Unlike simple interest, later interest can include interest on earlier interest.

Let P be the initial principal and r% the annual rate. With annual compounding, one period is one year, so the annual rate is also the period rate. After one year, A1 = P × (1 + r/100). Applying the same multiplier again gives A2 = P × (1 + r/100)^2.

After n complete years at a constant rate:

A = P × (1 + r/100)^n; total compound interest CI = A - P.

The exponent counts repeated multiplications; it does not multiply the percentage rate. Interest earned in a particular year is that year’s opening balance × r/100, not necessarily P × r/100.

If annual rates vary, multiply by each year’s own factor in sequence. The constant-rate power formula no longer describes all years with a single stated rate.

Assume positive principal, nonnegative annual rates, all interest retained, and no extra deposits, withdrawals, interim payments or fees. This lesson uses complete annual periods. Although 12 months make a year, an incomplete annual period needs its own stated rule; do not automatically use a fractional exponent. A day-based simple-interest extension, if explicitly specified, uses a stated year convention rather than an invented compounding period.

Keep every intermediate value exact. Round final money only to the nearest ₹0.01 when needed, rounding half a paise upward.

Worked examples

Example 1 — Two annual steps. ₹7,500 earns 4% compounded annually for 2 years. First-year interest = 7500 × 0.04 = ₹300; balance = ₹7,800. Second-year interest = 7800 × 0.04 = ₹312; balance = ₹8,112. Thus CI = 8112 - 7500 = ₹612.

Example 2 — Three annual steps. ₹12,500 earns 8% for 3 years. Successive balances are 12500 × 1.08 = ₹13,500, then 13500 × 1.08 = ₹14,580, then 14580 × 1.08 = ₹15,746.40. Therefore A = ₹15,746.40 and CI = 15746.40 - 12500 = ₹3,246.40.

Example 3 — Different yearly rates. ₹16,000 grows at 6%, 5% and 8% in three successive years. Balances become 16000 × 1.06 = ₹16,960, then 16960 × 1.05 = ₹17,808, then 17808 × 1.08 = ₹19,232.64. Total CI = 19232.64 - 16000 = ₹3,232.64. Each rate acts on its own opening balance.

Common mistakes

Do not report the amount as interest. Do not use the original principal for every year’s compound interest. Rates cannot simply be added, and a constant-rate shortcut requires a constant rate.

Practice questions

1. Find amount and compound interest on ₹5,400 at 10%, compounded annually, for 2 years.
2. Find amount and compound interest on ₹20,000 at 5%, compounded annually, for 3 years.
3. ₹11,000 earns 6% compounded annually. Find the second year’s interest and the amount after 2 years.
4. ₹18,000 earns 5% in the first year and 6% in the second, compounded annually. Find amount and total interest.

Worked answers

1. Balances are 5400 × 1.10 = ₹5,940 and 5940 × 1.10 = ₹6,534. CI = 6534 - 5400 = ₹1,134.
2. Balances are ₹21,000, ₹22,050 and ₹23,152.50, multiplying by 1.05 each year. CI = 23152.50 - 20000 = ₹3,152.50.
3. First balance = 11000 × 1.06 = ₹11,660. Second-year interest = 11660 × 0.06 = ₹699.60; amount = 11660 + 699.60 = ₹12,359.60.
4. Balances are 18000 × 1.05 = ₹18,900 and 18900 × 1.06 = ₹20,034. Total interest = 20034 - 18000 = ₹2,034.