In a reducing-balance loan, interest is charged each period on the balance still owing, then the repayment is subtracted. A repayment table tracks each period: opening balance, interest = balance × r, principal repaid = repayment − interest, closing balance = opening balance + interest − repayment.

As a recurrence relation: Vₙ₊₁ = (1 + r)Vₙ − M, with V₀ = the amount borrowed.

Worked example. $5000 at 12% p.a. compounded monthly (r = 0.01), repaid at $300 per month. Month 1: interest = $50.00, principal = $250.00, balance = $4750.00. Month 2: interest = $47.50, principal = $252.50, balance = $4497.50.

The regular repayment for a loan of P over n periods at rate r per period is M = [frac Pr/1 − (1 + r)[sup −n]]. Round M to the cent, then total interest = M × n − P.

The balance owing after k repayments is Vₖ = P(1 + r)[sup k] − M × [frac (1 + r)[sup k] − 1/r]: what the loan would have grown to, minus what the repayments have grown to.

Worked example. $10 000 at 12% p.a. compounded monthly over 1 year: M = 10 000 × 0.011 − 1.01⁻¹² = $888.49, and total interest = 888.49 × 12 − 10 000 = $661.88.

To find how many repayments are needed, set Vₙ = 0 and solve: (1 + r)ⁿ (M − Pr) = M, so n = [frac ln(M ÷ (M − Pr))/ln(1 + r)]. Round up: the last repayment is smaller and equals the remaining balance plus one period of interest.

Extra repayments and lump sums reduce the balance early, so every later period is charged less interest. A rate rise increases the repayment needed to finish on time.

Worked example. $10 000 at 12% p.a. compounded monthly, repaid at $500 per month: n = ln(500 ÷ 400)ln 1.01 = 22.4…, so 23 repayments (22 full ones and a smaller final one).