2.9 Installments — Repaying a Sum in Equal Annual Installments

Figure: Each installment chips away at the outstanding principal.

(a) Under Simple Interest. Suppose a debt of P is repaid in n equal annual installments of Rs. x each (at the end of each year), at simple interest rate R% p.a. The standard method used in SSC-level books equates the amount the original principal would have grown to (at SI, over n years) with the accumulated value of all the installments (each installment, once paid, is assumed to “earn” SI for the remaining years until year n):

P(1+(nR)/(100)) = x[n + (n(n-1))/(2)×(R)/(100)]

This gives: P = (x[n + (n(n-1)R)/(200)])/(1+(nR)/(100))

Solved Example 2.9.1: A sum is repaid in 2 equal annual installments of Rs. 1,200 each at 10% SI p.a. Find the sum borrowed.

Solution: n = 2, R = 10, x = 1200. Bracket term = n + n(n−1)R/200 = 2 + (2×1×10)/200 = 2 + 0.1 = 2.1 P(1 + 20/100) = 1200 × 2.1 → P × 1.2 = 2520 → P = Rs. 2,100.

Solved Example 2.9.2: A loan is repaid in 3 equal yearly installments of Rs. 800 each at 20% SI p.a. Find the loan amount.

Solution: n = 3, R = 20, x = 800. Bracket = 3 + (3×2×20)/200 = 3 + 0.6 = 3.6 P(1 + 60/100) = 800 × 3.6 → P × 1.6 = 2880 → P = Rs. 1,800.

(b) Under Compound Interest. Here the correct method is the present value (discounting) method: each future installment is worth less today because money today could grow via compounding. The sum borrowed equals the sum of the present values of all installments:

P = (x)/((1+(R)/(100))^1) + (x)/((1+(R)/(100))^2) + … + (x)/((1+(R)/(100))^n)

Solved Example 2.9.3: A sum is repaid in 2 equal annual installments of Rs. 1,210 each, interest being compounded annually at 10% p.a. Find the sum borrowed.

Solution: P = 1210/1.1 + 1210/1.1² = 1100 + 1000 = Rs. 2,100.

Notice: For the identical principal Rs. 2,100 at 10% over 2 years, the SI installment (Example 2.9.1) is Rs. 1,200, while the CI installment (Example 2.9.3) is Rs. 1,210 — the CI-based installment is always slightly higher because the lender is compensated for compounding by asking for marginally larger payments.

Solved Example 2.9.4: A sum is repaid in 3 equal annual installments of Rs. 1,331 each at 10% p.a. CI. Find the sum.

Solution: P = 1331/1.1 + 1331/1.1² + 1331/1.1³ = 1210 + 1100 + 1000 = Rs. 3,310.

Solved Example 2.9.5: A sum is repaid in 4 equal annual installments of Rs. 770 each at 10% p.a. simple interest. Find the sum borrowed (edge case: n = 4, beyond the usual n = 2 or 3).

Solution: n = 4, R = 10, x = 770. Bracket term = n + n(n−1)R/200 = 4 + (4×3×10)/200 = 4 + 120/200 = 4 + 0.6 = 4.6. P(1 + 40/100) = 770 × 4.6 = 3,542. P × 1.4 = 3542 → P = 3542/1.4 = Rs. 2,530. The method scales to any n exactly the same way — only the bracket term gets bigger.

Solved Example 2.9.6: A sum is repaid in 2 equal annual installments of Rs. 1,250 each, interest being compounded annually at 25% p.a. Find the sum borrowed (edge case: a much higher CI rate than usual, to test that the present-value method still applies cleanly).

Solution: P = 1250/1.25 + 1250/(1.25)² = 1250/1.25 + 1250/1.5625 = 1000 + 800 = Rs. 1,800. Even at a high rate like 25%, the discounting method (Trick 10) applies identically — factor out 1/(1+r) if you want the two-step shortcut instead of computing both fractions separately.