Annuities and Loan Repayment: Formulas and Naira Examples

By Engr. Henry Akpan

Money Paid in Regular Instalments

Car loans, mortgages, school fee plans, pensions and monthly savings plans all involve equal payments made at regular intervals. A sequence of equal payments like this is called an annuity. The maths behind it answers practical questions: how much will I have if I save ₦20,000 every month for three years? What is the monthly repayment on a ₦1 million loan? How much of each payment is interest? This tutorial derives the annuity formulas from geometric series, shows how loan repayment instalments are calculated, builds an amortisation schedule, and covers sinking funds, with worked examples in naira.

Compound Interest: The Starting Point

If P is invested at an interest rate r per period (as a decimal), compounded each period, then after n periods it grows to

A = P(1 + r)ⁿ.

Annual rates are usually quoted, but payments are often monthly. A loan at 24% per year compounded monthly has r = 0.24/12 = 0.02 per month. Always match r to the payment period.

The reverse question, how much money now is worth A in n periods' time, gives the present value: PV = A/(1 + r)ⁿ = A(1 + r)⁻ⁿ. Money today is worth more than the same amount later, because it can earn interest in between.

Future Value of an Annuity (Saving Up)

Suppose you pay R at the end of each period for n periods, at rate r. The last payment earns no interest, the one before it earns one period's interest, and so on, back to the first payment, which earns interest for n − 1 periods. The total is

R + R(1 + r) + R(1 + r)² + … + R(1 + r)ⁿ⁻¹.

This is a geometric series with first term R, common ratio (1 + r) and n terms. Using the sum formula a(qⁿ − 1)/(q − 1):

FV = R × [(1 + r)ⁿ − 1] / r.

Example 1. Tolu saves ₦20,000 at the end of every month for 3 years in an account paying 12% per year, compounded monthly. How much will she have?

r = 0.12/12 = 0.01 and n = 36. FV = 20,000 × (1.01³⁶ − 1)/0.01 = 20,000 × 43.0769 = ₦861,537.57.

She paid in 36 × ₦20,000 = ₦720,000, so ₦141,537.57 is interest.

Present Value of an Annuity

The present value of n payments of R is the sum of each payment's present value:

R(1 + r)⁻¹ + R(1 + r)⁻² + … + R(1 + r)⁻ⁿ.

Summing this geometric series gives

PV = R × [1 − (1 + r)⁻ⁿ] / r.

Example 2. A scholarship pays ₦50,000 at the end of each year for 5 years. If money earns 10% per year, what is the scholarship worth today?

PV = 50,000 × (1 − 1.1⁻⁵)/0.1 = 50,000 × 3.79079 = ₦189,539.34. A sponsor would need to invest this amount today to fund all five payments.

The Loan Repayment Formula

A loan is an annuity in reverse: the bank gives you the present value P now, and you repay it with n equal instalments R. Setting P equal to the present value of the instalments and solving for R:

R = P × r / [1 − (1 + r)⁻ⁿ].

Example 3. Emeka borrows ₦1,000,000 at 24% per year, compounded monthly, to be repaid in 12 equal monthly instalments. Find the monthly repayment and the total interest.

r = 0.02, n = 12. (1.02)⁻¹² = 0.78849, so 1 − 0.78849 = 0.21151.

R = 1,000,000 × 0.02 / 0.21151 = ₦94,559.60 per month.

Total repaid = 12 × 94,559.60 = ₦1,134,715.16, so the total interest is ₦134,715.16. Notice this is much less than 24% of ₦1,000,000 (₦240,000), because Emeka is paying the loan down every month and so pays interest on a shrinking balance.

The Amortisation Schedule

Each instalment is split into interest (on the balance still owed) and principal (which reduces the balance). For Emeka's loan:

• Month 1: interest = 2% of ₦1,000,000 = ₦20,000. Principal repaid = 94,559.60 − 20,000 = ₦74,559.60. New balance = ₦925,440.40.
• Month 2: interest = 2% of 925,440.40 = ₦18,508.81. Principal = ₦76,050.79. Balance = ₦849,389.61.
• Month 3: interest = ₦16,987.79. Principal = ₦77,571.80. Balance = ₦771,817.81.

The pattern continues: the interest part falls each month and the principal part rises, until the final payment brings the balance to zero. Early payments are mostly interest; late payments are mostly principal. On a long mortgage the effect is dramatic, which is why paying extra early saves so much interest.

Outstanding balance after k payments. Without building the whole table, use

Balance = P(1 + r)ᵏ − R × [(1 + r)ᵏ − 1]/r.

For Emeka after 6 payments: 1,000,000 × 1.02⁶ − 94,559.60 × (1.02⁶ − 1)/0.02 = ₦529,669.05. Halfway through the term, he still owes more than half the loan, because early payments went mostly to interest.

Sinking Funds

A sinking fund is money set aside regularly to reach a target amount, for example to replace a generator or to pay final-year fees. It uses the future value formula solved for R:

R = FV × r / [(1 + r)ⁿ − 1].

Example 4. A school wants ₦5,000,000 in 5 years to buy a new bus. The fund earns 8% per year. What should it deposit at the end of each year?

R = 5,000,000 × 0.08 / (1.08⁵ − 1) = 400,000 / 0.46933 = ₦852,282.27 per year. The school pays in about ₦4.26 million and interest makes up the rest.

Comparing Loan Offers

• A lower rate or a shorter term reduces total interest; a longer term lowers the monthly payment but increases the total paid.
• Watch for "flat rate" loans, where interest is charged on the original amount for the whole term. A 24% flat-rate loan of ₦1,000,000 over a year costs ₦240,000 in interest, far more than the ₦134,715 of the reducing-balance loan in Example 3, even though both are advertised as "24%".
• Fees and charges add to the true cost. Compare loans using the annual percentage rate (APR), which includes them.

Practice Questions

1. Find the future value of ₦10,000 saved at the end of each month for 2 years at 6% per year compounded monthly. Hint: r = 0.005, n = 24; FV = 10,000 × (1.005²⁴ − 1)/0.005 ≈ ₦254,320.
2. What monthly payment repays ₦500,000 over 24 months at 18% per year compounded monthly? Hint: r = 0.015, n = 24; R ≈ ₦24,962.
3. Explain why the interest part of each loan repayment falls over time. Answer: interest is charged on the outstanding balance, which falls with every payment.

Summary

• An annuity is a series of equal payments at regular intervals; match the rate r to the payment period.
• Future value: FV = R[(1 + r)ⁿ − 1]/r. Present value: PV = R[1 − (1 + r)⁻ⁿ]/r. Both come from geometric series.
• Loan repayment: R = Pr/[1 − (1 + r)⁻ⁿ]. Sinking fund deposit: R = FV × r/[(1 + r)ⁿ − 1].
• In an amortisation schedule, interest falls and principal rises with each payment.
• Reducing-balance loans cost far less than flat-rate loans at the same quoted rate.