B. Exponential Functions · Financial Mathematics: Compound Interest and Annuities

Finding the Regular Payment: Saving Goals, Loans and Amortization

B3.7About 75 minutes

Learning goals

I can find the regular payment needed to reach a savings goal.
I can find the regular payment needed to repay a loan, and the total interest paid.
I can read and build an amortization table showing the interest and principal in each payment.
I can compare the total interest on a loan with the original principal for different terms.

Words to know

amortization
Paying off a loan with regular payments that cover the interest and gradually reduce the principal.

amortization table (schedule)
A table showing, for each payment, how much is interest, how much reduces the principal, and the balance left.

balance owing
The principal still owed after a payment.

term (of a loan)
The length of time over which a loan is repaid, such as 3, 5 or 7 years.

sinking fund
A savings plan of regular deposits made to reach a set amount by a set date.

Turning the question around

Most real money questions sound like this: "I need $3 000 in 2 years. How much do I have to save each month?" or "I'm borrowing $15 000. What will my monthly payment be?" The unknown is the regular payment, R. You already have the two annuity formulas. You just solve them for R.

Two payment formulas

Saving for a future amount (you know FV):

FV = R[(1 + i)ⁿ − 1] ÷ i, so R = FV × i ÷ [(1 + i)ⁿ − 1]

Repaying a loan (you know PV, the amount borrowed):

PV = R[1 − (1 + i)⁻ⁿ] ÷ i, so R = PV × i ÷ [1 − (1 + i)⁻ⁿ]

Which one? Ask: is the known amount at the end (a goal: use FV) or at the start (a loan or a fund: use PV)?

Technology: in a TVM Solver, fill in everything except PMT and solve for PMT. In a spreadsheet use =PMT(rate, nper, pv, fv).

Worked example

Example 1: A savings goal

Hana wants $3 000 in 2 years for a school trip. Her account pays 2.4% compounded monthly. How much must she deposit at the end of each month?

The goal is at the end, so use FV. FV = 3000, i = 0.024 ÷ 12 = 0.002, n = 24.
R = 3000 × 0.002 ÷ [(1.002)²⁴ − 1].
(1.002)²⁴ ≈ 1.0491204, so the denominator is about 0.0491204. R ≈ 6 ÷ 0.0491204 ≈ 122.15.
Check: she deposits 24 × 122.15 = $2 931.60, and interest supplies the remaining ≈ $68.40.

Hana must deposit $122.15 at the end of each month. (TVM: N = 24, I% = 2.4, PV = 0, FV = 3000, P/Y = C/Y = 12, solve PMT ≈ −122.15.)

Worked example

Example 2: A car loan

Diego borrows $15 000 for a used car at 7.2% compounded monthly, to be repaid with monthly payments over 5 years. Find the payment and the total interest. Compare the interest with the principal.

The loan is at the start, so use PV. PV = 15 000, i = 0.072 ÷ 12 = 0.006, n = 60.
R = 15 000 × 0.006 ÷ [1 − (1.006)⁻⁶⁰] = 90 ÷ [1 − 0.6984272…] ≈ 90 ÷ 0.3015728 ≈ 298.44.
Total paid ≈ 60 × 298.44 = $17 906.40.
Total interest ≈ 17 906.40 − 15 000 = $2 906.40, which is about 19% of the principal.

Diego pays $298.44 per month and about $2 906 in interest (19% on top of what he borrowed). In practice the last payment is adjusted by a few cents ($298.08 here), so the exact interest is $2 906.04.

Inside each payment: the amortization table

Each loan payment does two jobs. First it pays the interest owed for that month: balance × i. The rest pays down the principal.

Payment 1: interest = 15 000 × 0.006 = $90.00, so principal = 298.44 − 90.00 = $208.44. New balance: $14 791.56.
Payment 2: interest = 14 791.56 × 0.006 ≈ $88.75, principal = $209.69. New balance: $14 581.87.

As the balance falls, each month's interest gets smaller, so more of each payment goes to principal.

Payment #	Payment	Interest (balance × 0.006)	Principal	Balance owing
0				$15 000.00
1	$298.44	$90.00	$208.44	$14 791.56
2	$298.44	$88.75	$209.69	$14 581.87
3	$298.44	$87.49	$210.95	$14 370.92
…	…	…	…	…
59	$298.44	$3.55	$294.89	$296.30
60	$298.08	$1.78	$296.30	$0.00

The start and end of Diego's amortization table. Early payments are about 30% interest; the last payments are almost all principal. The final payment is adjusted so the balance ends at exactly $0.

Each bar is one year of payments (about $3 581). In year 1, $996 goes to interest; by year 5 it is only $136. More and more of each payment pays down the loan.

The balance owing falls slowly at first and faster later, so it stays above the straight dashed line. After 3 years (36 payments) Diego still owes $6 651.90, more than 40% of the loan.

Worked example

Example 3: Longer term, lower payment, more interest

Compare Diego's $15 000 loan at 7.2% compounded monthly over 3, 5 and 7 years.

3 years (n = 36): R ≈ $464.53; total paid ≈ $16 723.08; interest ≈ $1 723.08.
5 years (n = 60): R ≈ $298.44; total ≈ $17 906.40; interest ≈ $2 906.40.
7 years (n = 84): R ≈ $227.86; total ≈ $19 140.24; interest ≈ $4 140.24.

Stretching the loan from 3 to 7 years cuts the payment by about half but more than doubles the interest (from about 11% to about 28% of the principal). A lower payment is not the same as a cheaper loan.

Just for fun

An amortization table is the only table where the bottom row is the happiest: balance owing, $0.00.

Hands-on activity

Build an amortization table

You need: a spreadsheet (or paper and a calculator for the first 6 rows).

Choose a loan: $5 000 for a used car or e-bike at 6% compounded monthly for 2 years.

Find the payment: R = 5000 × 0.005 ÷ [1 − (1.005)⁻²⁴], or =PMT(0.005, 24, -5000).
Columns: Payment #, Payment, Interest, Principal, Balance. Row 0: balance 5000.
Each row: Interest = previous balance × 0.005 (round to cents); Principal = Payment − Interest; Balance = previous balance − Principal.
Fill down to payment 24. The balance should be within a few cents of $0.
Add up the Interest column, and compare with 24 × R − 5000.

What to look for: R ≈ $221.60; first-month interest $25.00 and principal $196.60; the balance after payment 24 is about $0.09, so the last payment is $221.69 and the total interest is about $318.49 (24 × 221.60 − 5000 = $318.40, plus those 9 cents).

Your turn

You want $10 000 in 5 years. Your account pays 4% compounded quarterly. Find the deposit needed at the end of each quarter.

A $2 500 loan at 12% compounded monthly is repaid in 1 year with monthly payments. Find the payment and total interest.

For the loan in the last question, find the interest and principal in the first payment.

Which formula do you use to find a monthly saving amount for a goal, and which for a loan payment?

Find the monthly deposit needed to have $5 000 in 3 years at 3% compounded monthly.

Write the TVM Solver entries to find the payment on an $8 000 loan at 6% compounded monthly for 2 years, and the answer.

More practice: warm-up

A balance of $4 000 is charged 0.5% interest this month. How much interest?

A payment is $250 and $40 of it is interest. How much reduces the principal?

Total paid on 48 payments of $375?

A $16 000 loan is repaid with total payments of $18 000. Total interest?

In an amortization table, does the interest part of each payment go up or down over time?

More practice: on your own

Find the annual deposit (end of each year) needed to have $20 000 in 10 years at 5% compounded annually.

A $25 000 loan for college costs is repaid monthly over 10 years at 6.5% compounded monthly. Find the payment and the total interest.

A $1 200 credit card purchase at 19.99% compounded monthly is paid off with equal monthly payments over 1 year. Find the payment and total interest.

For Diego's loan (payment $298.44), how much of payment 3 is interest? Use the table.

Explain why a 7-year car loan costs more interest than a 5-year loan for the same amount and rate, even though the payment is smaller.

Challenge

A $400 000 mortgage is repaid with monthly payments over 25 years. Using 4.25% compounded monthly (a simplification; see the note), find the payment and compare total interest with the principal.

Hana (Example 1) starts her $3 000 goal 6 months late, so she has only 18 months. What monthly deposit does she need now, and how much more is that per month?

A loan payment is $500 a month, and the first payment's interest is $150 at 0.75% per month. What was the original loan amount? How much principal does the first payment repay?

Check your understanding

Quiz

To find the monthly deposit needed to reach a savings goal, use:

A loan balance is $6 000 and i = 0.005. Payment = $265. How much of this payment is principal?

Over the life of an amortized loan, the principal part of each payment:

Lengthening a loan's term (same amount and rate) makes:

A $10 000 loan is repaid with 36 payments of $310. Total interest:

In a spreadsheet, which function finds a loan payment?

For parents and teachers

Big idea (B3.7): Students find the regular payment of an ordinary simple annuity, both for savings goals and for loans, and (as the curriculum suggests) use a spreadsheet to find the total interest paid over the life of a loan and compare it with the original principal.

How to help: If you have a car loan, mortgage or line of credit, look at a statement together (hide account numbers) and find the interest and principal parts of a payment. Discuss the trade-off of a longer term: a lower payment but more interest.

Common difficulties:

Choosing the wrong formula. Ask: is the known amount now (a loan, use PV) or later (a goal, use FV)?
Calculating interest on the original loan every month instead of on the current balance.
Rounding inside the formula.

Canadian notes (checked October 2026): Section 6 of the federal Interest Act requires blended-payment mortgages to state the rate calculated yearly or half-yearly (laws-lois.justice.gc.ca), which is why Canadian fixed-rate mortgages are usually quoted with semi-annual compounding. Mortgages with monthly payments are therefore general annuities, beyond MCF3M; the challenge question simplifies this. Rates in this lesson are examples, not current offers.