Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 8 · pp. 482–549
This chapter shows how money grows or shrinks over time, and how sequences and functions describe that change. It starts with simple interest and compound interest, then moves to present and future value, and then to annuities, loans, amortization schedules, and tools such as spreadsheets and graphing calculators. In this part, you meet the basic vocabulary of interest and work with simple interest.
Interest is money earned on an investment, or the cost of borrowing money. If you deposit money, the bank pays you interest. If you borrow, you pay interest to the lender. For example, 120 in interest. (p. 486)
The principal, , is the sum of money that is invested or borrowed. The amount, , is the total value of the investment or loan. It is the principal plus the interest:
With and , the amount is . (p. 486)
An interest rate is per year unless the problem says otherwise. The notation %/a means "per annum," or per year. So 5%/a means 5% interest is earned each year. If a time is given in months or days, change it to years before using a formula. For example, 6 months is . (pp. 487, 491)
Simple interest is paid or charged only on the original principal. The interest already earned is never added to the base for later calculations. So the same dollar amount of interest is paid every period. For example, 120 every year, whether it is year 1 or year 10. (pp. 486, 490)
Each period earns , where is the rate written as a decimal. After periods the total interest is
The time must be in the same units as the rate. With a yearly rate, is in years. (pp. 488, 491)
Example. Allen invests at 2.4%/a, so . One year of interest is . (pp. 487–488)
To get the amount, add the interest to the principal and factor out :
For Allen after years, . The interest earned is . (pp. 488, 491)
Every year adds the same interest, , to the amount. This gives the following results.
For Allen, the amount is . A scatter plot of amount against year lies on a straight line. The slope is 3240 principal. For A(t) = 500 + 20tI(t) = 20t$. (pp. 488, 490, 501)
A spreadsheet can list a simple-interest loan year by year. Each row adds the same yearly interest to the previous amount. Tina borrows 15 000(0.068) = 102016 020, $17 040, and so on. After 10 years she owes
The check with the formula is . (p. 489)
When money is borrowed or invested for less than a year, write the time as a fraction of a year. For days, use days over 365. Then use .
Example. Philip borrows r = 0.26t = \frac{85}{365}$.
The interest is . (p. 489)
To find the time needed to reach a target amount, substitute into and solve for .
Example. Tanya invests 8000. The yearly interest is , so
The 8 means 8 full years. The decimal part must be changed to days: . Rounding up to a whole day gives about 8 years and 200 days. (p. 490)
Some savings accounts pay interest on the minimum monthly balance. The interest for a month is figured on the lowest balance held during that month. Money deposited during the month earns nothing until the next month. (p. 485)
Example. Mark starts with 50 each week into an account paying 0.35% on the minimum monthly balance. His goal is 200 he started with, and the interest is . His balance at the end of month 1 is . The interest is small, but it adds up. It shortens the time to reach $1895 by a little. To track this, build a table with starting balance, deposits, interest earned, and final balance for each month. (p. 485)
With compound interest, the interest is added to the principal before the next interest is worked out. So you earn interest on the principal and also on interest you already earned (p. 493).
Example: Mena invests 2000 × 0.06 = 2120, which is $127.20. The interest grows each year because the balance grows (pp. 493, 499).
The compounding period is how often the interest is calculated and added. Annually means 1 time per year, semi-annually 2, quarterly 4 and monthly 12 (p. 493). A 1000 × 0.02 (p. 499).
The formulas need the rate per compounding period and the number of compounding periods. To get them:
For 4.8%/a compounded quarterly for 21 years: and . For monthly compounding, divide by 12 and multiply the years by 12 (pp. 496, 499).
A common mistake is to put the annual rate and the number of years straight into the formula when compounding is more often than yearly. That gives a very different answer. Here you must use 0.012 and 84, not 0.048 and 21 (pp. 496, 499).
Future value is the total amount, , of an investment after a certain length of time (pp. 494, 499). A 1050 after one year (p. 502).
Each period, the balance is multiplied by . This gives
where is the principal, is the rate per period and is the number of periods (pp. 494–495, 499).
Example: Tim borrows 5300(1.046) = $5543.805543.80(1.046) \approx $5798.815300(1.046)^{10} \approx $8309.84$ (p. 494).
The total interest is the future value minus the principal. Factoring out gives
For Tim, I = 8309.84 - 5300 = \3009.84$. Use this form when a question asks for the interest and not the final amount (pp. 495, 499).
The balances at the end of each period are This is a geometric sequence with common ratio . The general term is an exponential function of time. For Tim's loan the ratio is 1.046 (pp. 494–495, 499, 501). The growth is a fixed percent each period, not a fixed dollar amount.
Lara's investment. Her grandparents invested i = 0.012n = 84$:
This is the same as earning 1.2%/a compounded annually for 84 years. Only the rate per period and the number of periods matter (p. 496).
Starting early. Trudy invests i = 0.08 \div 12n = 600538 781.94. Lina invests the same amount for 20 years () and gets 489 513.91 more. More compounding periods give interest more chances to earn interest (pp. 496–497).
Simple interest is modelled by a linear function. Compound interest is modelled by an exponential function. For A(t) = 500 + 20tA(t) = 500(1.04)^t520 after year 1. After 20 years, simple interest gives 1095.56 (pp. 501–502).
On a graph, simple interest is a straight line with a constant rate of change. Compound interest is a curve that starts at the same point and equals the line after one period. Then it climbs above the line, with a rate of change that keeps growing (pp. 498–499, 502).
Doubling time shows the difference. At 5%/a simple interest, 1000 = 1000(0.05)tt = 20i = 0.025n = 281000(1.025)^{28} \approx $1996.50n = 402685.06 (pp. 497–499).
Present value, , is the principal you must invest now to reach a given future value in a given time. It answers "how much should I invest today?" The symbol is used because already means principal (pp. 493, 503).
Start from and divide both sides by :
Example: Anton's parents want $15 000 in 10 years at 6%/a compounded annually. Then
(pp. 494, 497, 503).
Future value and present value are inverse problems with one formula. Future value multiplies by and present value divides by it. If you grow the present value forward, you get the future value back (pp. 495, 502).
Loan limits are present-value questions. Monica will repay at most i = 0.016n = 20\frac{20,000}{1.016^{20}} \approx $14,559.8120,000 - 14,559.81 = $5440.19$ (p. 495).
To find in , rearrange to . Then take the th root and subtract 1.
Example: Tony wants 50 000 in 40 years. Then , so and per year. You can also graph and and find where they meet (pp. 496–497).
The Rule of 72 is a quick estimate. An investment earning %/a compounded annually takes about years to double (p. 510).
At 8%/a it predicts years. The formula gives 1000(1.08)^9 \approx \1999$, which is very close to double (p. 510).
You can use the rule repeatedly, because each doubling period doubles the value again. At 8%/a, money doubles about every 9 years. In 45 years there are doublings, so 1000 \times 2^5 = $32,0001000(1.08)^{45}31 920, so the estimate is close (p. 510).
An annuity is a series of equal payments or investments made at regular intervals. In a simple ordinary annuity, the payment dates match the compounding periods, and each payment is made at the end of its period. Christine puts $1000 at the end of each year into a bond paying 6.5%/a compounded annually. That is a simple ordinary annuity. So is a fixed monthly deposit into an account that compounds monthly. (pp. 504–505)
Two terms are useful. The payment interval is the time between successive payments. The term is the time from the first payment to the last payment. (p. 520)
The future value of an annuity is the sum of all the payments plus the interest they earn, measured at the end of the term. To find it, work out what each payment is worth at the end of the term, then add the values. A timeline helps. (pp. 504, 506)
Because deposits come at the end of each period, the timing works like this:
(pp. 505–507)
Listing the values from the last payment back to the first gives
These values form a geometric sequence. The first term is and the common ratio is . Each earlier payment has had one more period of interest, so its value is times the value of the payment after it. The annuity's total value is therefore a geometric series. (pp. 505–506, 520)
Example: Hans deposits i = 0.048 \div 2 = 0.0241000,\ 1000(1.024),\ \dots,\ 1000(1.024)^{39}$. The future value is the sum of these 40 terms. (pp. 505–506)
The sum of a geometric series is . Here and , so . This gives the formula for a simple ordinary annuity:
(pp. 506–508, 520)
To get , divide the annual rate by the number of periods per year. To get , multiply the years by the periods per year.
Example 1: Hans has and . Then
He deposits only 40 00025 927 or so is interest. (pp. 505–506)
Example 2: Chie invests i = 0.052 \div 4 = 0.013n = 25 \times 4 = 100101 487.91. Her first payment is worth at the end, and her last is worth just $500. (pp. 506–508)
Sometimes the future value is known and the payment is not. Substitute , and into the formula, then solve for :
Example: Sam wants i = 0.096 \div 12 = 0.008n = 360$. Evaluate the bracket first:
This number is the annuity factor. It tells you how many "payments' worth" the account holds at the end. Now , so R = 500\,000 \div 2076.413 \approx \240.80$ per month. Round to the nearest cent. (pp. 508, 520)
In an ordinary annuity, the first deposit earns interest for periods, not . It is made at the end of period 1, so it has only periods left. The last deposit is made at the end of the term, so it earns no interest at all. This is why the exponents in the series stop at . (pp. 505–507)
A loan with equal payments can be treated as a schedule. Build a spreadsheet and update the balance after every payment until it is about zero. Each row follows three steps:
Only the part of the payment left after interest reduces the debt. (p. 520)
Example: Nahid borrows 750 each month. The monthly rate is . The first interest is 95\,000 \times 0.0045 = \427.50750 − 322.50$ reduces the balance. As the balance falls, the interest shrinks and more of each payment goes to principal. The balance is near zero after payment 188. (p. 520)
The number of payments is not the same as the number of years. Divide by the payments per year. Then change the decimal part of a year into months by multiplying by 12.
For Nahid, years. The decimal part gives months. The loan takes about 15 years and 8 months to repay. (p. 520)
An annuity is a series of equal payments made at regular intervals. The payment interval is the time between payments, and the term runs from the first payment to the last. In these problems the payment interval equals the compounding period. The future value of an annuity is the total of all the payments plus all the interest they earn, measured at the end of the term (p. 521).
In an ordinary annuity, each payment is made at the end of its period. So the last payment earns no interest and is worth just . The one before it earns interest once, so it is worth . The first payment earns interest for periods. For 40 payments of 250 earns interest 39 times and the last earns none (pp. 521, 542).
Listing the values from last to first gives a geometric series:
The first term is and the common ratio is (pp. 521, 532). The geometric sum formula then gives:
Here is the payment per period, is the rate per compounding period as a decimal, and is the number of periods (pp. 521, 532).
Example. Eric invests i=0.082n=25250 annuity above gives $15 100.50 (p. 542).
The formulas need the rate and the number of periods per compounding period. Divide the annual rate by the number of periods per year, and multiply the years by the number of periods per year. For 6.6%/a compounded monthly over 20 years:
Plugging the annual rate or the number of years straight into the formula gives wrong answers (pp. 528, 534).
Paying more often. Suppose the same yearly total, 100 monthly, 600 semi-annually, or $1200 yearly. Each plan compounds at the same frequency as its payments. The monthly plan puts money in earlier and compounds it more often, so it ends with the largest future value (p. 522).
Finding the number of payments. Kenny invests 6500. Here . Solve
Isolate , then use logarithms or technology to find . Always round up to a whole payment, because a partial payment cannot be made (pp. 522, 532).
The present value of an annuity is the lump sum needed at the start of the term to fund all the regular payments. It is found by adding the present values of the individual payments (pp. 523, 529).
A payment of due periods from now needs invested today. Each later payment is worth times the one before it. So the present values form a geometric sequence with first term and ratio (pp. 523–525).
Kew's scholarships. Kew wants \frac{1000}{1.055}\frac{1000}{1.055^2}\frac{1}{1.055}$ times the previous one (p. 523).
**Worked series, a=500(1.083)^{-1}r=1.083^{-1}n=103310.11. That is far less than the $5000 that will be paid out, because the invested money keeps earning interest while it waits (pp. 524–525).
The formula. Apply the geometric sum to and . Multiplying the numerator and denominator by turns the denominator into . This gives:
For this formula, is the payment, the rate per period, and the number of periods (pp. 526, 529).
A common mistake is to think the present value of an annuity is just the sum of the payments. For example, someone might claim that 1 000 000 today. But money received later is worth less than money in hand, since money in hand could be invested and grow (p. 527).
To compare options, find the present value of the annuity at the given rate and compare it with the lump sum. In a spreadsheet, divide each payment by and add the results. For Sharon's lottery at 5%/a, the 20 payments of 623 110.52. A lump sum of that amount is worth the same as the annuity (p. 527).
The amount borrowed equals the present value of all the payments. To find the payment, set equal to the loan and solve for (p. 528).
Len's loan. Len borrows i=0.0055n=240$. Then
so R \approx \1502.94$ per month (p. 528).
Total interest. Let be the sum of all payments. Then . Len pays about 360 706.60 − 160 706.60 (p. 528).
Scaling a loan. The present value is proportional to . If the loan amount is multiplied by some factor with the term and rate fixed, the payment is multiplied by the same factor. A loan twice as large needs payments twice as large. The rate is different. It sits inside in a non-linear way, so doubling the interest rate does not double the payment (p. 541).
A retirement plan has two annuities joined together. In the saving stage, monthly deposits grow to a future value at retirement. In the spending stage, that amount acts as the present value of the monthly withdrawals. Set the two equal and solve (pp. 522, 532, 537).
Charmaine's plan. She wants i=0.0075n=180246 500. Then find the monthly deposit over 25 years () whose future value is that amount. It comes to about $220 per month (p. 522).
An amortization schedule is a table with one row per payment on a loan or investment. Each row shows the payment, the interest paid, the principal paid, and the remaining balance. The interest is the previous balance times the rate per period. The principal paid is the payment minus that interest. The new balance is the old balance minus the principal paid. Reading down the table shows how the debt shrinks, and you can look up when the balance drops below any target, such as half the loan (pp. 533, 539).
The regular payment on an amortized loan is fixed. It is not an average, and it is not principal only. Each payment covers the interest first, and what is left reduces the principal. The interest part is big at the start and shrinks. The principal part grows, but the total payment stays the same (pp. 539, 549).
Three tools can find an unknown interest rate. Take Tina's stereo, which costs 29.95 a month for 2 years (pp. 533, 526):
The TVM Solver asks for the number of periods (N), present value (PV), payment (PMT), future value (FV), and payments per year. Leave the unknown blank and solve for it. A lump-sum investment has no regular payments, so set PMT = 0. For 50 000 goal, N comes out slightly above 460 months (pp. 526, 533).
Guess-and-check works on . Here (p. 534). Try values of and narrow in on the goal:
On a graphing calculator, graph and the horizontal line . The intersection near shows when the goal is reached. The same method works for the other bank's function (p. 535).
A higher rate can win even with less frequent compounding. Key Bank pays 5%/a compounded quarterly, so . The 50 000 in 130 quarters, or 32 years 6 months. That is almost 6 years sooner than the 4.2% monthly account. The extra compounding does not make up for the lower rate (pp. 534, 535, 526). More interest is earned when the rate is higher or when there are more compounding periods per year at the same rate (p. 539).
Bigger payments. Lia borrows i = 0.0041000 a month takes 319 months. Paying $1500 takes 164 months, almost 13 years sooner. The extra money goes straight to principal, so less interest builds up (pp. 526, 527, 539).
Spotting the payoff row. The loan is paid off in the row where the balance first goes negative. Lia's balance at payment 164 is -\286.99$. This means the last payment only needs to be smaller than the others (p. 527).
Equal chunks of debt take unequal time. Take a 861.50 payments. Interest is charged on the balance, so early payments are mostly interest. The schedule shows (pp. 538, 539):
Paying more often. John deposits 2400 yearly. Both earn 4%/a with matching compounding. After 5 years John has 12 999.17, a gap of $260.63. John's money starts earning interest sooner (p. 538).
One-time extra payment. Take Nadia's 15 000 extra. Find the balance after 36 payments and subtract $15 000. Then keep the same payment and solve for the new number of payments. Interest saved is the old total interest minus the new total (p. 540).
Rate change. Natalie's 924.86. After 5 years the rate becomes 7.5%. Use the balance at that moment as the new present value. Then do either of these (p. 541):
To find a future value in a spreadsheet, list each payment's value after growing for the periods it has been invested. Then add them. For 40 payments of 15 100.50. The formula gives the same result: (p. 542).
To find a present value, divide each payment by for its period and add. For 20 payments of 50/1.00550/1.005^250/1.005^{20}949.37$, matching the formula (p. 543).
Three formulas to keep straight (p. 544):
You can tell the interest type from amounts owed in successive years. Equal differences mean simple interest. Equal ratios mean compound interest. Loan #1 (3942, 146 each year, so it is simple. Loan #2 (1036.18, $1098.35) has ratio about 1.06, so it is compound at about 6% (p. 546).
The common ratio of a future value series is . If with monthly compounding, then per month. Multiply by 12 to get 0.06, or 6%/a. Do not answer 0.5% or 12% (p. 548).
