81
4 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Amortization and Sinking Funds Funds Arithmetic and Geometric Arithmetic and Geometric Progressions Progressions Mathematics of Finance Mathematics of Finance

4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

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Page 1: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

44

Compound InterestCompound Interest AnnuitiesAnnuities Amortization and Sinking FundsAmortization and Sinking Funds Arithmetic and Geometric Arithmetic and Geometric

ProgressionsProgressions

Mathematics of FinanceMathematics of Finance

Page 2: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

4.14.1Compound InterestCompound Interest

(4)(3)

12

1

0.081000 1

4

1000(1.02)

1268.24

mtr

A Pm

(4)(3)

12

1

0.081000 1

4

1000(1.02)

1268.24

mtr

A Pm

eff

1

1 1

0.081 1

1

1.08 1

0.08

mr

rm

eff

1

1 1

0.081 1

1

1.08 1

0.08

mr

rm

Page 3: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Simple Interest FormulasSimple Interest Formulas

Simple interestSimple interest is the interest that is is the interest that is computed on the computed on the original principaloriginal principal only. only.

If If II denotes the denotes the interest interest on aon a principal principal PP (in dollars) at an (in dollars) at an interest rateinterest rate of of rr per year per year for for t t yearsyears, then we have, then we have

I I == Prt Prt

The The accumulated amountaccumulated amount AA, the , the sumsum of the of the principalprincipal and and interestinterest after after tt yearsyears is given by is given by

AA = = PP + + II = = PP + + PrtPrt

= = PP(1 + (1 + rtrt))

and is a and is a linear functionlinear function of of tt..

Page 4: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

A bank pays A bank pays simple interestsimple interest at the at the raterate of of 8%8% per year for per year for certain deposits. certain deposits.

If a customer If a customer depositsdeposits $1000$1000 and makes and makes no withdrawalsno withdrawals for for 33 years, what is the years, what is the total amounttotal amount on deposit at the end on deposit at the end of of three yearsthree years??

What is the What is the interest earnedinterest earned in that period? in that period?SolutionSolution Using the Using the accumulated amount accumulated amount formula with formula with PP = 1000 = 1000, ,

rr = 0.08 = 0.08, and , and tt = 3 = 3, we see that the , we see that the total amounttotal amount on on deposit at the end of deposit at the end of 33 years is given by years is given by

or or $1240$1240..

(1 )

1000[1 (0.08)(3)] 1240

A P rt

(1 )

1000[1 (0.08)(3)] 1240

A P rt

Page 5: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

A bank pays A bank pays simple interestsimple interest at the at the raterate of of 8%8% per year for per year for certain deposits. certain deposits.

If a customer If a customer depositsdeposits $1000$1000 and makes and makes no withdrawalsno withdrawals for for 33 years, what is the years, what is the total amounttotal amount on deposit at the end on deposit at the end of of three yearsthree years??

What is the What is the interest earnedinterest earned in that period? in that period?SolutionSolution The The interest earnedinterest earned over the over the three year periodthree year period is given by is given by

or or $240$240..

1000(0.08)(3) 240

I Prt 1000(0.08)(3) 240

I Prt

Page 6: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Trust Funds Trust Funds

An An amountamount of of $2000$2000 is is investedinvested in a in a 1010-year trust fund -year trust fund that pays that pays 6%6% annual annual simple interestsimple interest..

What is the What is the total amounttotal amount of the trust fund at the end of of the trust fund at the end of 1010 yearsyears??

SolutionSolution The The total amounttotal amount is given by is given by

or or $3200$3200..

(1 )

2000[1 (0.06)(10)] 3200

A P rt

(1 )

2000[1 (0.06)(10)] 3200

A P rt

Page 7: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compound InterestCompound Interest

Frequently, interest earned is Frequently, interest earned is periodicallyperiodically added to the added to the principal and thereafter principal and thereafter earns interest itselfearns interest itself at the same at the same rate. This is called rate. This is called compound interestcompound interest..

Suppose Suppose $1000$1000 (the principal) is deposited in a bank for a (the principal) is deposited in a bank for a term of term of 33 years, earning interest at the rate of years, earning interest at the rate of 8%8% per year per year compounded annually.compounded annually.

Using the Using the simple interest formulasimple interest formula we see that the we see that the accumulated amount after the first year isaccumulated amount after the first year is

or or $1080$1080..

1 (1 )

1000[1 0.08(1)]

1000(1.08) 1080

A P rt

1 (1 )

1000[1 0.08(1)]

1000(1.08) 1080

A P rt

Page 8: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compound InterestCompound Interest

To find the accumulated amount To find the accumulated amount AA22 at the end of the at the end of the

second year, we use the second year, we use the simple interest formulasimple interest formula again, this again, this time with time with PP = = AA11, obtaining:, obtaining:

or approximately or approximately $1166.40$1166.40..

2 1

2 2

(1 ) (1 )

1000[1 0.08(1)][1 0.08(1)]

1000(1 0.08) 1000(1.08) 1166.40

A P rt A rt

2 1

2 2

(1 ) (1 )

1000[1 0.08(1)][1 0.08(1)]

1000(1 0.08) 1000(1.08) 1166.40

A P rt A rt

Page 9: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compound InterestCompound Interest

We can use the We can use the simple interest formulasimple interest formula yet againyet again to find to find the accumulated amount the accumulated amount AA33 at the end of the third year: at the end of the third year:

or approximately or approximately $1259.71$1259.71..

3 2

2

3 3

(1 ) (1 )

1000[1 0.08(1)] [1 0.08(1)]

1000(1 0.08) 1000(1.08) 1259.71

A P rt A rt

3 2

2

3 3

(1 ) (1 )

1000[1 0.08(1)] [1 0.08(1)]

1000(1 0.08) 1000(1.08) 1259.71

A P rt A rt

Page 10: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compound InterestCompound Interest

Note that the Note that the accumulated amountsaccumulated amounts at the end of each year at the end of each year have the following form:have the following form:

These observations suggest the following These observations suggest the following general rulegeneral rule::✦ If If PP dollars are dollars are investedinvested over a term of over a term of tt yearsyears earning earning

interestinterest at the rate of at the rate of rr per year per year compounded annuallycompounded annually, , then the then the accumulated amountaccumulated amount is is

1

22

33

1000(1.08)

1000(1.08)

1000(1.08)

A

A

A

1

22

33

1000(1.08)

1000(1.08)

1000(1.08)

A

A

A

1

22

33

(1 )

(1 )

(1 )

A P r

A P r

A P r

1

22

33

(1 )

(1 )

(1 )

A P r

A P r

A P r

or:

(1 )tA P r (1 )tA P r

Page 11: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compounding More Than Once a YearCompounding More Than Once a Year

The formula The formula

was derived under the was derived under the assumptionassumption that interest was that interest was compoundedcompounded annuallyannually..

In practice, however, interest is usually In practice, however, interest is usually compoundedcompounded more more than once a yearthan once a year..

The interval of time between successive interest The interval of time between successive interest calculations is called the calculations is called the conversion periodconversion period..

(1 )tA P r (1 )tA P r

Page 12: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compounding More Than Once a YearCompounding More Than Once a Year

If interest at a nominal rate of If interest at a nominal rate of rr per year is per year is compoundedcompounded mm times a year on a times a year on a principalprincipal of of PP dollars, then the dollars, then the simple simple interest rate interest rate perper conversion periodconversion period is is

For example, if the For example, if the nominal interestnominal interest rate is rate is 8%8% per year, per year, and interest is and interest is compounded quarterlycompounded quarterly, then, then

or or 2%2% per period. per period.

Annual interest rate

Periods per year

ri

m

Annual interest rate

Periods per year

ri

m

0.080.02

4

ri

m

0.080.02

4

ri

m

Page 13: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compounding More Than Once a YearCompounding More Than Once a Year

To find a general formula for the accumulated amount, we To find a general formula for the accumulated amount, we applyapply

repeatedlyrepeatedly with the interest rate with the interest rate i i == r/m r/m.. We see that the We see that the accumulated amountaccumulated amount at the at the end of each end of each

periodperiod is as follows: is as follows:

(1 )tA P r (1 )tA P r

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )

A P iA A i P i i P iA A i P i i P i

First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )

A P iA A i P i i P iA A i P i i P i

First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

Page 14: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Compound Interest FormulaCompound Interest Formula

Where Where n n == mt mt, and, and

AA == Accumulated amount at the end of Accumulated amount at the end of tt years years

PP == Principal Principal

rr == Nominal interest rate per year Nominal interest rate per year

mm == Number of conversion periods per year Number of conversion periods per year

tt == Term (number of years) Term (number of years)

1n

rA P

m 1

nr

A Pm

There are There are n n == mt mt periods in periods in tt years, so the years, so the accumulated accumulated amountamount at at the end ofthe end of tt yearsyears is given by is given by

Page 15: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the Find the accumulated amountaccumulated amount after after 3 3 years if years if $1000$1000 is is invested at invested at 8%8% per year per year compoundedcompoundeda.a. AnnuallyAnnuallyb.b. SemiannuallySemiannuallyc.c. QuarterlyQuarterlyd.d. MonthlyMonthlye.e. DailyDaily

Page 16: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolutiona.a. Annually. Annually.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and mm = 1 = 1. . Thus, Thus, i = ri = r = 0.08 = 0.08 and and nn = 3 = 3, so, so

or or $1259.71$1259.71..

3

3

1

0.081000 1

1

1000(1.08)

1259.71

nr

A Pm

3

3

1

0.081000 1

1

1000(1.08)

1259.71

nr

A Pm

Page 17: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolutionb.b. Semiannually. Semiannually.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and mm = 2 = 2. . Thus, Thus, and and nn = (3)(2) = 6 = (3)(2) = 6, so, so

or or $1265.32$1265.32..

6

6

1

0.081000 1

2

1000(1.04)

1265.32

nr

A Pm

6

6

1

0.081000 1

2

1000(1.04)

1265.32

nr

A Pm

0.082i 0.082i

Page 18: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolutionc.c. Quarterly. Quarterly.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and mm = 4 = 4..Thus, Thus, and and nn = (3)(4) = 12 = (3)(4) = 12, so, so

or or $1268.24$1268.24..

12

12

1

0.081000 1

4

1000(1.02)

1268.24

nr

A Pm

12

12

1

0.081000 1

4

1000(1.02)

1268.24

nr

A Pm

0.084i 0.084i

Page 19: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolutiond.d. Monthly. Monthly.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and mm = 12 = 12..Thus, Thus, and and nn = (3)(12) = 36 = (3)(12) = 36, so, so

or or $1270.24$1270.24..

36

36

1

0.081000 1

12

0.081000 1

12

1270.24

nr

A Pm

36

36

1

0.081000 1

12

0.081000 1

12

1270.24

nr

A Pm

0.0812i 0.0812i

Page 20: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolutione.e. Daily. Daily.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and mm = 365 = 365..Thus, Thus, and and nn = (3)(365) = 1095 = (3)(365) = 1095, so, so

or or $1271.22$1271.22..

1095

1095

1

0.081000 1

365

0.081000 1

365

1271.22

nr

A Pm

1095

1095

1

0.081000 1

365

0.081000 1

365

1271.22

nr

A Pm

0.08365i 0.08365i

Page 21: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

One question arises on compound interest: One question arises on compound interest:

✦ What happens to the What happens to the accumulated amountaccumulated amount over a over a fixed period of time if the interest is fixed period of time if the interest is compoundedcompounded more and more frequentlymore and more frequently??

We’ve seen that We’ve seen that the more oftenthe more often interest is interest is compoundedcompounded, , the largerthe larger the the accumulated amountaccumulated amount..

But does the But does the accumulated amountaccumulated amount approach aapproach a limitlimit when interest is computed more and more when interest is computed more and more frequently?frequently?

Continuous Compounding of InterestContinuous Compounding of Interest

Page 22: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Recall that in the Recall that in the compound interestcompound interest formula formula

the number ofthe number of conversion periods conversion periods is is mm.. So, we should let So, we should let mm get get larger and largerlarger and larger (approach (approach

infinity) and see what happens to the infinity) and see what happens to the accumulated accumulated amountamount AA..

1mt

rA P

m 1

mtr

A Pm

Continuous Compounding of InterestContinuous Compounding of Interest

Page 23: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

uu

10 10 2.593742.59374

100100 2.704812.70481

10001000 2.716922.71692

10,00010,000 2.718152.71815

100,000100,000 2.718272.71827

1,000,0001,000,000 2.718282.71828

11

u

u

If we let If we let u u == m m//rr so that so that m m == ru ru, then the above formula , then the above formula becomesbecomes

1 11 1 or

rturt u

A P A Pu u

1 11 1 or

rturt u

A P A Pu u

The table shows us that when The table shows us that when u u gets gets larger larger andand larger larger the expression the expression

approachesapproaches 2.718282.71828 (rounding to (rounding to five decimal places).five decimal places).

It can be shown that as It can be shown that as u u gets gets largerlarger and and largerlarger, the value of , the value of the expression the expression approachesapproaches the the irrational numberirrational number 2.71828…2.71828… which we denote by which we denote by ee..

11

u

u

Continuous Compounding of InterestContinuous Compounding of Interest

Page 24: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Continuous Compound Interest FormulaContinuous Compound Interest Formula

A A == Pe Pertrt

wherewhere

PP == PrincipalPrincipal

rr == Annual interest rate Annual interest rate compounded compounded

continuouslycontinuously

tt == Time in yearsTime in years

AA == Accumulated amount at the end Accumulated amount at the end of of tt years years

Continuous Compounding of InterestContinuous Compounding of Interest

Page 25: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExamplesExamples

Find the Find the accumulated amountaccumulated amount after after 33 years if years if $1000$1000 is is invested at invested at 8%8% per year compounded per year compounded (a)(a) daily, and daily, and (b)(b) continuously. continuously.

SolutionSolution

a.a. Using the Using the compound interest compound interest formula with formula with PP = 1000 = 1000, , rr = 0.08 = 0.08, , mm = 365 = 365, and , and tt = 3 = 3, we find, we find

b.b. Using the Using the continuous compound interestcontinuous compound interest formula with formula with PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and tt = 3 = 3, we find , we find

A A == Pe Pertrt = 1000= 1000ee(0.08)(3)(0.08)(3) ≈≈ 1271.25 1271.25

Note that the two solutions are Note that the two solutions are very close to each othervery close to each other..

(365)(3)0.08

1 1000 1 1271.22365

mtr

A Pm

(365)(3)0.08

1 1000 1 1271.22365

mtr

A Pm

Page 26: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Effective Rate of InterestEffective Rate of Interest

The last example demonstrates that the The last example demonstrates that the interest actually interest actually earnedearned on an investment on an investment depends on the frequencydepends on the frequency with with which the interest is which the interest is compoundedcompounded..

For clarity when comparing interest rates, we can use For clarity when comparing interest rates, we can use what is called the what is called the effective rateeffective rate (also called the (also called the annual annual percentage yieldpercentage yield): ): ✦ This is the This is the simple interest ratesimple interest rate that would produce the that would produce the

same accumulated amountsame accumulated amount in in 11 year as the year as the nominal rate nominal rate compoundedcompounded mm timestimes a year. a year.

We want to We want to derive a relationderive a relation between the between the nominal nominal compounded ratecompounded rate and the and the effective rateeffective rate. .

Page 27: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Effective Rate of InterestEffective Rate of Interest

The accumulated amount after The accumulated amount after 11 year at a year at a simple interestsimple interest raterate RR per year is per year is

The accumulated amount after The accumulated amount after 11 year at a year at a nominal interest nominal interest raterate rr per year per year compounded compounded mm times times a year is a year is

Equating the two expressions givesEquating the two expressions gives

(1 )A P R (1 )A P R

1m

rA P

m 1

mr

A Pm

(1 ) 1

1 1

m

m

rP R P

m

rR

m

(1 ) 1

1 1

m

m

rP R P

m

rR

m

Page 28: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Effective Rate of Interest FormulaEffective Rate of Interest Formula

Solving the last equation for Solving the last equation for RR we obtain the formula for we obtain the formula for computing the effective rate of interest:computing the effective rate of interest:

eff 1 1m

rr

m

eff 1 1m

rr

m

wherewhere

rreffeff == Effective rate of interestEffective rate of interest

rr == Nominal interest rate per yearNominal interest rate per year

mm == Number of conversion periods per yearNumber of conversion periods per year

Page 29: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the Find the effectiveeffective rate of interest rate of interest corresponding to a corresponding to a nominal ratenominal rate of of 8%8% per year per year compoundedcompoundeda.a. AnnuallyAnnuallyb.b. SemiannuallySemiannuallyc.c. QuarterlyQuarterlyd.d. MonthlyMonthlye.e. DailyDaily

Page 30: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolution

a.a. Annually. Annually.

Let Let rr = 0.08 = 0.08 and and mm = 1 = 1. Then. Then

or or 8%8%..

1

eff

0.081 1

1

1.08 1

0.08

r

1

eff

0.081 1

1

1.08 1

0.08

r

Page 31: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolution

b.b. Semiannually. Semiannually.

Let Let rr = 0.08 = 0.08 and and mm = 2 = 2. Then. Then

or or 8.16%8.16%..

2

eff

0.081 1

2

1.0816 1

0.0816

r

2

eff

0.081 1

2

1.0816 1

0.0816

r

Page 32: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolution

c.c. Quarterly. Quarterly.

Let Let rr = 0.08 = 0.08 and and mm = 4 = 4. Then. Then

or or 8.243%8.243%..

4

eff

0.081 1

4

1.08243 1

0.08243

r

4

eff

0.081 1

4

1.08243 1

0.08243

r

Page 33: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolution

d.d. Monthly. Monthly.

Let Let rr = 0.08 = 0.08 and and mm = 12 = 12. Then. Then

or or 8.300%8.300%..

12

eff

0.081 1

12

1.08300 1

0.08300

r

12

eff

0.081 1

12

1.08300 1

0.08300

r

Page 34: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

SolutionSolution

e.e. Daily. Daily.

Let Let rr = 0.08 = 0.08 and and mm = 365 = 365. Then. Then

or or 8.328%8.328%..

365

eff

0.081 1

365

1.08328 1

0.08328

r

365

eff

0.081 1

365

1.08328 1

0.08328

r

Page 35: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Effective Rate Over Several YearsEffective Rate Over Several Years

If the effective rate of interest If the effective rate of interest rreffeff is known, is known,

then the accumulated amount after then the accumulated amount after tt years on years on an investment of an investment of PP dollars can be more readily dollars can be more readily computed by using the formulacomputed by using the formula

eff(1 )tA P r eff(1 )tA P r

Page 36: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Present ValuePresent Value

Consider the compound interest formula:Consider the compound interest formula:

The The principalprincipal PP is often referred to as the is often referred to as the present valuepresent value, , and the and the accumulated valueaccumulated value AA is called the is called the future valuefuture value, , since it is realized at a future date.since it is realized at a future date.

On occasion, investors may wish to determine how much On occasion, investors may wish to determine how much money they should money they should invest nowinvest now, at a fixed rate of interest, , at a fixed rate of interest, so that they will so that they will realize a certain sumrealize a certain sum at some at some future datefuture date..

This problem may be solved by This problem may be solved by expressingexpressing PP in terms ofin terms of AA..

1mt

rA P

m 1

mtr

A Pm

Page 37: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Present ValuePresent Value

Present value formula for compound interestPresent value formula for compound interest

Where Where andand

1n

P A i 1

nP A i

ri

m n mt

Page 38: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExamplesExamples

How much moneyHow much money should be should be depositeddeposited in a bank paying a in a bank paying a yearly interest rate of yearly interest rate of 6%6% compounded monthlycompounded monthly so that so that after after 33 years the years the accumulated amountaccumulated amount will be will be $20,000$20,000??

SolutionSolution Here, Here, AA = 20,000 = 20,000, , rr = 0.06 = 0.06, , mm = 12 = 12, and , and tt = 3 = 3.. Using the Using the present valuepresent value formula we get formula we get

(12)(3)

1

0.0620,000 1

12

16,713

mtr

P Am

(12)(3)

1

0.0620,000 1

12

16,713

mtr

P Am

Page 39: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExamplesExamples

Find the Find the present valuepresent value of of $49,158.60$49,158.60 duedue in in 55 years at an years at an interest rateinterest rate of of 10%10% per year compounded per year compounded quarterlyquarterly..

SolutionSolution Here, Here, AA = 49,158.60 = 49,158.60, , rr = 0.1 = 0.1, , mm = 4 = 4, and , and tt = 5 = 5.. Using the Using the present valuepresent value formula we get formula we get

(4)(5)

1

0.149,158.60 1

4

30,000

mtr

P Am

(4)(5)

1

0.149,158.60 1

4

30,000

mtr

P Am

Page 40: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Present Value Present Value wwith Continuously Compounded Interestith Continuously Compounded Interest

If we solve the If we solve the continuouscontinuous compound interestcompound interest formula formula

A A == Pe Pertrt

for for PP, we get, we get

P P == Ae Ae–rt–rt

This formula gives the This formula gives the present valuepresent value in terms of the in terms of the futurefuture (accumulated) (accumulated) valuevalue for the case of for the case of continuous continuous compoundingcompounding..

Page 41: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Real Estate Investment Real Estate Investment

Blakely Investment Company owns an office building Blakely Investment Company owns an office building located in the commercial district of a city.located in the commercial district of a city.

As a result of the continued success of an urban renewal As a result of the continued success of an urban renewal program, local business is enjoying a mini-boom.program, local business is enjoying a mini-boom.

The The market valuemarket value of Blakely’s of Blakely’s propertyproperty is is

where where VV((tt) ) is measured in dollars and is measured in dollars and tt is the is the timetime in in years from the present.years from the present.

If the expected If the expected rate of appreciationrate of appreciation is is 9%9% compounded compounded continuouslycontinuously for the next for the next 1010 years, find an years, find an expressionexpression for the for the present valuepresent value PP((tt) ) of the market price of the of the market price of the property that will be valid for the next property that will be valid for the next 1010 years. years.

Compute Compute PP(7)(7), , PP(8)(8), and , and PP(9)(9), and then , and then interpretinterpret your your results.results.

/2( ) 300,000 tV t e /2( ) 300,000 tV t e

Page 42: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Real Estate Investment Real Estate Investment

SolutionSolution Using the Using the present valuepresent value formula for formula for continuous continuous

compoundingcompoundingP P == Ae Ae–rt–rt

with with AA = = VV((tt)) and and rr = 0.09 = 0.09, we find that the , we find that the present valuepresent value of of the market price of the property the market price of the property tt yearsyears from now is from now is

LettingLetting tt = 7 = 7, we find that, we find that

or or $599,837$599,837..

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09(7) 7 / 2(7) 300,000 599,837P e 0.09(7) 7 / 2(7) 300,000 599,837P e

Page 43: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Real Estate Investment Real Estate Investment

SolutionSolution Using the Using the present valuepresent value formula for formula for continuous continuous

compoundingcompoundingP P == Ae Ae–rt–rt

with with AA = = VV((tt)) and and rr = 0.09 = 0.09, we find that the , we find that the present valuepresent value of of the market price of the property the market price of the property tt yearsyears from now is from now is

LettingLetting tt = 8 = 8, we find that, we find that

or or $600,640$600,640..

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09(8) 8 / 2(8) 300,000 600,640P e 0.09(8) 8 / 2(8) 300,000 600,640P e

Page 44: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Real Estate Investment Real Estate Investment

SolutionSolution Using the Using the present valuepresent value formula for formula for continuous continuous

compoundingcompoundingP P == Ae Ae–rt–rt

with with AA = = VV((tt)) and and rr = 0.09 = 0.09, we find that the , we find that the present valuepresent value of of the market price of the property the market price of the property tt yearsyears from now is from now is

LettingLetting tt = 9 = 9, we find that, we find that

or or $598,115$598,115..

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09

0.09 / 2

( ) ( )

300,000 (0 10)

t

t t

P t V t e

e t

0.09(9) 9 / 2(9) 300,000 598,115P e 0.09(9) 9 / 2(9) 300,000 598,115P e

Page 45: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Real Estate Investment Real Estate Investment

SolutionSolution From these results, we see that the From these results, we see that the present valuepresent value of the of the

property’s market price seems to property’s market price seems to decreasedecrease after a certain after a certain period of growth.period of growth.

This suggests that there is an This suggests that there is an optimal timeoptimal time for the owners for the owners to to sellsell..

You can show that the You can show that the highest present valuehighest present value of the of the property’s market value is property’s market value is $600,779$600,779, and that it occurs , and that it occurs at at timetime tt ≈ 7.72≈ 7.72 years, by years, by sketchingsketching the graph of the the graph of the functionfunction PP..

Page 46: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Example: Example: Using Logarithms in Financial ProblemsUsing Logarithms in Financial Problems

How long will it take How long will it take $10,000$10,000 to to growgrow to to $15,000$15,000 if the if the investment earns an investment earns an interest rateinterest rate of of 12%12% per year per year compounded quarterlycompounded quarterly??

SolutionSolution Using the compound interest formulaUsing the compound interest formula

with with AA = 15,000 = 15,000, , PP = 10,000 = 10,000, , rr = 0.12 = 0.12, and , and mm = 4 = 4, we obtain, we obtain

1n

rA P

m 1

nr

A Pm

40.12

15,000 10,000 14

t

40.12

15,000 10,000 14

t

4 15,000(1.03) 1.5

10,000t 4 15,000

(1.03) 1.510,000

t

Page 47: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Example: Example: Using Logarithms in Financial ProblemsUsing Logarithms in Financial Problems

How long will it take How long will it take $10,000$10,000 to to growgrow to to $15,000$15,000 if the if the investment earns an investment earns an interest rateinterest rate of of 12%12% per year per year compounded quarterlycompounded quarterly??

SolutionSolution We’ve gotWe’ve got 4(1.03) 1.5t 4(1.03) 1.5t

4ln(1.03) ln1.5t 4ln(1.03) ln1.5t

4 ln1.03 ln1.5t 4 ln1.03 ln1.5t

ln1.54

ln1.03t

ln1.54

ln1.03t

ln1.53.43

4ln1.03t

ln1.53.43

4ln1.03t

So, it will take about So, it will take about 3.4 3.4 yearsyears for the investment for the investment to to growgrow from from $10,000$10,000 to to $15,000$15,000..

Taking logarithms Taking logarithms on both sideson both sides

loglogbbmmnn = = nnloglogbbmm

Page 48: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

4.24.2AnnuitiesAnnuities

12(1 ) 1 (1 0.01) 1100 1268.25

0.01

niS R

i

12(1 ) 1 (1 0.01) 1100 1268.25

0.01

niS R

i

361 (1 ) 1 (1 0.01)400 12,043

0.01

niP R

i

361 (1 ) 1 (1 0.01)400 12,043

0.01

niP R

i

Page 49: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Future Value of an AnnuityFuture Value of an Annuity

The The future valuefuture value S S of an annuityof an annuity of of nn paymentspayments of of RR dollars each, paid at the dollars each, paid at the endend of each investment of each investment periodperiod into an account that earns into an account that earns interestinterest at the at the rate of rate of ii per period, is per period, is

(1 ) 1niS R

i

(1 ) 1niS R

i

Page 50: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the amount of an Find the amount of an ordinaryordinary annuityannuity consisting of consisting of 1212 monthlymonthly paymentspayments of of $100$100 that earn that earn interestinterest at at 12%12% per per year year compounded monthlycompounded monthly..

SolutionSolution Since Since ii is the is the interest rateinterest rate per period and since interest is per period and since interest is

compounded monthlycompounded monthly in this case, we have in this case, we have

Using the Using the future value of an annuityfuture value of an annuity formula, with formula, with RR = 100 = 100, , nn = 12 = 12, and , and ii = 0.01 = 0.01, we have, we have

or or $1268.25$1268.25..

12(1 ) 1 (1 0.01) 1100 1268.25

0.01

niS R

i

12(1 ) 1 (1 0.01) 1100 1268.25

0.01

niS R

i

0.120.01

12i

0.120.01

12i

Page 51: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Present Value of an AnnuityPresent Value of an Annuity

The The present valuepresent value P P of an annuityof an annuity consisting of consisting of nn paymentspayments of of RR dollars each, paid at the dollars each, paid at the endend of of each investmenteach investment period period into an account that earns into an account that earns interestinterest at the rate of at the rate of ii per period, is per period, is

1 (1 ) niP R

i

1 (1 ) niP R

i

Page 52: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the Find the present value of an ordinary annuitypresent value of an ordinary annuity consisting of consisting of 2424 monthly paymentsmonthly payments of of $100$100 each and earning each and earning interestinterest of of 9%9% per year per year compounded monthlycompounded monthly..

SolutionSolution Here, Here, RR = 100 = 100, , i i = = rr//mm = 0.09 = 0.09//12 = 0.007512 = 0.0075, and , and nn = 24 = 24, so , so

or or $2188.91$2188.91..

241 (1 ) 1 (1 0.0075)100 2188.91

0.0075

niP R

i

241 (1 ) 1 (1 0.0075)100 2188.91

0.0075

niP R

i

Page 53: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example: Applied Example: Saving for a College EducationSaving for a College Education

As a savings program towards Alberto’s college education, As a savings program towards Alberto’s college education, his parents decide to his parents decide to depositdeposit $100$100 at the at the endend of every of every monthmonth into a bank account paying into a bank account paying interestinterest at the rate of at the rate of 6%6% per year per year compounded monthlycompounded monthly..

If the savings program If the savings program beganbegan when Alberto was when Alberto was 66 years years old, how much money would have old, how much money would have accumulatedaccumulated by the by the time he turns time he turns 1818??

Page 54: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example: Applied Example: Saving for a College EducationSaving for a College Education

SolutionSolution By the time the child turns By the time the child turns 1818, the parents would have , the parents would have

mademade

depositsdeposits into the account, so into the account, so nn = 144 = 144.. Furthermore, we have Furthermore, we have RR = 100 = 100, , rr = 0.06 = 0.06, and , and mm = 12 = 12, so, so

Using the Using the future value of an annuityfuture value of an annuity formula, we get formula, we get

or or $21,015$21,015..

144(1 ) 1 (1 0.005) 1100 21,015

0.005

niS R

i

144(1 ) 1 (1 0.005) 1100 21,015

0.005

niS R

i

0.060.005

12i

0.060.005

12i

(18 6) 12 144 (18 6) 12 144

Page 55: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Financing a Car Financing a Car

After making a After making a down paymentdown payment of of $4000$4000 for an automobile, for an automobile, Murphy paid Murphy paid $400$400 per monthper month for for 3636 monthsmonths with with interestinterest charged at charged at 12%12% per year per year compounded monthlycompounded monthly on the on the unpaid balanceunpaid balance..

What was the What was the original costoriginal cost of the car? of the car? What What portionportion of Murphy’s total car payments went of Murphy’s total car payments went

toward toward interest chargesinterest charges??

Page 56: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Financing a Car Financing a Car

SolutionSolution The loan taken up by Murphy is given by the The loan taken up by Murphy is given by the present present

value of the annuity value of the annuity formulaformula

or or $12,043$12,043.. Therefore, the Therefore, the original costoriginal cost of the automobile is of the automobile is $16,043$16,043

(($12,043$12,043 plus the plus the $4000$4000 down paymentdown payment).). The The interest chargesinterest charges paid by Murphy are given by paid by Murphy are given by

(36)(400) – 12,043 = 2,357(36)(400) – 12,043 = 2,357

or or $2,357$2,357..

361 (1 ) 1 (1 0.01)400 12,043

0.01

niP R

i

361 (1 ) 1 (1 0.01)400 12,043

0.01

niP R

i

Page 57: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

4.34.3Amortization and Sinking FundsAmortization and Sinking Funds

8

(0.025)(30,000)3434.02

(1 ) 1 (1 0.025) 1n

iSR

i

8

(0.025)(30,000)3434.02

(1 ) 1 (1 0.025) 1n

iSR

i

360

(120,000)(0.0075)965.55

1 (1 ) 1 (1 0.0075)n

PiR

i 360

(120,000)(0.0075)965.55

1 (1 ) 1 (1 0.0075)n

PiR

i

Page 58: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Amortization FormulaAmortization Formula

The The periodic paymentperiodic payment RR on a on a loanloan of of P P dollars dollars to be to be amortizedamortized over over n n periodsperiods with with interestinterest charged at the rate of charged at the rate of ii per period is per period is

1 (1 ) n

PiR

i 1 (1 ) n

PiR

i

Page 59: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Home Mortgage Payment Home Mortgage Payment

The Blakelys The Blakelys borrowedborrowed $120,000$120,000 from a bank to help from a bank to help finance the purchase of a house.finance the purchase of a house.

The bank charges The bank charges interest at a rate of at a rate of 9%9% per year on the per year on the unpaid balanceunpaid balance, with interest , with interest computationscomputations made at the made at the endend of eachof each month month..

The Blakelys have agreed to repay the loan in The Blakelys have agreed to repay the loan in equalequal monthly installmentsmonthly installments over over 3030 years. years.

How muchHow much should should each paymenteach payment be if the loan is to be be if the loan is to be amortized at the end of term?amortized at the end of term?

Page 60: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Home Mortgage Payment Home Mortgage Payment

SolutionSolution Here, Here, PP = 120,000 = 120,000, , ii = = r r//mm = 0.09 = 0.09//12 = 0.007512 = 0.0075, and , and

nn = (30)(12) = 360 = (30)(12) = 360..

Using the Using the amortizationamortization formula we find that the formula we find that the sizesize of of each each monthly installmentmonthly installment required is given by required is given by

or or $965.55$965.55..

360

(120,000)(0.0075)965.55

1 (1 ) 1 (1 0.0075)n

PiR

i 360

(120,000)(0.0075)965.55

1 (1 ) 1 (1 0.0075)n

PiR

i

Page 61: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Home Affordability Home Affordability

The Jacksons have determined that, after making a The Jacksons have determined that, after making a down down paymentpayment, they could , they could affordafford at most at most $2000$2000 for a for a monthly monthly house paymenthouse payment..

The bank charges The bank charges interestinterest at a rate of at a rate of 7.2%7.2% per year on the per year on the unpaid balanceunpaid balance, with interest computations made at the , with interest computations made at the endend of each month. of each month.

If the loan is to be If the loan is to be amortizedamortized in in equal monthly equal monthly installmentsinstallments over over 3030 years, what is the years, what is the maximum amountmaximum amount that the Jacksons canthat the Jacksons can borrow borrow from the bank? from the bank?

Page 62: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Home Affordability Home Affordability

SolutionSolution We are required to find We are required to find PP, given , given ii = = r r//mm = 0.072 = 0.072//12 = 0.00612 = 0.006, ,

nn = (30)(12) = 360 = (30)(12) = 360, and , and RR = 2000 = 2000.. We first We first solve solve for for PP in the in the amortizationamortization formula formula

SubstitutingSubstituting the numerical values for the numerical values for RR, , nn, and , and ii we obtain we obtain

Therefore, the Jacksons can Therefore, the Jacksons can borrow at mostborrow at most $294,643$294,643..

1 (1 ) nR iP

i

1 (1 ) nR i

Pi

3602000 1 (1 0.006)294,643

0.006P

3602000 1 (1 0.006)

294,6430.006

P

Page 63: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Sinking Fund PaymentSinking Fund Payment

The The periodic paymentperiodic payment RR required to required to accumulate accumulate a sum of a sum of SS dollars over dollars over n n periodsperiods with with interestinterest charged at the rate of charged at the rate of ii per period is per period is

(1 ) 1n

iSR

i

(1 ) 1n

iSR

i

Page 64: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Sinking Fund Sinking Fund

The proprietor of Carson Hardware has decided to set up The proprietor of Carson Hardware has decided to set up a a sinking fundsinking fund for the purpose of for the purpose of purchasing a truckpurchasing a truck in in 2 2 years’ time.years’ time.

It is expected that the truck will cost It is expected that the truck will cost $30,000$30,000.. If the fund earns If the fund earns 10%10% interest per year compounded interest per year compounded

quarterly, quarterly, determine the sizedetermine the size of each (equal) quarterly of each (equal) quarterly installmentinstallment the proprietor should pay into the fund. the proprietor should pay into the fund.

Page 65: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Sinking Fund Sinking Fund

SolutionSolution Here, Here, SS = 30,000 = 30,000, , i i = = rr//mm = 0.1 = 0.1//4 = 0.0254 = 0.025, and , and nn = (2)(4) = 8 = (2)(4) = 8.. Using the Using the sinking fund paymentsinking fund payment formula we find that the formula we find that the

required required size of each quarterly paymentsize of each quarterly payment is given by is given by

or or $3434.02$3434.02..

8

(0.025)(30,000)3434.02

(1 ) 1 (1 0.025) 1n

iSR

i

8

(0.025)(30,000)3434.02

(1 ) 1 (1 0.025) 1n

iSR

i

Page 66: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

4.44.4Arithmetic and Geometric ProgressionsArithmetic and Geometric Progressions

5

5

1,000,000(1 (1.1) )6,105,100

1 1.1S

5

5

1,000,000(1 (1.1) )6,105,100

1 1.1S

20

20[2 2 (20 1)5] 990

2S 20

20[2 2 (20 1)5] 990

2S

Page 67: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Arithmetic ProgressionsArithmetic Progressions

An arithmetic progression is a An arithmetic progression is a sequence of numberssequence of numbers in in which each term after the first is obtained by which each term after the first is obtained by adding a adding a constantconstant dd to the preceding term. to the preceding term.

The The constantconstant dd is called the is called the common differencecommon difference.. An arithmetic progression is An arithmetic progression is completely determinedcompletely determined if the if the

first termfirst term and the and the common differencecommon difference are are knownknown..

Page 68: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

nnth Term of an Arithmetic Progressionth Term of an Arithmetic Progression

The The nnthth term of an arithmetic progression term of an arithmetic progression with with first termfirst term a a and and common differencecommon difference dd is given byis given by

aann = = a a + ( + (nn – 1) – 1)dd

Page 69: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the Find the twelfth termtwelfth term of the of the arithmetic progressionarithmetic progression

2, 7, 12, 17, 22, … 2, 7, 12, 17, 22, …

SolutionSolution The The first termfirst term of the arithmetic progression is of the arithmetic progression is aa11 = = aa = 2 = 2, ,

and the and the common differencecommon difference is is d d = 5= 5.. So, upon setting So, upon setting n n = 12= 12 in the arithmetic progression in the arithmetic progression

formula, we findformula, we findaann == a a + ( + (nn – 1) – 1)d d

aa1212 = 2 + (12 – 1)5 = 2 + (12 – 1)5 = 57= 57

Page 70: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Sum of Terms in an Arithmetic ProgressionSum of Terms in an Arithmetic Progression

The The sumsum of the first of the first nn terms of an terms of an arithmetic arithmetic progressionprogression with with first termfirst term a a and and common common differencedifference dd is given by is given by

[2 ( 1) ]2n

nS a n d [2 ( 1) ]

2n

nS a n d

Page 71: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

ExampleExample

Find the sum of the first Find the sum of the first 2020 terms of the terms of the arithmetic arithmetic progressionprogression

2, 7, 12, 17, 22, … 2, 7, 12, 17, 22, …

SolutionSolution Letting Letting aa = 2 = 2, , dd = 5 = 5, and , and nn = 20 = 20 in the in the sum of termssum of terms

formula, we getformula, we get

20

[2 ( 1) ]220

[2 2 (20 1)5] 9902

n

nS a n d

S

20

[2 ( 1) ]220

[2 2 (20 1)5] 9902

n

nS a n d

S

Page 72: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Company Sales Company Sales

Madison Electric Company had Madison Electric Company had salessales of of $200,000$200,000 in its in its first yearfirst year of operation. of operation.

If the If the sales increasedsales increased by by $30,000$30,000 per year thereafter, find per year thereafter, find Madison’s sales in the Madison’s sales in the fifth yearfifth year and its and its total salestotal sales over the over the first first 55 years of operation. years of operation.

SolutionSolution Madison’s yearly sales follow an Madison’s yearly sales follow an arithmetic progressionarithmetic progression, ,

with with aa = 200,000 = 200,000 and and dd = 30,000 = 30,000.. The sales in the The sales in the fifth yearfifth year are found by using the are found by using the nnthth term term

formulaformula with with nn = 5 = 5. . Thus,Thus,

or or $320,000$320,000..

aann == a a + ( + (nn – 1) – 1)d d

aa55 = 200,000 + (5 – 1)30,000 = 200,000 + (5 – 1)30,000

= 320,000= 320,000

Page 73: 4 Compound Interest Compound Interest Annuities Annuities Amortization and Sinking Funds Amortization and Sinking Funds Arithmetic and Geometric Progressions

Applied Example:Applied Example: Company Sales Company Sales

Madison Electric Company had Madison Electric Company had salessales of of $200,000$200,000 in its in its first yearfirst year of operation. of operation.

If the If the sales increasedsales increased by by $30,000$30,000 per year thereafter, find per year thereafter, find Madison’s sales in the Madison’s sales in the fifth yearfifth year and its and its total salestotal sales over the over the first first 55 years of operation. years of operation.

SolutionSolution Madison’s Madison’s total salestotal sales over the first over the first 5 5 years of operation are years of operation are

found by using the found by using the sum of termssum of terms formula. formula. Thus,Thus,

or or $1,300,000$1,300,000..

[2 ( 1) ]25

[2(200,000) (5 1)30,000]21,300,000

n

nS a n d

[2 ( 1) ]25

[2(200,000) (5 1)30,000]21,300,000

n

nS a n d

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Geometric ProgressionsGeometric Progressions

A A geometric progressiongeometric progression is a is a sequence of numberssequence of numbers in in which each term after the first is obtained by which each term after the first is obtained by multiplying the preceding term by a the preceding term by a constantconstant..

The The constantconstant is called the is called the common ratiocommon ratio.. A geometric progression is A geometric progression is completely determinedcompletely determined if the if the

first termfirst term aa and the and the common ratiocommon ratio rr are given. are given.

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nnth Term of a Geometric Progressionth Term of a Geometric Progression

The The nnthth termterm of a of a geometric progressiongeometric progression with with first termfirst term aa and and common ratiocommon ratio r r is given byis given by

1nna ar 1nna ar

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ExampleExample

Find the Find the eighth termeighth term of a of a geometric progressiongeometric progression whose whose first five terms are first five terms are 162162, , 5454, , 1818, , 66, and , and 22..

SolutionSolution The The common ratiocommon ratio is found by taking the ratio of any term is found by taking the ratio of any term

other than the first to the preceding term.other than the first to the preceding term. Taking the ratio of the Taking the ratio of the fourth termfourth term to the to the third termthird term, for , for

example, gives example, gives

Using the Using the nnthth term formulaterm formula to find the to find the eighth termeighth term gives gives

6 1

18 3r

6 1

18 3r

8 11

8

1 2162

3 27so, n

na ar a

8 11

8

1 2162

3 27so, n

na ar a

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Sum of Terms in a Geometric ProgressionSum of Terms in a Geometric Progression

The The sumsum of the first of the first n n termsterms of a of a geometric geometric progressionprogression with with first termfirst term aa and and common common ratioratio r r is given byis given by

(1 )if 1

1if 1

n

n

a rr

S rna r

(1 )if 1

1if 1

n

n

a rr

S rna r

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ExampleExample

Find the Find the sumsum of the of the first six termsfirst six terms of the of the geometric geometric progression progression 33, , 66, , 1212, , 2424, … , …

SolutionSolution Using the Using the sum of termssum of terms formula formula

withwith aa = 3 = 3, , rr = 6/3 = 2 = 6/3 = 2, and , and nn = 6 = 6, gives, gives

(1 )if 1

1if 1

n

n

a rr

S rna r

(1 )if 1

1if 1

n

n

a rr

S rna r

6

6

3(1 2 )189

1 2S

6

6

3(1 2 )189

1 2S

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Applied Example:Applied Example: Company Sales Company Sales

Michaelson Land Development Company had Michaelson Land Development Company had salessales of of $1$1 million in its million in its first yearfirst year of operation. of operation.

If If sales increasedsales increased by by 10%10% per year thereafter, find per year thereafter, find Michaelson’s Michaelson’s salessales in the in the fifth yearfifth year and its and its total salestotal sales over theover the first first 55 years years of operation. of operation.

SolutionSolution TheThe sales sales follow a follow a geometric progressiongeometric progression, with , with first termfirst term

aa = 1,000,000 = 1,000,000 and and common ratiocommon ratio rr = 1.1 = 1.1.. The The salessales in the in the fifth yearfifth year are thus are thus

or or $1, 464,100$1, 464,100..

45 1,000,000(1.1) 1,464,100a 45 1,000,000(1.1) 1,464,100a

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Applied Example:Applied Example: Company Sales Company Sales

Michaelson Land Development Company had Michaelson Land Development Company had salessales of of $1$1 million in its million in its first yearfirst year of operation. of operation.

If If sales increasedsales increased by by 10%10% per year thereafter, find per year thereafter, find Michaelson’s Michaelson’s salessales in the in the fifth yearfifth year and its and its total salestotal sales over theover the first first 55 years years of operation. of operation.

SolutionSolution TheThe sales sales follow a follow a geometric progressiongeometric progression, with , with first termfirst term

aa = 100,000,000 = 100,000,000 and and common ratiocommon ratio rr = 1.1 = 1.1.. The The total salestotal sales over the over the first first 55 years years of operation are of operation are

or or $6,105,100$6,105,100..

5

5

1,000,000 1 (1.1)6,105,100

1 1.1S

5

5

1,000,000 1 (1.1)6,105,100

1 1.1S

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End of End of Chapter Chapter