Jacques I. Mathematics for Economics and Business

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All of the problems considered in Section 3.2 involved a single lump-sum payment into an

investment account. The task was simply to determine its future value after a period of time

when it is subject to a certain type of compounding. In this section, we extend this to include

multiple payments. This situation occurs whenever individuals save regularly or when busi-

nesses take out a loan that is paid back using ﬁxed monthly or annual instalments. To tackle

these problems we need to be able to sum (that is, to add together) consecutive terms of a geo-

metric progression. Such an expression is called a geometric series. Suppose that we want to

sum the ﬁrst six terms of the geometric progression given by the sequence

2, 6, 18, 54,... (*)

The easiest way of doing this is to write down these six numbers and add them together to get

2 + 6 + 18 + 54 + 162 + 486 = 728

There is, however, a special formula to sum a geometric series which is particularly useful

when there are lots of terms or when the individual terms are more complicated to evaluate. It

can be shown that the sum of the ﬁrst n terms of a geometric progression in which the ﬁrst term

is a, and the geometric ratio is r, is equal to

a (r ≠ 1)

Use of the symbol r to denote both the interest rate and the geometric ratio is unfortunate but

fairly standard. In practice, it is usually clear from the context what this symbol represents, so

no confusion should arise.

− 1

r − 1

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210

Example

Which of the following sequences are geometric progressions? For those sequences that are of this type,

write down their geometric ratios.

(a) 1000, −100, 10, −1,... (b) 2, 4, 6, 8,... (c) a, ar, ar

, ar

,...

Solution

(a) 1000, −100, 10, −1,...is a geometric progression with geometric ratio, − .

(b) 2, 4, 6, 8,...is not a geometric progression because to go from one term to the next you add 2. Such a

sequence is called an arithmetic progression and is of little interest in business and economics.

, ar

,...is a geometric progression with geometric ratio, r.

Practice Problem

1 Decide which of the following sequences are geometric progressions. For those sequences that are of

this type, write down their geometric ratios.

(a)

3, 6, 12, 24,... (b) 5, 10, 15, 20,... (c) 1, −3, 9, −27,...

(d) 8, 4, 2, 1,

/2,... (e) 500, 500(1.07), 500(1.07)

,...

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A proof of this formula is given in Practice Problem 11 at the end of this section. As a check,

let us use it to determine the sum of the ﬁrst six terms of sequence (*) above. In this case the

ﬁrst term a = 2, the geometric ratio r = 3 and the number of terms n = 6, so the geometric series

is equal to

2 = 3

− 1 = 728

which agrees with the previous value found by summing the terms longhand. In this case there

is no real beneﬁt in using the formula. However, in the following example, its use is almost

essential. You might like to convince yourself of the utility of the formula by evaluating this

series longhand and comparing the computational effort involved!

− 1

3 − 1

3.3 • Geometric series

211

Example

Evaluate the geometric series

500(1.1) + 500(1.1)

+ 500(1.1)

+ ...+ 500(1.1)

Solution

We have a = 500(1.1), r = 1.1, n = 25, so the geometric series is equal to

500(1.1) = 54 090.88

(1.1)

− 1

1.1 − 1

Practice Problem

2 (a) Write down the next term in the sequence

1, 2, 4, 8,...

and hence find the sum of the first five terms. Check that this agrees with the value obtained using

(b) Evaluate the geometric series

100(1.07) + 100(1.07)

+ ...+ 100(1.07)

− 1

r − 1

There are two particular applications of geometric series that we now consider, involving

savings and loans. We begin by analysing savings plans. In the simplest case, an individual

decides to invest a regular sum of money into a bank account. This is sometimes referred to

as a sinking fund and is used to meet some future ﬁnancial commitment. It is assumed that

he or she saves an equal amount and that the money is put into the account at the same time

each year (or month). We further assume that the interest rate does not change. The latter

may not be an entirely realistic assumption, since it can ﬂuctuate wildly in volatile market

conditions. Indeed, banks offer a variety of rates of interest depending on the notice required

for withdrawal and on the actual amount of money saved. Practice Problem 9 at the end of

this section considers what happens when the interest rate rises as the investment goes above

certain threshold levels.

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212

Example

A person saves $100 in a bank account at the beginning of each month. The bank offers a return of 12%

compounded monthly.

(a) Determine the total amount saved after 12 months.

(b) After how many months does the amount saved ﬁrst exceed $2000?

Solution

(a) During the year a total of 12 regular savings of $100 are made. Each $100 is put into an account that

gives a return of 12% compounded monthly, or equivalently, a return of 1% each month. However,

each payment is invested for a different period of time. For example, the ﬁrst payment is invested for

the full 12 months, whereas the ﬁnal payment is invested for 1 month only. We need to work out the

future value of each payment separately and add them together.

The ﬁrst payment is invested for 12 months, gaining a monthly interest of 1%, so its future value is

100(1.01)

The second payment is invested for 11 months, so its future value is

100(1.01)

Likewise, the third payment yields

100(1.01)

and so on. The last payment is invested for 1 month, so its future value is

100(1.01)

The total value of the savings at the end of 12 months is then

100(1.01)

+ 100(1.01)

+ ...+ 100(1.01)

If we rewrite this series in the order of ascending powers, we then have the more familiar form

100(1.01)

+ ...+ 100(1.01)

+ 100(1.01)

This is equal to the sum of the ﬁrst 12 terms of a geometric progression in which the ﬁrst term is

100(1.01) and the geometric ratio is 1.01. Its value can therefore be found by using

with a = 100(1.01), r = 1.01 and n = 12, which gives

$100(1.01) = $1280.93

(b) In part (a) we showed that after 12 months the total amount saved is

100(1.01) + 100(1.01)

+ ...+ 100(1.01)

Using exactly the same argument, it is easy to see that after n months the account contains

100(1.01) + 100(1.01)

+ ...+ 100(1.01)

(1.01)

− 1

1.01 − 1

− 1

r − 1

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The formula for the sum of the ﬁrst n terms of a geometric progression shows that this is the same as

100(1.01) = 10 100(1.01

− 1)

The problem here is to ﬁnd the number of months needed for total savings to rise to $2000.

Mathematically, this is equivalent to solving the equation

10 100(1.01

− 1) = 2000

for n. Following the strategy described in Section 2.3 gives

1.01

− 1 = 0.198

1.01

= 1.198

log(1.01)

= log(1.198)

n log(1.01) = log(1.198)

n =

= 18.2

It follows that after 18 months savings are less than $2000, whereas after 19 months savings exceed this

amount. The target ﬁgure of $2000 is therefore reached at the end of the 19th month.

log(1.198)

log(1.01)

1.01

− 1

1.01 − 1

3.3 • Geometric series

213

Practice Problem

3 An individual saves $1000 in a bank account at the beginning of each year. The bank offers a return

of 8% compounded annually.

(a) Determine the amount saved after 10 years.

(b) After how many years does the amount saved first exceed $20 000?

We now turn our attention to loans. Many businesses ﬁnance their expansion by obtaining

loans from a bank or other ﬁnancial institution. Banks are keen to do this provided that they

receive interest as a reward for lending money. Businesses pay back loans by monthly or annual

repayments. The way in which this repayment is calculated is as follows. Let us suppose that

interest is calculated on a monthly basis and that the ﬁrm repays the debt by ﬁxed monthly

instalments at the end of each month. The bank calculates the interest charged during the ﬁrst

month based on the original loan. At the end of the month, this interest is added on to the ori-

ginal loan and the repayment is simultaneously deducted to determine the amount owed. The

bank then charges interest in the second month based on this new amount and the process is

repeated. Provided that the monthly repayment is greater than the interest charged each

month, the amount owed decreases and eventually the debt is cleared. In practice, the period

during which the loan is repaid is ﬁxed in advance and the monthly repayments are calculated

to achieve this end.

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214

Example

Determine the monthly repayments needed to repay a $100 000 loan which is paid back over 25 years when

the interest rate is 8% compounded annually.

Solution

In this example the time interval between consecutive repayments is 1 month, whereas the period during

which interest is charged is 1 year. This type of ﬁnancial calculation typiﬁes the way in which certain types

of housing loan are worked out. The interest is compounded annually at 8%, so the amount of interest

charged during the ﬁrst year is 8% of the original loan: that is,

× 100 000 = 8000

This amount is added on to the outstanding debt at the end of the ﬁrst year. During this time, 12 monthly

repayments are made, so if each instalment is $x, the outstanding debt must decrease by 12x. Hence, at the

end of the ﬁrst year, the amount owed is

100 000 + 8000 − 12x = 108 000 − 12x

In order to be able to spot a pattern in the annual debt, let us write this as

100 000(1.08) − 12x

where the ﬁrst part simply reﬂects the fact that 8% interest is added on to the original sum of $100 000. At

the end of the second year, a similar calculation is performed. The amount owed rises by 8% to become

[100 000(1.08) − 12x](1.08) = 100 000(1.08)

− 12x(1.08)

and we deduct 12x for the repayments to get

100 000(1.08)

− 12x(1.08) − 12x

This is the amount owed at the end of the second year. Each year we multiply by 1.08 and subtract 12x, so

at the end of the third year we owe

[100 000(1.08)

− 12x(1.08) − 12x](1.08) − 12x = 100 000(1.08)

− 12x(1.08)

− 12x(1.08) − 12x

and so on. These results are summarized in Table 3.17. If we continue the pattern, we see that after 25 years

the amount owed is

100 000(1.08)

− 12x(1.08)

− ...− 12x

= 100 000(1.08)

− 12x[1 + 1.08 + (1.08)

+ ...+ (1.08)

]

where we have taken out a common factor of 12x and rewritten the powers of 1.08 in ascending order. The

ﬁrst term is easily evaluated using a calculator to get

100 000(1.08)

= 684 847.520

100

Table 3.17

End of year Outstanding debt

1 100 000(1.08)

− 12x

2 100 000(1.08)

− 12x(1.08)

− 12x

3 100 000(1.08)

− 12x(1.08)

− 12x

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The geometric series inside the square brackets can be worked out from the formula

The ﬁrst term a = 1, the geometric ratio r = 1.08, and we are summing the ﬁrst 25 terms, so n = 25. (Can

you see why there are actually 25 terms in this series rather than 24?) Hence

[1 + 1.08 + (1.08)

+ ...+ (1.08)

] ==73.106

The amount owed at the end of 25 years is therefore

684 847.520 − 12x(73.106) = 684 847.520 − 877.272x

In this expression, x denotes the monthly repayment, which is chosen so that the debt is completely cleared

after 25 years. This will be so if x is the solution of

684 847.520 − 877.272x = 0

Hence

x ==$780.66

The monthly repayment on a 25-year loan of $100 000 is $780.66, assuming that the interest rate remains

ﬁxed at 8% throughout this period.

It is interesting to substitute this value of x into the expressions for the outstanding debt given in Table

3.17. The results are listed in Table 3.18. What is so depressing about these ﬁgures is that the debt only falls

by about $1500 to begin with, in spite of the fact that over $9000 is being repaid each year!

684 847.520

877.272

1.08

− 1

1.08 − 1

− 1

r − 1

3.3 • Geometric series

215

Table 3.18

End of year Outstanding debt

1 $98 632.08

2 $97 154.73

3 $95 559.18

Practice Problem

4 A person requests an immediate bank overdraft of $2000. The bank generously agrees to this, but

insists that it should be repaid by 12 monthly instalments and charges 1% interest every month on the

outstanding debt. Determine the monthly repayment.

The mathematics used in this section for problems on savings and loans can be used for

other time series. Reserves of non-renewable commodities such as minerals, oil and gas con-

tinue to decline, and geometric series can be used to estimate the year in which these stocks are

likely to run out.

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Practice Problem

5 It is estimated that world reserves of oil currently stand at 2625 billion units. Oil is currently extracted

at an annual rate of 45.5 billion units and this is set to increase by 2.6% a year. After how many years

will oil reserves run out?

Example

Total reserves of a non-renewable resource are 250 million tonnes. Annual consumption, currently at

20 million tonnes per year, is expected to rise by 2% a year. After how many years will stocks be exhausted?

Solution

In the ﬁrst year, consumption will be 20 million tonnes. In the second year, this will rise by 2%, so con-

sumption will be 20(1.02) million tonnes. In the third year, this will again rise by 2% to become 20(1.02)

million tonnes. The total consumption (in millions of tonnes) during the next n years will be

20 + 20(1.02) + 20(1.02)

+ ...+ 20(1.02)

n−1

This represents the sum of n terms of a geometric series with ﬁrst term a = 20 and geometric ratio r = 1.02,

so is equal to

20 = 1000(1.02

− 1)

Reserves will run out when this exceeds 250 million, so we need to solve the equation

1000(1.02

− 1) = 250

for n. This is easily solved using logarithms:

1.02

− 1 = 0.25 (divide both sides by 1000)

1.02

= 1.25 (add 1 to both sides)

log(1.02

) = log(1.25) (take logs of both sides)

n log(1.02) = log(1.25) (rule 3 of logs)

n = (divide both sides by log(1.02))

= 11.27

so the reserves will be completely exhausted after 12 years.

log(1.25)

log(1.02)

1.02

− 1

1.02 − 1

Example

Current annual extraction of a non-renewable resource is 40 billion units and this is expected to fall at a rate

of 5% each year. Estimate the current minimum level of reserves if this resource is to last in perpetuity

(that is, for ever).

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Solution

In the ﬁrst year 40 billion units are extracted. In the second year this falls by 5% to 40(0.95) billion units. In

the third year this goes down by a further 5% to 40(0.95)

. After n years the total amount extracted will be

40 + 40(0.95) + 40(0.95)

+ ...+ 40(0.95)

n−1

Using the formula for the sum of a geometric progression gives

= 40

= 800(1 − 0.95

)

To see what happens in perpetuity we need to investigate the behaviour 0.95

as n tends to inﬁnity. Now

since the magnitude of 0.95 is less than unity, it is easy to see that 0.95

converges to zero and so the total

amount will be 800 billion units.

0.95

− 1

−0.05

0.95

− 1

0.95 − 1

3.3 • Geometric series

217

Arithmetic progression A sequence of numbers with a constant difference between con-

secutive terms; the nth term takes the form, a + bn.

Geometric progression A sequence of numbers with a constant ratio between consecutive

terms; the nth term takes the form, ar

n−1

Geometric ratio The constant multiplier in a geometric series.

Geometric series A sum of the consecutive terms of a geometric progression.

Sinking fund A ﬁxed sum of money saved at regular intervals which is used to fund some

future ﬁnancial commitment.

Key Terms

Practice Problems

6 Find the value of the geometric series

1000 + 1000(1.03) + 1000(1.03)

+ ...+ 1000(1.03)

7 A regular saving of $500 is made into a sinking fund at the start of each year for 10 years. Determine

the value of the fund at the end of the tenth year on the assumption that the rate of interest is

(a) 11% compounded annually

(b) 10% compounded continuously

8 Monthly sales figures for January are 5600. This is expected to fall for the following 9 months at a rate

of 2% each month. Thereafter sales are predicted to rise at a constant rate of 4% each month. Estimate

total sales for the next 2 years (including the first January).

9 A bank has three different types of account in which the interest rate depends on the amount invested.

The ‘ordinary’ account offers a return of 6% and is available to every customer. The ‘extra’ account

offers 7% and is available only to customers with $5000 or more to invest. The ‘superextra’ account



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offers 8% and is available only to customers with $20 000 or more to invest. In each case, interest is

compounded annually and is added to the investment at the end of the year.

A person saves $4000 at the beginning of each year for 25 years. Calculate the total amount saved

on the assumption that the money is transferred to a higher-interest account at the earliest opportunity.

10 Determine the monthly repayments needed to repay a $50 000 loan that is paid back over 25 years

when the interest rate is 9% compounded annually. Calculate the increased monthly repayments

needed in the case when

(a) the interest rate rises to 10%

(b) the period of repayment is reduced to 20 years

11 If

= a + ar + ar

+ ...+ ar

n−1

write down an expression for rS

and deduce that

− S

= ar

− a

Hence show that the sum of the first n terms of a geometric progression with first term a and geo-

metric ratio r is given by

provided that r ≠ 1.

12 A prize fund is set up with a single investment of $5000 to provide an annual prize of $500. The fund

is invested to earn interest at a rate of 7% compounded annually. If the first prize is awarded 1 year

after the initial investment, find the number of years for which the prize can be awarded before the

fund falls below $500.

13 The current extraction of a certain mineral is 12 million tonnes a year and this is expected to fall at a

constant rate of 6% each year. Estimate the current minimum level of world reserves if the extraction

is to last in perpetuity.

14 At the beginning of a month, a customer owes a credit card company $8480. In the middle of the

month, the customer repays $A, where A < $8480, and at the end of the month the company adds

interest at a rate of 6% of the outstanding debt. This process is repeated with the customer continu-

ing to pay off the same amount, $A, each month.

(a) Find the value of A for which the customer still owes $8480 at the start of each month.

(b) If A = 1000, calculate the amount owing at the end of the eighth month.

ment is given by

A = where R = 1.06

(d) Find the value of A if the debt is to be paid off exactly after 2 years.

(Excel) A bank decides to produce a simple table for its customers, indicating the monthly repayments

of a $5000 loan that is paid back over different periods of time. Produce such a table, with 13 rows

corresponding to monthly interest rates of 0.5%, 0.525%, 0.55%, 0.575%, ..., 0.8%, and

9 columns corresponding to a repayment period of 12, 18, 24, ..., 60 months.

16 (Excel) Determine the monthly repayments needed to repay a $500 000 loan that is paid back over

25 years when the interest rate is 7.25% compounded annually. Use a spreadsheet to tabulate the

outstanding debt at the end of 1, 2, ..., 25 years.

8480R

n−1

(R − 1)

− 1

r − 1

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17 (Excel) Estimates of reserves of an oil field are 60 billion barrels. Current annual production of 4 bil-

lion barrels is set to rise at a constant rate of r% a year. Show that the value of r required to exhaust

this oil over 10 years (including the current year) satisfies the equation

1 +

− 0.15r − 1 = 0

By tabulating values of the left-hand side for r = 8.00, 8.05, 8.10, 8.15, . . . calculate the value of

r correct to 1 decimal place.

100

3.3 • Geometric series

219

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