THE SUM OF A GEOMETRIC SERIES - TSFX - The School · PDF file · 2012-09-06THE SUM...

© The School For Excellence 2011 Trial Exam Preparation Lectures – 2 Unit Maths – Book 3 Page 21

THE SUM OF A GEOMETRIC SERIES

1 2 3 1...n n nS T T T T T

2 2 1... n na ar ar ar ar _____ (1)

Multiplying line (1) by r gives …

nrS 2 2 1... n n nar ar ar ar ar _____ (2)

(2) – (1) nn nrS S ar a

Making nS the subject:

( 1) ( 1)nnS r a r

( 1)1

n

n

a rS

r

The above form of nS is most commonly used where r > 1. Or, multiplying top and bottom

by –1 gives (1 )1

n

n

a rS

r

which is most commonly used where r < 1.

Note: The above guide does not need to be adhered to. Either formulae is suitable in each case, as they are equivalent.

( 1) , 11

n

n

a rS r

r

(1 ) , 11

n

n

a rS r

r

© The School For Excellence 2011 Trial Exam Preparation Lectures – 2 Unit Maths – Book 3 2

QUESTION 16

Find the sum of the geometric series 1 1 1log log log3 9 81 to 8 terms.

Solution

31loga

2

1

1log log391 log3log3

r

=

and 8n

( 1)1

n

n

a rS

r


ARITHMETIC AND GEOMETRIC MEANS If n number of arithmetic means (let’s call them

21, , ... , nm m m ) are inserted between two

existing numbers x and y , the resulting sequence of 21, , , ... , ,nx m m m y is an A.P.,

containing 2n terms. Similarly, if n number of geometric means are inserted between x and y , a G.P. is

formed, again with 2n terms.

QUESTION 17 Insert 5 arithmetic means between 15 and -21. Solution

The resulting A.P. has 7 terms 5 2 , where 1 15T and 7 21T .

i.e. 15a _____ (1) and 6 21a d _____ (2)

(Note that 7 7 1 6T a d a d .)

Substitute (1) into (2):

15 6 21d

6 36d

6d

1 15T

2 315 6 9, 9 6 3T T and so on.

Hence the five arithmetic means are 9,3, 3, 9, 15 .


QUESTION 18

Insert 4 geometric means between 26

and 34 .

Solution

The resulting G.P. has 6 terms 4 2 , where 26

1 T and 6 4 3T .

i.e. 6

2a (1)

and 5 4 3ar (2)

(Note that 6 1 56T ar ar ).


LIMITING SUM

Any geometric series with a common ratio r such that –1 < r < 1 i.e. 1r , has a limiting

sum. This means that there is some finite sum (called S ) that cannot be exceeded, even if

the series is added forever.

Proof:

r

raS

n

n

1)1(lim

But if 1 1r then as n , 0nr

Hence r

a

r

aS

11

)01(

Alternative Proof:

Let 2 3 ...S a ar ar ar

i.e. 2 ...S a r a ar ar

Note that although this series has one less term than the previous line, one less than infinity is still infinity! i.e. S =

r

aS

1

only where –1 < r < 1


QUESTION 19 A young tree of height 120 cm grows 30 cm in its first year after planting, 20 cm in its second year, and two thirds of its previous year’s growth in each subsequent year. Find the maximum height of the tree. Solution

The limiting height is the infinite sum of 1120 30 20 13 ...3

(cm).

Note however that the 120 does not belong to the remaining pattern, which is a geometric

series with 30a and 23

r .

Limiting height = 1120 30 20 13 ...3

= 120 S

= 30120 21

3

= 120 90 = 210cm

QUESTION 20

Write 0.015

as an infinite series of fractions, and hence express 0.015

in rational form.

(Note: “rational form” is p

q where p and q are integers with no common factor other than 1.)

Solution


FINANCIAL APPLICATIONS

COMPOUND INTEREST Imagine that $300 was to be increased by 5 % in interest each year for 7 years. In all but the first year, we would gain interest not just on our original investment of $300, but also on any interest earned in the meantime. This is called “compound interest” (as opposed to simple interest where the interest is calculated only on the original investment, for any year). The most convenient way of increasing an amount by 5 % is to multiply by 1.05. Why? Because …

100 5 % 105% 1.05

We need to do this each year for 7 years. i.e. $300 x 1.05 x 1.05 x … x 1.05 7 years = $300 x 1.057 This can be expressed in the general case by: Where nA = amount of final balance

P = principal (original) investment r = interest rate as a percent n = number of calculations

If the interest was calculated every 6 months (half yearly) instead, then r becomes 5% 2 2.5% and the number of calculations is 7 2 14n , i.e. $300 x 1.02514. The two answers are then $422.13 and $423.89 respectively (i.e. for yearly compounding and then for six monthly compounding). Note the difference.

WATCH OUT!

Make sure that the interest rate you use is consistent with the frequency of calculations e.g 12% p.a. should appear as 12% if compounded yearly,

but as 6% if compounded half yearly, 3% if compounded quarterly and so on.

(1 )100

nn

rA P


SUPERANNUATION “Superannuation” questions, in general, are concerned with a situation where a person invests a set amount each time period (normally deposited at the beginning of each year), with interest being compounded, and the accumulated total then being withdrawn at the end of the required length of time. We can track the progress of each individual deposit, as it accumulates interest, independent of the other deposits, almost as if each deposit was assigned its own account.

QUESTION 21 $1000 was to be deposited at the beginning of each year, for 10 years, earning interest at the rate of 7 % per annum. Find the total value of these deposits at the end of the 10 years. Solution

100 7 % 107 % 1.07

Adding these individual amounts right to left (for convenience): Let nA be the value of the nth investment.

1A = 1000 (1.07)10

2A =

. . . 10A =

2 10(1000 1.07) (1000 1.07 ) (1000 1.07 )nA

2 3 101000[1.07 1.07 1.07 1.07 ]

This is a G.P. with 1.07, 1.07, 10a r n


101.07 1.07 11000

1.07 1nA

1000 14.783

$14,783.60 Note also that the interest earned is:

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

QUESTION 22 Consider a similar situation to the above, except after 3 years the interest rate is increased from 7 % p.a. to 8 % p.a. Find now the total value of the investments. Solution


TIME PAYMENTS There are many variations possible in time payments – not just in the numbers of course, but in the frequency of payments, frequency of interest calculations and so on. Consider these examples.

QUESTION 23 A loan for $30,000 is to be paid off with equal instalments of $400, paid at the end of each month. Interest is calculated at a rate of 12 % p.a. (i.e. 1 % per month), based on the reducing balance, and debited at the beginning of each month. Find the balance owing after 1, 2, 3 months. Solution


QUESTION 24 Now imagine similar circumstances to the above, except where the monthly repayments are $R , but where we wish to fully repay the loan in 10 years. i.e. 120 months. How do we find R ? Solution

100 1 % 101% 1.01

0A = 30000

1A = 30000 1.01 R

2A = 30000 1.01 1.01R R

=

=

(Note the reversal in order of the last two terms – this gives an increasing series which I think looks more manageable).

3A = 230000 1.01 1.01 1.01R R R

(Note here the use of the second last line of A2, not the last line, for the balance carried forward to the start of the third month).

=

=

(Again, I reversed the order.) It is safe to predict that 120A will be:

120 2 119120 30000 1.01 1 1.01 1.01 ... 1.01A R

Note that although the last term’s power is 119, there are still _____ terms in the bracket,

since 01 1.01 and counting from 0 119 means 120n . Furthermore, each term in the

bracket corresponds to a ___________________, and 120 months means 120 payments.

After 120 months, the balance is to be 0. That is, the loan is to be fully repaid.

120 2 1190 30000 1.01 1 1.01 1.01 ... 1.01R

2 119 1201 1.01 1.01 ... 1.01 30000 1.01R

This is a geometric series with 1, 1.01, 120a r n


1201201(1.01 1) 30000 1.01

1.01 1R

120120(1.01 1) 30000 1.01

0.01R

120

120

0.01 30000 1.011.01 1

R

$430.41 ( )nearest cent Note then that the total interest charged on this loan would be:

$430.41 120 $30000I $21649.20 Because we are paying the same instalment each month, yet the interest each month is calculated on the reducing balance, less of the instalment for successive months is used to offset the interest, and more therefore is used to reduce the principal. Hence, if we were to graph the balance nA against the number of months n , the graph would look like:

Now, imagine a similar situation to the previous example, but where the first six months are interest free. Then:

0A = $30,000

1A =

2A =

=

6A =

nA

n


7 30000 6 1.01A R R

30000 1.01 6 1.01R R

30000 1.01 6 1.01 1R

8 30000 1.01 6 1.01 1.01A R R R

2 230000 1.01 6 1.01 1.01R R R

230000 1.01 ______________________R

2 29 30000 1.01 6 1.01 1.01 1.01A R R R R

3 3 230000 1.01 6 1.01 1.01 1.01R R R R

330000 1.01 _____________________________________R

It is safe to predict that 120A will be:

114 114 2 113120 30000 1.01 6 1.01 1 1.01 1.01 ... 1.01A R

Again, let 120 0A :

114 114 1130 30000 1.01 6 1.01 1 1.01 ... 1.01R

114 113 1146 1.01 1 1.01 ... 1.01 30000 1.01R

This is a G.P. with 1, 1.01, 114a r n

114114 1141(1.01 1)6 1.01 30000 1.01

1.01 1R

114

114 114

30000 1.016 1.01 100 1.01 1

R

Note that dividing by 0.01 is equivalent to multiplying by 100.

R $406.30 nearest cent


FURTHER QUESTIONS QUESTION 25 (HSC 1992)

(i) For what value of r does the geometric series 2 ...a ar ar have a limiting sum? For these values of r write down the limiting sum.

(ii) Find a geometric series with common ratio 1w

that has a limiting sum 1

1 w.


QUESTION 26 (HSC 1992) A timber worker is stacking logs. The logs are stacked in layers, where each layer contains one log less than the layer below. There are five logs in the top layer, six logs in the next layer, and so on. There are n layers altogether. (i) Write down the number of logs in the bottom layer.

(ii) Show that there are 1 ( 9)2n n logs in the stack.

Solution


QUESTION 27 (HSC 2004)

Consider the geometric series 2 41 ....tan tan (i) When the limiting sum exists, find its value in simplest form.

(ii) For what value of in the interval 2 2 does the limiting sum of the series

exist? Solution


QUESTION 28 (HSC 1993) A tap and n water troughs are in a straight line. The tap is first in line, 2 metres from the first trough, and there is 3 metres between consecutive troughs. A stable hand fills the troughs by carrying a bucket of water from the tap to each trough and then returning to the tap. Thus she walks 2 + 2 = 4 metres to fill the first trough, 10 metres to fill the second trough, and so on. (i) How far does the stable hand walk to fill the k th trough? (ii) How far does the stable hand walk to fill all n troughs? (iii) The stable hand walks 1220 metres to fill all the troughs. How many water troughs are there?

First Trough

Second Trough

2m

Third Trough

3m 3m

Tap


QUESTION 29 (HSC 1998) A fish farmer began business on 1 January 1998 with a stock of 100 000 fish. He had a contract to supply 15 400 fish at a price of $10 per fish to a retailer in December each year. In the period between January and the harvest in December each year, the number of fish increases by 10 %. (i) Find the number of fish just after the second harvest in December 1999. (ii) Show that nF , the number of fish just after the n th harvest, is given by

154000 54000(1.1)nnF .


(iii) When will the farmer have sold all his fish, and what will his total income be?


(iv) Each December the retailer offers to buy the farmer's business by paying $15 per fish for his entire stock. When should the farmer sell to maximise his total income?

THE SUM OF A GEOMETRIC SERIES - TSFX - The School · PDF file · 2012-09-06THE SUM...

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