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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 n.
THE SUM OF A GEOMETRIC SERIES S n  T1  T2  T3  ...  Tn 1  Tn

 



 

 a   ar   ar 2  ...  ar n  2  ar n 1



_____ (1)

Multiplying line (1) by r gives …

rSn 

 ar    ar 2   ...   ar n2    ar n1    ar n 

_____ (2)

(2) – (1) rS n  S n  ar n  a Making S n the subject:

S n (r  1)  a(r n  1) a(r n  1) Sn  r 1 The above form of S n is most commonly used where r > 1. Or, multiplying top and bottom

a(1  r n ) by –1 gives S n  which is most commonly used where r < 1. 1 r Note: The above guide does not need to be adhered to. Either formulae is suitable in each case, as they are equivalent.

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Sn 

a (r n  1) , r 1

r 1

Sn 

a(1  r n ) , 1 r

r 1

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 21

QUESTION 16

1 3

Find the sum of the geometric series log  log

1 1  log  to 8 terms. 9 81

Solution

1 log 1 9 a  log r  1 3 log 3 and n  8



log 32 log 31

=

a(r n  1) Sn  r 1

© The School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 22

ARITHMETIC AND GEOMETRIC MEANS If n number of arithmetic means (let’s call them m1 , m2 , ... , mn ) are inserted between two





existing numbers x and y , the resulting sequence of x, m1 , m2 , ... , mn , y is an A.P., containing n  2 terms. Similarly, if n number of geometric means are inserted between x and y , a G.P. is formed, again with n  2 terms. QUESTION 17 Insert 5 arithmetic means between 15 and -21. Solution The resulting A.P. has 7 terms  5  2  , where T1  15 and T7  21 . i.e. a  15 and a  6d  21

_____ (1) _____ (2)

(Note that T7  a   7  1 d  a  6d .) Substitute (1) into (2):

15   6d  21 6d  36 d  6

T1  15 T2  15  6  9, T3  9  6  3 and so on. Hence the five arithmetic means are 9,3, 3, 9, 15 .

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 23

QUESTION 18 Insert 4 geometric means between

6 and 4 3 . 2

Solution The resulting G.P. has 6 terms  4  2  , where T1  i.e.

a

6 2

and ar 5  4 3

6 and T6  4 3 . 2

(1) (2)

(Note that T6  ar 6 1  ar 5 ).

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 24

LIMITING SUM Any geometric series with a common ratio r such that –1 < r < 1 i.e. r  1 , 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.

S 

a 1 r

only where –1 < r < 1

Proof:

a (1  r n ) n 1 r

S  lim

But if 1  r  1 then as n   , r n  0 Hence S  

a (1  0) a  1 r 1 r

Alternative Proof: Let S  a  ar  ar 2  ar 3  ...



i.e. S  a  r a  ar  ar 2  ...



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

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 25

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

1 3

The limiting height is the infinite sum of 120  30  20  13  ... (cm). Note however that the 120 does not belong to the remaining pattern, which is a geometric series with a  30 and r   Limiting height

2 . 3  

1 3

 

= 120   30  20  13  ... 

= 120  S = 120 

30 2 1 3

= 120  90 = 210 cm QUESTION 20  

 

Write 0.015 as an infinite series of fractions, and hence express 0.015 in rational form. (Note: “rational form” is

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

Solution

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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:

An  P(1 

r n ) 100

Where An = 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 n  7  2  14 , 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.

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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 An be the value of the nth investment.

A1 = 1000 (1.07)10 A2 = . . .

A10 =

A

n

 (1000 1.07)  (1000 1.07 2 )   (1000  1.0710 )

 1000[1.07  1.07 2  1.073    1.0710 ] This is a G.P. with a  1.07,

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r  1.07, n  10

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 28



 

1.07 1.0710  1  An 1000  1.07  1 



 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

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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 A0 = 30000 A1 = 30000 1.01  R A2 =  30000 1.01  R  1.01  R = = (Note the reversal in order of the last two terms – this gives an increasing series which I think looks more manageable).





A3 = 30000 1.012  R 1.01  R 1.01  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 A120 will be:



A120  30000 1.01120  R 1  1.01  1.012  ...  1.01119



Note that although the last term’s power is 119, there are still _____ terms in the bracket, since 1  1.010 and counting from 0  119 means n  120 . 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.



 0  30000  1.01120  R 1  1.01  1.012  ...  1.01119







R 1  1.01  1.012  ...  1.01119  30000 1.01120 This is a geometric series with a  1,

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r  1.01, n  120

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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1(1.01120  1)  120 R    30000 1.01  1.01 1   R(1.01120  1)  30000 1.01120 0.01 R

0.01 30000 1.01120 1.01120  1

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

I  $430.41120  $30000  $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 An against the number of months n , the graph would look like:

An

n

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

A0 = $30,000 A1 = A2 = =

A6 =

© The School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 32

A7   30000  6 R  1.01  R  30000  1.01  6 R  1.01  R

 30000  1.01  R  6  1.01  1

A8  30000 1.01  6 R  1.01  R  1.01  R

 30000 1.012  6 R 1.012  R 1.01  R  30000  1.012  R  ______________________ 

A9  30000  1.012  6 R 1.012  R 1.01  R  1.01  R  30000  1.013  6 R  1.013  R 1.012  R  1.01  R  30000  1.013  R  _____________________________________  It is safe to predict that A120 will be:

A120  30000 1.01114  R  6 1.01114  1  1.01  1.012  ...  1.01113  Again, let A120  0 :

0  30000 1.01114  R 6 1.01114  1  1.01  ...  1.01113  R 6 1.01114  1  1.01  ...  1.01113   30000 1.01114 This is a G.P. with a  1,

r  1.01, n  114

 1(1.01114  1)  114 114 R 6 1.01    30000 1.01 1.01  1  

R 

30000  1.01114 6  1.01114  100 1.01114  1







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

R  $406.30  nearest cent 

© The School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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FURTHER QUESTIONS QUESTION 25 (HSC 1992) (i)

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

(ii)

Find a geometric series with common ratio

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1 1 that has a limiting sum . w 1 w

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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 n(n  9) logs in the stack. 2

Solution

© The School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 35

QUESTION 27 (HSC 2004) Consider the geometric series 1  tan 2  tan 4  .... (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

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 36

QUESTION 28 (HSC 1993) First Trough

Tap

2m

Second Trough

3m

Third Trough

3m

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?

© The School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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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 Fn , the number of fish just after the n th harvest, is given by

Fn  154 000  54 000(1.1) n .

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

Page 38

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

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Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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(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 School For Excellence 2011

Trial Exam Preparation Lectures – 2 Unit Maths – Book 3

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