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Series (7.1 – 7.3) and Series applications (7.5)

 

7.1        Arithmetic series. Formulae for the nth term and sum of n terms. Applications of arithmetic series.

7.2        Geometric series. Formulae for the nth term and sum of n terms. Applications of geometric series: compound interest, simplified hire purchase and repayment problems.

7.3        Geometric series with a ratio between -1 and 1. The limit of xn, as n → ∞, for < 1, and the concept of limiting sum for a geometric series. Applications to recurring decimals.” (syllabus)

7.4  Series applications

 

For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here

 

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SEQUENCES & SERIES

 

Arithmetic Progression (A.P.): A sequence of numbers in which each term after the first, is the sum of the previous term and a common difference, d. Tn = Tn-1 + d

 

The common difference, d, of an arithmetic progression is the difference between any two consecutive terms of the A.P.            d = Tn – Tn-1 = Tn+1 - Tn

 

The nth term of an arithmetic progression where the first term is a and the common difference is d, is given by:

Tn = a + (n – 1)d

 

The sum of n terms of an A.P. is given by:                  

Sn =                          or         Sn = where  is the last term.

 

Geometric Progression (G.P.): A sequence of numbers in which each term after the first, is the product of the previous term and a common ratio, r.

 

The common ratio, r, of a geometric progression consists of any term divided by the one before it.        r =

 

The nth term of a geometric progression where the first term is a, and the common ratio is r, is given by: Tn = arn-1

 

The sum of n terms of a G.P. is given by:

Sn =              if  > 1           or         Sn =              if  < 1

 

Limiting Sum: If  < 1 then as n becomes very large, the sum of a G.P. approaches a particular value.

 

Compound Interest: A =  where A = amount, P = principal, r = percent interest per compounding period, n = number of compounding periods.

 

Sigma Notation: The notation is used to find the sum of an A.P. or G.P. from the ath term to the bth term.    = Sb - Sa-1

 

 

 

Arithmetic Progressions

 

1.         Write down the next three terms of the sequence: 2,5,8, ………

2.         Find the first three terms of the sequence in which Tn = 3n

3.         Determine whether or nor the following numbers form an arithmetic sequence:

            (i) 18, 25, 32,

            (ii) -3, 1, 5,

            (iii) -1, 1, 2,

            (iv) -1, -2, -3,

            (v) -2, 0, 2

4.         Find the 24th member of the sequence: 8, 10, 12, …….

5.         Find the 18th member of the sequence: 24, 21, 18, …….

6.         Is 242 a member of the sequence: 2, 5, 8, ……?          If so, which term is it?

7.         Is 584 a member of the sequence: 2, 9, 16, ……?        If so, which term is it?

8.         Find the sum of the first 10 terms of the arithmetic sequence in which the first term is 8 and the common difference is 3.

9.         Find the sum of the first 16 terms of the sequence: 3, 8, 13, …….

10.       Find

11.       The 12th term of an A.P. is 38 and the 20th term is 62. Find the first term and the common difference.

12.       Find the sum of all the integers from 1 to 100 inclusive.

13.       Amanda was starting a bird sanctuary. She bought 14 swans in January, 18 in February, 22 in March and so on, with the number increasing by 4 swans every month for the year.

            (i) How many swans did she buy in December?

            (ii) How many swans did she buy throughout the year?

14.       Samantha’s parents opened a bank account for her when she was born and deposited $200 in it. They deposited $250 on her first birthday, $300 on her second birthday and so on, with the amounts increasing by $50 each year until her 21st birthday.

            (i) How much is deposited into her account on her 21st birthday?

            (ii) What is the total amount deposited into her account up to and including her 21st birthday?

 

Answers:

1.         11, 14, 17,

2.         3, 6, 9,             Since 9-6 = 6-3 then it is an A.P.

3.         (i) yes   (ii) yes  (iii) no     (iv) yes   (v) yes        

4.         54                    5.         -27                   6.         Yes. 81st term

7. No.  term. n is not an integer 584 not a member of the series.

8. 215                          9. 648              10. 70             

11. a = 5, d = 3            12. 5050          13. (i) 58          (ii) 432

14. (i) $1250    (ii) $15 950

 

 

Geometric Progressions

1.         Write down the next three terms of the sequence: 2,6,18, ………

2.         Find the first three terms of the sequence in which Tn = 3n

3.         Determine whether or nor the following numbers form a geometric sequence:

            (i) 8, 16, 32,

            (ii) -3, 0, 3,

            (iii) -2, 6, -18,

            (iv) 12, 22, 32,

            (v) 53, 54, 55

4.         Find the 7th member of the sequence: 7, 21, 63, …….

5.         Find the 9th member of the sequence: -5, 30, -180, …….

6.         Is 242 a member of the sequence: 2, 8, 32, ……?        If so, which term is it?

7.         Is 584 a member of the sequence: 4, 12, 36, ……?      If so, which term is it?

8.         Find the sum of the first 10 terms of the geometric sequence in which the first term is 8 and the common ratio is 3.

9.         Find the sum of the first 12 terms of the sequence: 3, 12, 48, …….

10.       Find

11.       The 6th term of a G.P. is 972 and the 12th term is 708 588. Find the first term and the common ratio.

12.       Find the sum of 41 + 42 + 43 + …….. + 420. Give your answer in scientific notation to 3 significant figures.

13.       Amanda was starting a bird sanctuary. She bought 14 swans in January, 2000. These had increased to 28 in July 2000 and 56 in January 2001. If the swan population continued to increase at the same rate, how many swans would Amanda have in January 2006?

14.       Samantha’s parents opened a bank account for her when she was born and deposited $1 in it. They deposited $2 on her first birthday, $4 on her second birthday and so on, with the deposits doubling each birthday. The parents discontinued making deposits after they made a deposit exceeding a million dollars.

            On what birthday did the deposit exceed a million dollars?

15.       A poker machine was programmed to return 80% to the player i.e. for every $100 inserted into the machine, $80 is returned to the player in prizes. Willie Gobroke put $50 into the poker machine and continued to play, inserting all prize money back into the poker machine until he had run out of money. If he maintained a consistent 80% return and each spin of the poker machine cost $1, how many spins would Willie get until he had run out of money?

16.       Use the limiting sum formula to evaluate the following recurring decimals:

            (i)            (ii)           (iii)           (iv)

 

Answers:        

1.         54, 162, 486,               2.         3, 9, 27           

3.         (i) yes   (ii) no   (iii) yes               (iv) no   (v) yes         

4.         15309                          5.         -8 398 080                   6.         No      

7.         No.                              8.         236 192                       9.         16 777 215     

10.       508                              11.       a = 4, r = 3                   12.       1.47 x 1012      

13.       114 688                       14.       19th                              15.       250

16.       (i)     (ii)              (iii)             (iv)

 

 

SEQUENCES & SERIES PROBLEMS

 

1.         In a variation of the potato race, tennis balls were placed at 10 metres, 20 metres, 30 metres, 40 metres and 50 metres from the start. Each competitor had to run from the start, collect a tennis ball from 10m and carry it back to the start, then run to collect a ball from 20m and return with it and so on until all 5 balls were back at the start.

            How far did each competitor run?

2.         7p, 8p +1, 9p + 2, are successive terms of an A.P. Determine the value of p and hence the three terms.

3.         The population of a certain town increases by 12% each year. If the town had 60 000 residents on 1st January 2000,

            (i) How many people would the town have on 1st January 2006?

            (ii) In what year would the population first exceed 200 000?

4.         Express the following recurring decimals in the form  where m and n are integers with no common factor.

            (i)               (ii)           (iii)         (iv)

5.         The fifth term of a G.P. is 9 times the third term. The sum of the first 3 terms and the first 5 terms is 1608. Determine the common ratio and the first term.

6.         Is the series log 2, log 4, log 8, log 16, …… an A.P. or a G.P.? Justify your answer and state the common difference or common ratio.

7.         Mugsey bought a car for $8 000. The car depreciated by 15% of its value each year. What is the car worth (to the nearest $1) after 5 years?

8.         Pat and Herb each invest $8 000 for 5 years at 6% compound interest. Pat’s fund compounds interest annually and Herb’s fund compounds monthly. How much more than Pat’s investment is Herb’s investment worth after 5 years?

9.         Olivia paid $1600 into a superannuation fund at the beginning of each year. The fund pays interest at the rate of 8% per annum, compounded annually. What would Olivia’s superannuation be worth (to the nearest $1) immediately after she made her 40th deposit?

10.       Toby and Sarah borrowed $300 000 for 25 years to buy a house. The interest was 8% per annum, compounded monthly. Calculate the monthly repayments.

 

Answers:

1.         300m

2.         p = 3    terms 21, 25, 29

3.         (i) 118 429       (ii) 2010           (10.6 years so approx July/Aug. 2010)

4.         (i)     (ii)              (iii)            (iv)

5.         r = 3, a = 12.

6.         A.P.     log 4 = 2log 2, log 8 = 3log 2, log 16 = 4log 2 d = log 2

7.         $ 3550

8.         $85.00 (Herb $10790.80 – Pat $10705.80)

9.         $414 490

10.       $2315.45

 

 

SEQUENCES & SERIES: MORE PROBLEMS

 

1.         In a variation of the potato race, tennis balls were placed at 5 metres, 10, 15, 20, 25, 30, 35, 40, 45 and 50 metres from the start. Each competitor had to run from the start, collect a tennis ball from 5m and carry it back to the start, then run to collect a ball from 10m and return with it and so on until all 10 balls were back at the start.

            How far did each competitor run?

2.         7p, 8p -1, 9p - 2, are successive terms of an A.P. Determine the value of p and hence the three terms.

3.         The population of a certain town increases by 9% each year. If the town had 40 000 residents on 1st January 2005,

            (i) How many people would the town have on 1st January 2010?

            (ii) In what year would the population first exceed 100 000?

4.         Express the following recurring decimals in the form  where m and n are integers with no common factor.

            (i)               (ii)           (iii)         (iv)

5.         The fifth term of a G.P. is 64 times the second term. The sum of the first 2 terms and the first 5 terms is 3114. Determine the common ratio and the first term.

6.         Show that log 3, log 9, log 81, are three terms of a G.P.  Write down the next term.

7.         Bluey bought a car for $12 000. The car depreciated by 18% of its value each year. What is the car worth (to the nearest $1) after 5 years?

8.         Frank and Elsie each invest $9 000 for 6 years at 8% compound interest. Frank’s fund compounds interest annually and Elsie’s fund compounds monthly. How much more than Frank’s investment is Elsie’s investment worth after 6 years?

9.         Pauline paid $900 into a superannuation fund every 6 months. The fund pays interest at the rate of 8% per annum, compounded monthly. What would Pauline’s superannuation be worth (to the nearest $1) immediately after she made her 40th deposit?

10.       Lawrence and Eva borrowed $250 000 for 20 years to buy a house. The interest was 7% per annum, compounded annually. Calculate the monthly repayments.

 

Answers:

1.         550m

2.         p = 4    terms 28, 31, 34

3.         (i) 61 545         (ii) 2015           (10.63 years so approx. Aug. 2015)

4.         (i)     (ii)   (iii)            (iv)

5.         r = 4, a = 19.

6.         G.P.     log 3, log 9, log 81 = log 3, 3log 3, 9 log 3                    r = 3

7.         $ 4449

8.         $239.65           (Elsie $14521.52 – Frank $14281.87)

9.         $86 892

10.       $1966.52

 

 

INTEREST

 

Simple Interest: I = (Prn)/100 where

I = Simple Interest

P = Principal,

r = rate per interest period (usually a year),

n = number of interest periods.

 

Example: What will be the value of $2000 invested for 5 years at 6% p.a. simple interest?

                        I = (Prn)/100    = 2000x6x5/100 = 600

                        Amount = Principal + Interest = $2000 + $600 = $2600

Exercises 1:

Q.1.     If I invest $600 at a simple interest rate of 7% p.a., how much will I have in the account after 10 years?

Q.2.     Ryan invested $50 in a bank account that paid 4% p.a. simple interest. How much would it be worth after 9 months?

Q.3.     What simple interest rate would be required to double the amount of the investment in 20 years?

 

Compound Interest:      A =          where

A = final amount,

P = Principal,

r = rate per interest period (usually a year),

n = number of interest periods.

Compound interest is where you earn interest on your interest.

 

Example: What will be the value of $2000 invested for 5 years at 6% p.a. compound interest?

                        A =          = 2000(1 + 6/100)5 = 2000(1.06)5 = $2676.45

 

Exercises 2:

Q.1.     If I invest $600 at a compound interest rate of 7% p.a., how much will I have in the account after 10 years?

Q.2.     Josh invested $50 in a bank account that paid 4% p.a. compound interest. How much would it be worth after 5 years?

Q.3.     Chris invested $5000 for 6 years at 6% p.a. compound interest. What was the final value of the account?

Q.4.     Karen invested $5000 for 6 years in an account where the 6% p.a. interest was compounded every 6 months. What was the final value of the account?

 

 

Answers:

Ex.1     Simple Interest:             Q.1. = $1020               Q.2. = $51.50              Q.3. = 5%

Ex. 2.   Compound Interest:      Q.1. = $1180.29          Q.2. = $60.83 

Q.3. = $7092.60          Q.4. = $7128.80

 

 

 

INTEREST 2

Compound Interest:      A =          where

A = final amount,

P = Principal,

r = rate per interest period (usually a year),

n = number of interest periods.

Compound interest is where you earn interest on your interest.

 

Example: What will be the value of $5000 invested for 10 years at 5% p.a. compound interest?

                        A =          = 5000(1 + 5/100)10 = 5000(1.05)10 = $8144.47

Exercises:

Q.1.     If I invest $200 at a compound interest rate of 8% p.a., how much will I have in the account after 10 years?

Q.2.     Josh invested $100 in a bank account that paid 5% p.a. compound interest. How much would it be worth after 4 years?

Q.3.     Chris invested $10 000 for 20 years at 6% p.a. compound interest. What was the final value of the account?

Q.4.     Karen invested $10 000 for 20 years in an account where the 6% p.a. interest was compounded every 6 months. What was the final value of the account?

Q.5.     Michelle invested $10 000 for 20 years in an account where the 6% p.a. interest was compounded every month. What was the final value of the account?

 

The compound interest formula can also be used to determine depreciation. In this case the rate is subtracted.

Depreciation:    A = P(1 - r/100)n         where

A = final value,

P = Initial cost,

r = rate per depreciation period (usually a year),

n = number of depreciation periods.

 

Example: A photocopier costs $4 000 and depreciates at the rate of 20% per year. How much is it worth after 5 years?

            A = P(1 – r/100)n         4000(1 – 20/100)5       = 4000(0.8)5    = $1310.72

Exercises.

Q.1.     Simone bought a walkman for $120. If it depreciates at the rate of 15% per year, how much is it worth after 3 years.

Q.2.     Emma bought a car for $8000. If it depreciates at the rate of 12% per year, how much will it be worth after 10 years.

Q.3.     Lauren bought a sound system for $3 800. If it depreciates at the rate of 1% per month, how much will it be worth after 2 years?

Answers:

Compound Interest:      Q.1. = $431.78            Q.2. = $121.55            Q.3. = $32071.35        Q.4. = $3260.38          Q.5. = $33102.04

Depreciation:    Q.1. = $73.70  Q.2. = $2228.01                      Q.3. = $2985.58

 

 

SUPERANNUATION

 

Example:  Mickey pays $1000 into a superannuation fund at the beginning of each year. If the interest rate is 8% compounded annually, what will Mickey’s superannuation be worth after 20 years.

 

Answer: The first $1000 earns 8% compound interest for 20 years, the second $1000 earns 8% compound interest for 19 years and so on.

            By applying the formula; A =

            The first $1000 is worth  1000(1.08)20

            The second $1000 is worth 1000 (1.08)19 ……………..

            The twentieth $1000 is worth 1000 (1.08)1

            Adding these gives: A = 1000 ( 1.08 + 1.082 + ………..+ 1.0819 + 1.0820)

            The brackets represent a G.P. with a = 1.08 and r = 1.08

            Sn =              A = 1000 []    = 1000[]

            = $49 422.92 (to nearest cent)

 

Exercise:

 

Q.1.     Shafia invested $1000 in a superannuation fund at the beginning of each year for 10 years at an interest rate of 6% per annum, compounded annually. What was her investment worth at the end of 10 years?

Q.2.     Mugsey invested $1000 in a superannuation fund at the beginning of each year for 10 years at an interest rate of 6% per annum, compounded monthly. What was his investment worth at the end of 10 years?

Q.3.     Juan paid $6000 into a superannuation fund every year. Interest was 9% per annum, compounded annually. What was his superannuation worth when he retired after 40 years?

Q.4.     Shakeeba paid $500 into a superannuation fund every month. Interest was 9% per annum, compounded monthly. What was her superannuation worth when she retired after 40 years?

Q.5.     Amanda paid $100 per week into a superannuation fund. The interest rate was 10.4% per annum, compounded weekly. After 10 years Amanda transferred her superannuation to a new fund because she changed jobs. How much superannuation did Amanda transfer?

Q.6.     Wafaa paid $800 per month into a superannuation fund for 15 years. The interest rate was 8% per annum, compounded monthly. After 15 years, Wafaa left her job and didn’t make any more payments to her superannuation, but had to wait a further 20 years before she could collect her superannuation. What was Wafaa’s superannuation worth at the end of the 35 years?

Answers:

 Q.1. $1397.64                                    Q.2. $14 104.51          Q3. $2 209 751.19

Q.4.  $2 358 215.09                Q.5. $91 496.63          Q.6. $1 372 981.94

 

 

TIME PAYMENT

 

One application of geometric progressions is the calculation of time payments.

Example:

Shafia borrowed $250 000 from the bank to buy a house. She agreed to pay back the loan over 25 years at an interest rate of 6% per annum, reducible monthly. Calculate Shafia’s monthly repayments.

 

Let the monthly repayments be R and the amount owing after n months be An

The rate of interest r is 6% p.a. = 0.5% per month.

At the end of the first month, after paying her first installment, Shafia owes

A1 = 250 000 x 1.005 – R

At the end of the second month she owes A2 = A1 x 1.005 – R

A2 = (250 000 x 1.005 – R)x 1.005 – R

            = 250 000 x 1.0052 – 1.005R -R

At the end of the third month she owes A3 = A2 x 1.005 – R

= (250 000 x 1.0052 – 1.005R –R) x 1.005 – R

= 250 000 x 1.0053 – 1.0052R – 1.005R – R

By continuing the pattern for n months

An = 250 000 x 1.005n – 1.005(n-1)R  - 1.005(n-2)R …………-R

An = 250 000 x 1.005n – R( 1 + 1.005 + 1.0052 + ………..+ 1.005(n-1))

When the loan has been fully repaid n = 25 x 12 = 300 and A300 = 0.

0 = 250 000 x 1.005300 – R( 1 + 1.005 + 1.0052 + 1.0053 + ……… + 1.005299)

The part inside the brackets is a geometric series with a = 1 and r = 1.005

0 = 1116242.453 – R

0 = 1116242.453 – R x 692.9939624

R =       = 1610.75

Repayments = $1610.75 per month.

 

Exercise.

Q.1.     James borrowed $10 000 to buy a car. The interest rate was 12% per annum, reducible monthly and the loan was to be repaid over 3 years. What were the monthly repayments?

Q.2.     Stephanie borrowed $50 000 to start a business. She didn’t make any repayments for the first year and then repaid the loan in monthly installments over the next 5 years. If the interest rate was 9%, compounded monthly, what were Stephanie’s monthly repayments?

Q.3.     Mary and Norm borrowed $100 000 over 15 years to buy a house. The interest rate was 8% per annum, compounded annually. What were their monthly repayments?

Q.4.     Mac and Sue wanted to borrow $200 000 to buy a house and pay it back over 20 years. One lender offered them a loan at 8˝ % per annum, compounded annually and another lender offered them a loan at 9% per annum, compounded monthly.

            Calculate the monthly repayments on each loan.

 

Answers:          Q.1. $332.14   Q.2. $1135.28             Q.3. $973.58

            Q.4. 8 ˝ % p.a. = $1761.18 p.m.        9% p.a. = $1799.45 p.m.

 

 (ii)       How much of the principal has Nathan paid back after 10 years?   (2 mks)

 

A10 = 200 000 x 1.005120 – R( 1 + 1.005 + 1.0052 + ………..+ 1.005119)

A10 = 363879.3468 – 1433

A10 = 363879.3468 – 234839.104 = 129040.24

Principal paid back = 200 000 – 129 040.24

= $70960 (nearest $1 based on repayments of $1433.00

 

(iii)       If Nathan increases his repayments to $1500 per month, how long will it take him to pay off the loan? (2 mks)

 

220 months, = 18 years & 4 months

 

0= 200 000 x 1.005n – 1500( 1 + 1.005 + 1.0052 + ………..+ 1.005(n-1))

            Dividing by 1500

            0 = 133.333333 x 1.005n -

            Multiplying by 0.005

            0.6666666667 x 1.005n – 1.005n +1 = 0

            0.3333333333 x 1.005n = 1

            Multiplying by 3

            1.005n = 3

            Taking logs of both sides

            n log 1.005 = log 3

            2.166 x 10-3 n = 0.47712

            n =  = 220.28

            = 220 months (to nearest month) = 18 years 4 months.

 

 

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