mathematic symbols & progression

Mathematic symbols & Progression

Week 13

Contents:

Mathematic symbols:1. Summation notation2. Factorial

Progressions1. Arithmetic Progressions and simple interest2. Geometric Progressions and compound interest

Mathematic symbols

• Summation notation (Sigma notation)

read as “the sum of xi as i goes from 1 to n.”Σ (capital sigma) sumi can be replaced by any other letter, j, or k.

n

1iin321 xx...xxx

• Summation notation (Sigma notation)

15

)5(3

14

)4(3

13

)3(3

12

)2(3

1i

i3

aaaaaa

7654321k

5

2i

543215

1k

k

33333337

1k

3

• Summation notation (Sigma notation)Example:Given x1=4, x2=5, x3=-1, and x4=2, find(a)

24

1kk )2x(

220994

)22()21()25()24(

)2x()2x()2x()2x(2222

24

23

22

21

• Summation notation (Sigma notation)Example: Given x1=4, x2=5, x3=-1, and x4=2,

find (b)

Thus, = 22 =4

24

1kk )2x(

)2x()2x()2x()2x()2x( 4321

4

1kk

20332

)22()21()25()24(

24

1kk )2x(

ARITHMETIC PROGRESSIONS (A.P.)

• A sequence of numbers in which each term after the first is obtained by adding a constant d (common difference) to the preceding term.

• Example: (1) 3, 6, 9, 12, …… (2) 250, 248, 246, 244, …., 202

• In Example (1), first term is 3, d = 3• In Example (2), first term is 250, d=-2

ARITHMETIC PROGRESSIONS (A.P.)• If a is the first term, and d the common difference of

an A.P., then the successive terms of the A.P. are a, a+d, a+2d, a+3d, …….

• The n term is given by the formula

• n=1, a1 = a + (1 – 1)d = a• n=2, a2 = a + (2 – 1)d = a + d• n=3, a3 = a + (3 – 1)d = a + 2d

an = a + (n – 1)d


Example 1 Find the 12th term of the 2, 7, 12, 17, 22, …..Solution: a = 2, d = 5, a12 = a + 11d = 2 + 11(5) = 57

Example 2Write the first five terms of an AP whose 3rd and 11th terms are 21 and 85, respectively.Solution: a3 = a + 2d = 21 ------ (1)

a11 = a + 10d = 85 ---- (2)(2) – (1): 8d = 64, d=8

a + 2(8) = 21, a = 21 – 16 = 5.The first 5 terms of an AP: 5, 13, 21, 29, 37, …..


Example 3: Ali’s monthly payments to the bank toward her loan form an A.P. If her sixth and tenth payments are RM345 and RM333, respectively, what will be her fifteenth payment to the bank?Solution:Let a be the first term and d be the common difference of the monthly payments of the A.P. a6 = a + 5d = 345 ----- (1)a10 = a + 9d = 333 ---- (2)(2) – (1): 4d = -12, d = -3

a = 345 – 5(-3) = 345 +15 = 360.Fifteen payment: a15 = a + 14d = 360 + 14(-3) = 360 – 42

= RM318.

ARITHMETIC PROGRESSIONS (A.P.)• Simple InterestLet a sum of money P be invested at an annual rate of interest of R per cent. In 1 year the amount of interest earned is given by

If the investment is at simple interest, then interest in succeeding years is paid only on the principal P and not on any earlier amounts of earned interest. Thus a constant amount I is added to the investment at the end of each year. After 1 year the total value is P + I, after 2 years it is P + 2I and so on. The sequence of annual values of the investment, P, P+I, P+2I, P+3I, …. After t years, the value is P+tI

100R

PI


Example:A sum of RM2,000 is invested at simple interest at a 3% annual rate of interest. Find an expression for the value of the investment t years after it was made. Compute the value after 6 years.Solution:P=2,000, R=3, I=2,000(3/100) = 60After t years, the total interest added is tI, value of the investment = P+tI = 2,000 + 60tAfter 6 years, value = 2,000 + 60(6) = RM2,360.

ARITHMETIC PROGRESSIONS (A.P.)Let Sn denote the sum of the first n terms of an A.P. with first term a1 = a and common difference d. Then Sn = a + (a+d) + (a+2d) + …..+ [a+(n-1)d] ------(1) Sn = [a+(n-1)d] + [a+(n-2)d] + ... + (a+d) + a ---- (2)(1)+(2):

2Sn = [2a+(n-1)d] + [2a+(n-1)d] +….+[2a+(n-1)d]2Sn = n[2a+(n-1)d]Sn = (n/2)[2a+(n-1)d]

The sum of the first n term of an A.P. with first term a and common difference d is given by:

Sn = (n/2)[2a+(n-1)d]


Example: Find the sum of the first 20 terms of the A. P. 2 + 5 + 8 + 11 + 14 + ….

Solution: a = 2, d = 3, n = 20Sn = (n/2)[2a+(n-1)d] = (20/2)[2(2)+(20-1)3]

= 10[4+57] = 610.


Example: Madison Electric Company had sales of $200,000 in its first year of operation. If the sales increased by $30,000 per year thereafter, find Madison’s sales in the fifth year and its total sales over the first 5 years of operation.Solution:a = 200,000, d = 30,000.Sales in the fifth year = 200,000 + 4(30,000) = $320,000.Total sales over the first 5 years of operation:

S5 = (5/2)[2(200,000)+(5-1)30,000] = 2.5[400,000 + 120,000] = $1,300,000

ARITHMETIC PROGRESSIONS (A.P.)Example: A man agrees to pay an interest-free debt of $5,800 in a number of installments, each installment (beginning with the second) exceeding the previous one by $20. If the first installment is $100, find how many installments will be necessary to repay the loan completely.Solution:The installments (in $) are: 100, 120, 140, 160, …..Sn = 5800 = (n/2)[2(100)+(n-1)20]

5800 = (n/2)[200+20n-20] = (n/2)[180+20n]5800 = 90n + 10n2

10n2 + 90 n + 5800 = 0 n2 + 9n + 580 = 0 (n-20)(n+29)=0 n = 20

Thus, 20 installments will be necessary to repay the loan

GEOMETRIC PROGRESSIONS (G.P.)

Is a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by a constant r. The constant r is called the common ratio.Example: 2, 6, 18, 54, 162, …… (r = 3)

1/3, -1/6, 1/12, -1/24,….. (r = -1/2)If a is the first term and r the common ratio, then successive terms of the G.P. are

a, ar, ar2, ar3, ……………. The n term is given by an = arn-1


Example: Find the eighth term of a G.P. whose first five terms are 162, 54, 18, 6 and 2.Solution:Common ratio r = 54/162 =1/3, a = 162Eighth term, a8 = 162(1/3)8-1 = 2/27

Example: Find the tenth term of a G.P. whose third term is 16 and whose seventh term is 1.a3 = ar3-1 = 16 ---(1)a7 = ar7-1 = 1 --- (2)(2)/(1): r6/r2 = 1/16 r4= 1/16, r = ½; a=16/(1/4) = 64a10 = 64(1/2)10-1 = 26(2-9) = 2-3 = 1/8


If a is the first term and r the common ratio of a G.P., then the sum Sn of n terms of the G.P. is given by

when r ≠ 1; and

Example: Find the sum of the first six terms of the G.P.: 3, 6, 12, 24a = 3, r = 2, S6 = 3(1-26)/(1-2) = 189

r1)r1(a

Sn

n

Sn = na when r =1


Example: Michaelson Land Development Company had sales of $1 million in its first year of operation. If sales increased by 10% per year thereafter, find Michaelson’s sales in the fifth year and its total sales over the first 5 years of operation.Solution:a = 1,000,000 r = 1.1a5 = 1,000,000(1.1)4 = 1,464,100

= 6,101,1001.11])1.1(1[000,000,1

S5

5

GEOMETRIC PROGRESSIONS (G.P.)Compound interestExample: Suppose $1,000 is deposited with a bank that calculate interest at the rate of 3% compounded annually. The value of this investment (in dollars) at the end of 1 year is equal to 1000 + 4% of 1000 = 1000(1+0.04) = 1000(1.04) = 1040If the investment is at compound interest, then during the second year interest is paid on this whole sum of $1040. Therefore, the value of the investment at the end of 2 years is 1040 + 4% of 1040 = 1040 + 0.04(1040) = 1040(1+0.04) = 1040(1.04) = 1000(1.04)(1.04) = 1000(1.04)2

Similarly, the value of the investment at the end of 3 years will be 1000(1.04)3

GEOMETRIC PROGRESSIONS (G.P.)Thus, the values of the investment at the end of 0 years, 1 year, 2 years, 3 years, and so on are:

1000, 1000(1.04), 1000(1.04)2, 1000(1.04)3, …..

Example: Each year a person invests $1000 in a savings plan that pays interest at the fixed rate of 3% per annum. What is the value of this savings plan on the tenth anniversary of the first investment? (Include the current payment into the plan)Solution: The first $1000 has been invested for 10 years, so it has increased in value to $1000(1+i)10 i = R/100 = 0.03, thus the value is $1000(1.03)10


The second $1000 was invested 1 year later; hence it has been in the plan for 9 years. Its value has therefore increased to $1000(1.03)9 . The third $1000 has been in the plan for 8 years and has the value of $1000(1.03)8 . It is continue until reach the tenth payment of $1000, which was made 9 years after the first. Its value 1 year later is $1000(1.03). Thus the total value of the plan on its tenth anniversary is:S =1000(1.03)10 +1000(1.03)9 + ....+1000(1.03) +1000(1.03)This is a G.P. with a=1000, r=1.03, and n=11. Therefore

= 1000(12.8078)=$12807.80

103.11)03.1(

1000S11

GEOMETRIC PROGRESSIONS (G.P.)Double declining-balance Method of Depreciation• Let C (in dollars) denote the original cost of the asset.• Let the asset be depreciated over N years.•Amount depreciated each year is 2/N times the value of the asset at the beginning of that year.•The amount that the asset is depreciated in its first year of us is given by 2C/N •Book value of the asset at the end of the first year,

•Similarly, at the end of the second year,

N2

1CNC2

C)1(V

2

N2

1CN2

1N2

1CN2

N2

1CN2

1C)2(V


n

N2

1C)n(V

nn

N2

11CN2

1CC)n(D


Example: A tractor purchased at a cost of $60,000 is to be depreciated by the double declining-balance method over 10-years. What is the book value of the tractor at the end of 5 years? By what amount has the tractor been depreciated by the end of the fifth year?Solution: C = 60,000 N = 10

The amount by which the tractor has been depreciated by the end of the fifth year = 60,000-19,660.80=$40,339.20

8.660,1954

000,60102

1000,60N2

1C)5(V555

20.339,40102

11000,60N2

11C)5(D5n

mathematic symbols & progression

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