Mathematic symbols & Progression

1. Mathematic symbols &
Progression
Week 13

2. Contents:
Mathematic symbols:
1. Summation notation
2. Factorial
Progressions
1. Arithmetic Progressions and simple interest
2. Geometric Progressions and compound interest

3. Mathematic symbols
• Summation notation (Sigma notation)
read as “the sum of xi as i goes from 1 to n.”
Σ (capital sigma) sum
i can be replaced by any other letter, j, or k.
∑
=
=++++
n
1i
in321 xx...xxx

4. • 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
−
+
−
+
−
+
−
=
−
++++=
++++++=
∑
∑
∑
=
=
=

5. • Summation notation (Sigma notation)
Example:
Given x1=4, x2=5, x3=-1, and x4=2, find
(a) 24
1k
k )2x( −∑
=
220994
)22()21()25()24(
)2x()2x()2x()2x(
2222
2
4
2
3
2
2
2
1
=+++=
−+−−+−+−=
−+−+−+−=

6. • Summation notation (Sigma notation)
Example: Given x1=4, x2=5, x3=-1, and x4=2,
find (b)
Thus,
= 22
=4
24
1k
k )2x( 



−∑
=
)2x()2x()2x()2x()2x( 4321
4
1k
k −+−+−+−=−∑
=
20332
)22()21()25()24(
=+−+=
−+−−+−+−=
24
1k
k )2x( 



−∑
=

7. 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

8. 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

9. ARITHMETIC PROGRESSIONS (A.P.)
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 2
Write 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, …..

10. ARITHMETIC PROGRESSIONS (A.P.)
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.

11. ARITHMETIC PROGRESSIONS (A.P.)
• Simple Interest
Let 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






=
100
R
PI

12. ARITHMETIC PROGRESSIONS (A.P.)
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) = 60
After t years, the total interest added is tI,
value of the investment = P+tI = 2,000 + 60t
After 6 years, value = 2,000 + 60(6) = RM2,360.

13. 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]

14. ARITHMETIC PROGRESSIONS (A.P.)
Example: Find the sum of the first 20 terms of the A. P.
2 + 5 + 8 + 11 + 14 + ….
Solution:
a = 2, d = 3, n = 20
Sn = (n/2)[2a+(n-1)d] = (20/2)[2(2)+(20-1)3]
= 10[4+57] = 610.

15. ARITHMETIC PROGRESSIONS (A.P.)
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

16. 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

17. 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

18. GEOMETRIC PROGRESSIONS (G.P.)
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 = 162
Eighth 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) = 64
a10 = 64(1/2)10-1
= 26
(2-9
) = 2-3
= 1/8

19. GEOMETRIC PROGRESSIONS (G.P.)
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, 24
a = 3, r = 2, S6 = 3(1-26
)/(1-2) = 189
r1
)r1(a
S
n
n
−
−
=
Sn = na when r =1

20. GEOMETRIC PROGRESSIONS (G.P.)
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 = 1.1
a5 = 1,000,000(1.1)4
= 1,464,100
= 6,101,1001.11
])1.1(1[000,000,1
S
5
5
−
−
=

21. GEOMETRIC PROGRESSIONS (G.P.)
Compound interest
Example: 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) = 1040
If 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

22. 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

23. GEOMETRIC PROGRESSIONS (G.P.)
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
103.1
1)03.1(
1000S
11
−
−
=

24. 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,






−=−=
N
2
1C
N
C2
C)1(V
2
N
2
1C
N
2
1
N
2
1C
N
2
N
2
1C
N
2
1C)2(V 





−=





−





−=





−−





−=

25. GEOMETRIC PROGRESSIONS (G.P.)
n
N
2
1C)n(V 





−=














−−=





−−=
nn
N
2
11C
N
2
1CC)n(D

26. GEOMETRIC PROGRESSIONS (G.P.)
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
8.660,19
5
4
000,60
10
2
1000,60
N
2
1C)5(V
555
=





=





−=





−=
20.339,40
10
2
11000,60
N
2
11C)5(D
5n
=














−−=














−−=