Arithmetic Sequences and Series
a = first term
d = common difference
To find a certain term in the sequence (n = term you want):
U n  a  n  1d
To find the sum of all terms up to n in a series:
S n  n 2 2a  n  1d 
Geometric Sequences and Series
r = common ratio = 2nd term/1st term = 3rd term/2nd term
a = first term
To find a certain term in the sequence (n = term you want): U n  ar n 1
To find the sum of all terms up to n in a series:
Sn 

a 1 rn
1 r
Example 1:

if r  1
Sn 


a r n 1
r 1
if r  1
In an arithmetic sequence u1  u3  12
a and d.
and
u 4  u 6  24 . Find the values of
u1  u 3  12  a  1  1d  a  3  1d  12  2a  2d  12
u 4  u 6  24  a  4  1d  a  6  1d  24  2a  8d  24
Solve using simultaneo us equations :
2a  2d  12

2a  2d  12
2a  8d  24
 2a  8d  24
 6d  12  d  2  a  4
Example 2:
In an arithmetic series, the sum of the first 2n terms is half the sum of the first 3n
terms. If a = 12 and d = 3, find the value of n.
S 2 n  2 n 2 212   2n  13  n24  6n  3  n21  6n  21n  6n 2
S 3n  3n 2 212  3n  13  3n 2 24  9n  3  3n 2 21  9n 
S 2n 

1
2
 21n  6n 2 
S 3n
 
4 21n  6n 2  3 21n  9n 2

1 3
2 2
21n  9n 
2
3
21n  9n 
2
2
 21n  6n 2 
 84n  24n 2  63n  27n 2
3
21n  9n 
2
4
 21n  3n 2  0
3n7  n  0  3n  0 or 7  n  0  n  7 or n  0
Example 3:
Find the sum of the first 8 terms of the series 2, 6, 18, 54.
6
ar n  1 23 n  1 23 n  1
a  2 r   3  Sn 


 3n  1
2
r 1
3 1
2
S 8  38  1  6560
Example 4:
In a geometric series the 5th term is 8 times the 2nd term and the sum of the 6th and
8th terms is 160. Determine:
a) Common Ratio
b) First Term
c) The sum of the 4th to 10th terms inclusive
a 
a, ar , ar 2 , ar 3 
U 5  ar 4 U 2  ar  ar 4  8ar  r 3  8  r  2
b 
U 6  ar 5 U 8  ar 7
 ar 5  ar 7  160
Using r  2 from (a) : a2   a2   160  32a  128a  160
5
160a  160  a  1
7
c 
Sn 


a r n 1
r 1
S10  S 3 

 
 
 
 
a r 10  1 a r 3  1 1 210  1 1 2 3  1



 210  1  2 3  1
r 1
r 1
2 1
2 1
 

 1024  8  1016
To find the sum to infinity of a series (NOTE: This can only be used when |r| < 1)
S 
Example 5:
a
1 r
when  1  r  1
If €400 is invested at compound interest of 10% per annum, determine:
a) Value after 9 years
b) The time correct to the nearest year it takes to reach more than €1500
a 
a  400 r  1.10  U n  ar n 1  4001.1
U 9  4001.1
9 1
b 
n 1
 4001.1  €857.44
U n  ar n 1  4001.1
8
n 1
 1500 
1.1n1  1500
Taking logs of both sides : ln 3.75  ln 1.1
400
n 1
1.1n1  3.75
 ln 3.75  n  1 ln 1.1
ln 3.75
 n  1  13.87  n  14.87
ln 1.1
 It will take 15 years to reach more than €1500
n 1 

Logic
Symbols for negative in logic A’ or ~A ;
Truth table for the logic relation
Derivatives
dy
dv
du
u v
dx
dx
dx
a) Product Rule
(Log Tables pg. 42)
b) Quotient Rule
dy
(Log Tables pg. 42 )

dx
c) Chain Rule
v
du
dv
u
dx
dx
v2
dy dy du

*
dx du dx
Find the Maximum and Minimum Turning Points
SLOPE = 0 at the TURNING POINTS 
dy
0
dx
To find the Turning Points:
dy
i) Find
dx
dy
ii) Let
= 0, and solve for value(s) of x
dx
iii) For each value of x, find a corresponding value of y
To determine if the Turning Points are Maximum or Minimum:
d2y
i) Find
dx 2
ii) Substitute in values of x found in turning points. If:
d2y
 Negative Number
dx 2
d2 y
 Positive Number
dx 2
 MAXIMUM
 MINIMUM
Probability Distributions
To Get mean of a set of values
where N is the number of samples taken. Next, the standard deviation simplifies to
And The variance is S 2
Example1: a list of numbers: 1, 3, 4, 6, 9, 19
mean: (1+3+4+6+9+19) / 6 = 42 / 6 = 7

sum of deviations:  ( x  x ) = (1-6) (1  6) 2  (3  6) 2 ......(19  6) 2 +36+16+9+1+4+144 = 210
=
Variance is equal to 42
210 /(6  1) =
42 about 6.48
Binomial Distribution:
Pr   nCr * p r q nr
Pr   e 
q=1-p
(In LOG TABLES pg 35)
r
For( n>30)
where   np
r!
Example 1 A bag contains a large number of marbles of which 25% are white and the rest are
green. Five marbles are drawn at random. Determine the probability:
Poisson Distribution:
a) 2 white & 3 green
b) At least 3 white
c) Not more than 3 green
a) PWhite   0.25  p
PNot White   0.75  q
n5
This is basically P(2 White), since the rest will be green
P2  5C 20.75 0.25  100.4218750.0625  0.2637
3
2
b) At least 3 white marbles implies 3 white or 4 white or 5 white
P3  5C 3 0.75 0.25  100.56250.015625  0.08789
2
3
PAt least 3 white   P3  P4  P5
c) Not more than 3 green implies 2 white or 3 white or 4 white or 5 white, so in
this case, we can add the answers to parts (a) and (b)
PNot More than 3 green   P2   P3  P4  P5
Example 2 If 3% of machines produced by a company are defective, determine the probability
that in a sample of 100 machines (a) 4 machines, (b) Not more than 3 machines
and (c) at least 2 machines will be defective.
a)
p  0.03
  0.03100  3
P4 
e 3 * 3 4
=
4!
b) PNot more than 3 defective   P0  P1  P2  P3
P0 
e 3 * 30
=
0!
.
.
.
.
c) PAt least two defective   P3  P4  ...  1  P0  P1
Matrices
Here is an example of matrix multiplication for two 2x2 matrices
Here is an example of matrices multiplication for a 3x3 matrix
Determinant of a 2x2 matrix
Assuming A is an arbitrary 2x2 matrix A, where the elements are given by:
then the determinant of a this matrix is as follows:
The inverse of a 2x2 matrix
Take for example a arbitury 2x2 Matrix A whose determinant (ad-bc) is not equal to zero
where a,b,c,d are numbers, The inverse is:
Inverse Matrix Method
DEFINITION: Assuming we have a square matrix A, which is non-singular ( i.e.
det(A) does not equal zero ), then there exists an nxn matrix A-1 which is called
the inverse of A, such that this property holds:
AA-1= A-1A = I where I is the identity matrix.
DEFINITION: The inverse matrix method uses the inverse of a matrix to help
solve a system of equations, such like the above Ax = b. By pre-multiplying both
sides of this equation by A-1 gives:
or alternatively this gives
Similarly for three simultaneous equations we would have:
Written in matrix form would look like
and by rearranging we would get that the solution would look like
Example 1
Encrypt the plaintext “TEST MY PROGRAM” using the matrix
 25 9 

H  
 11 4 
i)
ii)
(i)
Giving the ciphertext in matrix form.
Find the inverse matrix H 1 and use it to decrypt the ciphertext obtained
above.
abcde fgh i j k l m n o p q r s t u v w x y z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
T E S T
20 5 19 20
M Y
13 25
P ROG RAM
16 18 15 7 18 1 13
 20 19 13 16 15 18 13 

Plaintext matrix p  
 5 20 25 18 7 1 0 
Ciphertext matrix C  HP is given by
 25
C  HP  
 11
 545 655
 
 240 289
(ii)
H 1 
9  20 19 13 16 15 18 13 


4  5 20 25 18 7 1 0 
550 562 438 459 325 

243 248 193 202 143 
 4  9  4  9
1



100  99   11 25    11 25 
 4  9  545 655 550 562 438 459 325 


H 1C  H 1 HP  P  H 1C  
  11 25  240 289 243 248 193 202 143 
 20 19 13 16 15 18 13 

 
 5 20 25 18 7 1 0 
Thus the plaintext is 20 5 19 20
T E S T
Logarithm:
Log Rules:
1) logb(mn)
= logb(m) + logb(n)
2) logb(m/n)
= logb(m) – logb(n)
3) logb(mn)
= n · logb(m)
13 25
M Y
16 18 15 7 18 1 13
P ROG RAM