Mathematics Magazine for Grades 112 

4/2004 



You should be more
afraid of a stupid man than of an evil one. Grade 11
Theory:Arithmetic ProgressionAn arithmetic progression is a sequence in which each term after the first is formed by adding a fixed amount, called the common difference, to the preceding term. If a is the first term, d is the common difference, and n is the number of terms an arithmetic progression, the successive terms are a, a + d, a + 2d, a + 3d,….a + (n1)d Thus, the last term (or nth term) l is given by l= a + (n  1)d The sum S of the n terms of this progression is given by or Arithmetic
Means The terms
between the first and last terms of an arithmetic progression are called
arithmetic means between these two terms. Thus to insert k arithmetic
means between two numbers is to form an arithmetic progression of (k+2)
terms having the two given numbers as the first and the last terms. Geometric
Progression A
geometric progression is a sequence in which each term after the first is
formed by multiplying the preceding term by a fixed number, called the
common ratio. If a is
the first term, r is the common ratio, and n is the number of terms, the
geometric progression is: a, a∙r, a∙r^{2}, ….. a∙r^{n1 } Thus, the
last ( or nth ) term l is given by l = a∙r^{n1}. The sum S
of the first n terms of the geometric progression is given by:
or
Geometric
Means The terms
between the first and the last terms of a geometric progression are called
geometric means between two terms. Thus, to insert k geometric means
between two numbers is to form a geometric progression of k+2 terms having
the two given numbers as the first and last terms. Solutions from the Previous Issue:1. Find the twentieth term and the sum of the first 20 terms of the arithmetic progression 4, 9, 14, 19…. Solution: For this progression a = 4, d = 5, and n = 20; The twentieth term is l = a + (n  1) ∙ d = 4 + 19 ∙ 5 = 99. And the sum of the first 20 terms is: = 1030 2. Insert five arithmetic means between 4 and 22. Solution: We have a = 4, l = 22, and n = 5 + 2 = 7. Then 22 = 4 + 6 ∙ d and d = 3. The first mean is 4 + 3 = 7, the second is 7 + 3 = 10, and so on. The required means are: 7, 10, 13, 16, 19 and the resulting progression is 4, 7, 10, 13, 16 19, 22 3. Find the arithmetic mean of the two numbers a and l. Solution: We seek the middle term of an arithmetic progression of three terms having a and l as first and third terms, respectively. If d is the common difference, then a + d = l – d and d = (l  a). The arithmetic mean is a + d = a + (l  a) = (l + a). 4. Find the ninth term and the sum of the first mine terms of the geometric progression 8, 4, 2, 1,…. Solution: Here a = 8, and r = ½, and n = 9; The ninth term is l = ar^{n1}= The sum of the first nine terms is:
Proposed exercises:Proposed by Mihai Rosu
Professor Toronto 1.
Find
vector and parametric equations of the line that passes through the point
P and direction vector m. a)
P(2,4),
b)
P(
,3),
2.
Find a
vector and parametric equations of the line
:
a)
,
b)
,
3.
Determine
the coordinates of three points corresponding to the parameters 0, 1, and
2 on each line: a)
b)
4.
Find
vector and parametric equations of the line that passes through the point
and is parallel to: a) the
vector
; b) the line
. 5. Show that both lines and contain the point . Find the acute angle of intersection of these lines, to the nearest degree.  