Mat 126 Week 1 Assignment
Real world applications
XXX
MAT126: Survey of Mathematical Methods
Instructor: XXX
May 20, 2012
In this assignment I would like to talk about arithmetic sequences and geometric sequences and want to give an example each how to calculate with those sequences. First I want to give a short definition of each sequence.
“An arithmetic sequence is a sequence of numbers in which each succeeding term differs from the preceding term by the same amount. This amount is known as the common difference.” (Bluman, A. G. 2500, page 221)
An example for an arithmetic sequence is:
1, 3, 5, 7, 9, 11, … (The common difference is 2. (Bluman, A. G. 2500, page 221)
“A geometric sequence is a sequence of terms in which each term after the first term is obtained by multiplying the preceding term by a nonzero number. This number is called the common ratio.” (Bluman, A. G. 2005, p. 225) Here you can see that there is always added 2.
1 + 2 = 3; 3 + 2 = 5; 5 + 2 = 7; 7 + 2 = 9; …
An example for a geometric sequence is:
2, 10, 50, 250, 1250, … (The common ratio is r = 5 (Bluman, A. G. 2005, p. 225)
Here you can see that the 2 is multiplied by 5, which is 10. Then the 10 is also multiplied by 5, which is 50 and so on.
2 x 5 = 10; 10 x 5 = 50; 50 x 5 = 250; 250 x 5 = 1250; …
In this assignment I have solved two exercises, one referring to arithmetic sequences and one referring to geometric sequences.
Exercise 35:
A person hired a firm to build a CB radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125, the next 10 feet will cost $150, etc. How much will it cost to build a 90-foot tower? (Bluman, A. G. 2005, page 230)
I have calculated it this way: n = the number of terms together; n = 9 (because the CB radio tower will be 90 feet high. One term are 10 feet, so 90 feet are 9 terms) d = common
XXX
MAT126: Survey of Mathematical Methods
Instructor: XXX
May 20, 2012
In this assignment I would like to talk about arithmetic sequences and geometric sequences and want to give an example each how to calculate with those sequences. First I want to give a short definition of each sequence.
“An arithmetic sequence is a sequence of numbers in which each succeeding term differs from the preceding term by the same amount. This amount is known as the common difference.” (Bluman, A. G. 2500, page 221)
An example for an arithmetic sequence is:
1, 3, 5, 7, 9, 11, … (The common difference is 2. (Bluman, A. G. 2500, page 221)
“A geometric sequence is a sequence of terms in which each term after the first term is obtained by multiplying the preceding term by a nonzero number. This number is called the common ratio.” (Bluman, A. G. 2005, p. 225) Here you can see that there is always added 2.
1 + 2 = 3; 3 + 2 = 5; 5 + 2 = 7; 7 + 2 = 9; …
An example for a geometric sequence is:
2, 10, 50, 250, 1250, … (The common ratio is r = 5 (Bluman, A. G. 2005, p. 225)
Here you can see that the 2 is multiplied by 5, which is 10. Then the 10 is also multiplied by 5, which is 50 and so on.
2 x 5 = 10; 10 x 5 = 50; 50 x 5 = 250; 250 x 5 = 1250; …
In this assignment I have solved two exercises, one referring to arithmetic sequences and one referring to geometric sequences.
Exercise 35:
A person hired a firm to build a CB radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125, the next 10 feet will cost $150, etc. How much will it cost to build a 90-foot tower? (Bluman, A. G. 2005, page 230)
I have calculated it this way: n = the number of terms together; n = 9 (because the CB radio tower will be 90 feet high. One term are 10 feet, so 90 feet are 9 terms) d = common