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Date Submitted: 07/13/2012 03:03 PM
Week One Assignment
Karen Childress
MAT 126
Mahmoud Mryyan
June 25, 2012
Week One Assignment
Our assignment was to solve two problems in The Real World Applications section of our text. Both of these problems are examples of how we can use formulas to solve every day problems. The first problem is an arithmetic sequence and the second problem is a geometric sequence. I will explain both processes using the formulas to solve these problems.
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 proceeding 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?
The firm charges $100 for the first ten feet and then charges $25 added to the previous price for each additional ten feet. The repeated addition of $25 tells us this is an arithmetic sequence. An arithmetic sequence is a sequence of numbers in which each succeeding term differs from the proceeding term by the same amount known as the common difference. To solve this problem we need to identify the following numbers:
n = the number of terms altogether n = 9
d = the common difference d = 25
a1 = the first term a1 = 100
an = the last term an = a9 (yet to be computed)
The next step is to find what a9 is by using the formula on page 222 of Mathematics in Our World, section 5-7. This is the formula to find the nth term of the sequence, or the 9th term in this problem.
an = a1 + (n – 1) d
a9 = 100 + (9 – 1) (25)
a9 = 100 + 8(25)
a9 = 100 + 200
a9 = 300
Now that we know what a9 is, we need to know what the sum of the sequence is from a1 to a9. The formula we need is on page 227 of Mathematics in Our World, section 5-7, right below the blue “Try This One” box.
Sn = n (a1 + a9)
2
S9 = 9(100 + 300)
2
S9 = 9(400)
2...