# Mathematics

Below are two real life application problems which are very similar to the problems required for the Week One Written Assignment. Among other things, these examples show how the math work and explanations could look. Similar types of sequences and numbers are used so the student can learn how to work these types of problems. Work submitted in the paper should not be identical to these examples. Keep in mind this is only the math work. Your paper will still need an introductory paragraph and a conclusion, along with the other usual mechanics of a paper in APA style. Example 1: The equipment cost to run a certain drilling rig for water wells is $200 to drill the first ten feet. After that, the cost for drilling increases $50 for each successive ten feet. In other words, drilling from 10 to 20 feet costs $250, drilling from 20 to 30 feet is $300, and so on. How much will it cost for the rig to drill a 180 foot well? Here is how we think about it: We see that there is a new price every ten feet down into the ground. Each new price is $50 added to the previous price. The repeated addition tells us this is an arithmetic sequence. First, we need to identify the following numbers: n = the number of terms altogether n = 18 d = the common difference d = 50 a1 = the first term a1 = 200 an = the last term an = a18 (yet to be computed) Next, we need to compute what a18 is. The formula we need is on page 271 of Mathematics in Our World, in the tan box. This is the formula to find the nth term of the sequence, or the 18th term in this case. an = a1 + (n – 1)d a18 = 200 + (18 – 1)(50) a18 = 200 + 17(50) a18 = 200 + 850 a18 = 1050 Now that we know what a18 is, we need to know what the sum of the sequence is from a1 to a18. The formula we need is on page 273 Mathematics in Our World, right below the blue “Try This One” box. Sn = n(a1 + a18) 2 S18 = 18(200 + 1050) 2 S18 = 18(1250) 2 S18 = 9(1250) = 11,250 Thus, the equipment cost for the rig to drill a 180 foot well is $11,250.

Example 2: A person deposits $1500 in a savings account that pays 3% annual interest compounded yearly. At the end of 12 years, what will be the balance in the savings account if no additional deposits or withdrawals are made on the account? Here is how we think about it: Each year 3% of the balance is added to the balance. If we let B = the balance, that would look like: B + (.03)B B(1 + .03) B(1.03) In other words, each year the existing balance is multiplied by 1.03. This repeated multiplication by the same number tells us we have a geometric sequence. First, we need to identify the following numbers: n = the number of terms n = 12 r = the common ratio r = 1.03 a1 = the first term a1 = 1500(1.03) = 1545, the balance at the end of

the first year, thus a1. In a savings account, the total balances at the end of each year form the sequence, so we don’t need to add up all the terms in the sequence. We just need to find out what the balance is at the end of 12 years, so we are looking for the value of a12. The formula we need is on page 275 of Mathematics in Our World, in the lavender box. an = a1(r

n-1) a12 = 1545(1.03

11) a12 = 1545(1.384233…)

a12 = 2138.64 Thus, the balance in the savings account at the end of 12 years will be $2138.64. References: Bluman, A. G. (2005). Mathematics in our world (Ashford University Custom Edition).

United States: McGraw-Hill.

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