Real World Application

Real World Applications

By: Sherry H Priester

Course: MAT126: Survey of Mathematical Methods

Instructor: Anne Gloag

Date: May 13, 2013

In this assignment, I will be solving two different problems, one problem that involves geometric sequences and the other involving arithmetic sequences. First, I would like to give a brief definition of a geometric sequence and an arithmetic sequence. A geometric sequence is a term 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 and the arithmetic sequence is a sequence of numbers in which each succeeding term differs from the preceding term by the same amount. The amount is known as the common difference (Bluman, A.G, 2005).

The first problem that I will solve is, 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?

N= the number of terms altogether

D= the common difference

A1= the first term

An= the last term

N= 9

D= 25

A1= 100

An= a9

An= a1 + (n-1) d

A9= 100 + (a-1)25

A9= 10 + (8)25

A9= 100 + 200

A9= 300

Sn+ (n/2) [2a + (n-1) d]

S9= (9/2) [2*100 + (9-1) (25)]

S9= 1800

The problem involves arithmetic sequence and deductive reasoning to solve the equation. Since the labor cost for each 10ft remains constant at $25, the arithmetic sequence of the labor cost is 100 + 125 + 150 + 175 + 200 + 225 + 275 + 300 = 1800. To find the cost for building a 90ft tower, sum the first 9 terms of the sequence a = 100, d = 25, n = 9. Sn = (n/2) [2a + (n-1) d] s9 + (9/2) [* 100 + (9-1) (25) ] = 1800. The final cost to build a 90-foot tower is $1800....