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# Problems Involving Sequences and Series, Coursework Example

Pages: 2

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Coursework

Problem 1

Some of the typical mathematical problems that are encountered by different industries are those that involve periodic increments. That is, there is a variable that increases by a certain amount under specific conditions. This is the phenomenon that is observed in the problem which is solved in this essay. The following paragraphs give the details of the problem and present the proposed answer with complete solution.

The problem is about a firm that was hired to build a 90-foot CB radio tower. The firm charges \$100 for the first 10 feet, but this price increases by \$25 for every succeeding 10 feet. The question is how much would it cost to build the tower?

The answer to this question requires the application of arithmetic series, which is the sum of a sequence of numbers that have a constant increment. Looking at the problem, we see that the first 10 feet would cost \$100, the second \$125, and so on. Since the tower is 90 feet long, the price would increment 8 times, where the price for the last 10 feet would be \$100 + 8 x 25 = \$300. Therefore, the total cost can be represented mathematically as follows…

Total cost = 10 (\$100)+10(125)+…10(300)

Rather than solving the equation directly, we employ the use of arithmetic series by first factoring out 10 from the expression to isolate the series.

Total cost = 10 (100 + 125 + … 300)

The sum of the arithmetic series can be computed by the formula below:

Sum = n(First term + Last term)/2 where n is the number of terms.

Since the tower is 90 feet and the increments come in 10’s, there are 90/10= 9 terms.

Thus the sum is 9(100+300)/2 = 1800.

Plugging the sum to the equation for the total cost…

Total cost = 10 (\$1800) = \$18000

Thus, the cost of building the tower if \$18,000.00. The arithmetic series was used because it was the most convenient method to solve the problem by hand. What I learned in this problem is that factoring can be used to make an arithmetic series visible. This knowledge can be applied to other real-life problems involving arithmetically incrementing numbers.

Problem 2

Some of the most common mathematical problems that one encounters in real life are those about finances. Among these are problems that deal with the application of compound interest. Such problems require the application of geometric sequence and one such problem is solved in the following paragraphs.

In the problem, a deposit of \$500 in a savings account that pays 5% annual interest compounded yearly was made. The question is how much the money in the account would be after 10 years.

The solution of the problem can be seen by viewing the interest gained it as a geometric sequence. The first term is 500 and the succeeding terms increase by 5% of the previous term.  Mathematically, this can be expressed as.

500, 500(1.05),500(1.05)(1.05),500(1.05)(1.05)(1.05)…

Our interest is not in the sum of the sequence, but only in a particular term. That is, the term that has been increased by 10 factors of 1.05. This is the 11th term, and we choose this term because it is the term wherein the money has remained in the bank for 10 years. The formula for any term in a geometric sequence is found by the following formula.

Term n = First term (common ratio)^(n-1)

Thus the 11th term is solved as follows.

Term 11 = 500 (1.05)^(11-1) = 814.45

Therefore, the amount in the account after 10 years is about \$814.45. This problem shows that the geometric sequence can be used to solve problems involving interest. I find this very applicable in determining how much one needs to save in order to afford some future expense.

Reference

Bluman, A. G. (2005). Mathematics in our world (Ashford University Custom Edition). United States: McGraw-Hill.

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