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# Free Custom «Math» Essay Paper

1. 1. A person hired a firm to build a CB radio tower. The firm charges \$100 for labor for the first 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?

This an arithmetic progression question. An arithmetic progression is defined as a series in which each term is a number larger or smaller than the previous term. These terms in the series are said to increase or decrease by a common difference, denoted by letter d. The first term in the series is denoted by a while the number of terms denoted by n (Ayres & Schmidt, 2003). For the above question the first term is \$100; the second term is \$125; and the third term is \$150. Every term in this sequence exceeds the previous by \$25. Every term in this case constitutes of 10 feet portion of work. Since a 90 foot tower is to be constructed, nine terms are to be considered. The formula for summing up all the terms until the nth term is given below.

Sn = ½ n [2a + (n-1) d]

The above equation can be substituted and calculated for the total cost of tower construction to be arrived at accurately.

S9 = ½ (9) [2(100) + (9-1) 25]

S9 = ½ X 9 [200 + (8) 25]

S9 = ½ X 9 (200 + 200)

S9 = ½ X 9 X 400

S9 = 9 X 200

S9 = 1,800

Therefore the total cost of building a 90 foot CB radio tower was \$1,800. It becomes very accurate and easier to use an arithmetic progression in calculating problems like the one above.

1. 2. A person deposited \$500 in a savings account that pays 5% annual interest that is compounded yearly. At the end of 10 years, how much money will be in the savings account?

This is a geometric progression question. A geometric progression series can be defined as a sum of number of terms that are either finite or infinite. Each proceeding term after the first term of a geometric sequence is a multiple of the preceding term by a fixed constant, r and this is a ratio (Ayres & Schmidt, 2003). In the above question the ratio is a percentage 5% which is equal to 5/100. The first term in the sequence is \$500 and this is usually denoted by letter a. The total number of terms is n terms and the last term is an nth term, in this question we are calculating for the 10th term since the 10 years represent 10 terms. We can use the formula for the geometric progression to work out for the nth term. The total amount of money in the savings after the 10th year can be calculated by using the formula below.

FV = a [1 + (r/100)] n where FV = Future value

a = Present value

r = rate

n = number of periods

The above equation can be substituted and calculated for the total savings after the 10th year to be arrived at accurately.

FV = 500 [1+ (5/100)] 10

FV = 500 (105/100) 10

FV = 500 (1.05) 10

FV = 500 (1.05) 10

FV = 814.4473134

After the period of 10 years an approximate of \$814 will be in the savings account. This formula is very important such that the result is obtained easily and accurately. It is applicable in the real life situation for instance in banking institutions where numerous commercial arithmetic problems are handled.

Arithmetic and geometric progression approaches are very important in the real life situation. Arithmetic progression was used to calculate the total cost of building a 90 foot CB radio tower while the second problem involved, calculating the total amount of money in the savings account after 100 years.

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