Geometric and Arithmetic progression
A series is defined as a set of number whose sum equals to a constant A. for example, 4 + 10 + 46 = 60. Sometimes, a series is also known as progression. There are two types of series: Arithmetic and geometric series. In arithmetic series, successive terms differ by the same constant, d. For example, b + (b +3d) + (b +6b) + (b+9b) + . . . + (b + (n-1)d. The sum of terms in an arithmetic series Sn =, where a = first term of the series. In geometric series, successive terms have a common ratio. In this case, the sum of first n terms Sn = b + br + + + . . . + + . The last term of the expression is (Bluman, 2005).
Problems in Real application
Problem 1
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35. 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?
Solution
The problem above is an arithmetic progression; therefore, it can be solved by identifying the terms:
n= 90/10= 9
Common difference = d d = 25
First term = a1 a1 = 100
Last term = an
an = a9
an = a1 + (n-1)d
a9 = 100 + (9-1)25
a9 = 100 + 8(25)
a9 = 300
Sn = 1800
The cost of building the tower is $1800.
Problem 2
37. 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?
Solution
In this case, the problem presents a case of geometric progression. To solve the problem, the first term (a1), the common ration (r), and the number of terms (n) terms must be identified.
n = 10
r = 1.05
a1 = 1500*1.05 = 1575, the amount after year 1
an = a1r(n-1)
a10 = 1575 * 1.05(10-1)
a10 = 1575 * 1.059
a10 = 2443.34
The amount of money in the savings after 10 years will be $2443.34
In conclusion, the problems were solved with different methods. In first method, arithmetic progression method was used. This is because the difference between the costs increased after 10 feet by a constant value of $25.therefore, a common difference of 25. In problem 2, the method used was geometric progression. This is because the amount increased by 5%, at a rate of 1.05, from year to year; the common ratio is1.05.
The method of geometric progression can be applied in calculation of future amounts when invested in shares, bonds, or fixed accounts deposits earning with an annual compound interest. On the other hand, arithmetic progression can be used to in calculating depreciation of property, if the yearly depreciation amount us know, and also in predicting events if the sequence of occurrence is know, such as eruption of a volcano when the time between eruption is constant.
References
Bluman, A.G. (2005). Mathematics in our world (Ashford University custom Edition). United
States: McGraw-Hill.
